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Unit 3 Electric, Gravitational, and Magnetic Fields ARE YOU READY? (Pages 314–315)

Knowledge and Understanding 1. (a) Particle electron proton neutron

Mass –31 9.11 ¯ 10 kg –27 1.67 ¯ 10 kg –27 1.67 ¯ 10 kg

Charge –e +e 0

Force weak strong strong

(b) Electrons are responsible for the conduction of electricity in solids. Since the force holding the outermost electrons is weaker, they are able to move more easily from atom to atom. Protons do not move as easily because they are held in the nucleus by a stronger force. (c) In an atom the number of electrons is equal to the number of protons. In a positive ion the number of protons is greater than the number of electrons. In a negative ion the number of electrons is greater than the number of protons. 2. (a) Charging by friction involves rubbing two objects together. One object gains electrons while the other loses the same number of electrons. If both objects are neutral before starting, one will end up with a positive charge and the other will end up with a negative charge.

Charging by contact involves placing a charged object into contact with a neutral object. If the charged object is negative, electrons are transferred to the neutral object. If the charged object is positive, electrons move from the neutral object. In both cases objects end up with the same type of charge as the charging object.

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Chapter 7 Electric Charges and Electric Fields 417

To charge by induction and grounding, a charged object is brought close to a neutral object. The charging object changes the charge distribution of the electrons but not the protons. Grounding the neutral objects allows electrons to leave if the inducing charge is negative, and to enter if the inducing charge is positive. Once the ground is removed the charge is moved away. The formally neutral object ends up with a charge opposite to that of the inducing object.

(b)

3.

Method of Charging initial charges on objects steps

Friction both neutral

final charges on objects

oppositely charged

rub two objects together

(a)

(b)

Contact one neutral and the other charged touch charged object to neutral object similarly charged

Induction and Grounding one neutral and the other charged bring charged object close to neutral grounded object; remove ground and then charged object oppositely charged (c)

4. (a) Rub the bar magnet with the pole of a strong magnet in one direction, or place it in a strong magnetic field and strike it. This aligns the magnetic fields of domains within the magnet, making the magnet stronger. (b) Increase the electric current in the conductor. (c) Increase the electric current in the coil, increase the number of windings, or place a substance with a high permeability in the core.

Inquiry and Communication 5.

Balance a charged ebonite rod on a hanger and bring another similarly charged ebonite rod close. Repeat with a glass rod rubbed with silk. Separately, bring both charged rods close to a neutral pith ball electroscope. In this case both rods will attract the neutral pith ball.

418 Unit 3 Electric, Gravitational, and Magnetic Fields

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6. (a) As long as the wire is not parallel to the magnetic field lines, the conductor will experience a force. The magnitude of the force will depend on the length of the wire, the magnitude of the current, and the external magnetic field. (b) Test each of the following factors one at a time on a suspended wire and measure the amount of deflection from equilibrium: (i) The angle between the wire and field. (ii) The length of wire passing through the field. (iii) The magnitude of the current. (iv) The magnitude of the external field.

Making Connections 7. (a) The speaker has a movable voice coil wrapped around a permanent magnet. As charge flows through the voice coil in either direction it causes a magnetic field. This field interacts with the magnetic field of the permanent magnet and either attracts or repels the voice coil, which in turn causes the speaker cone to move back and forth and to rapidly produce sound waves. (b) Charge flows from the battery through the circuit wrapped around the armature (B), through the graphite brush (D), into the two-piece commutator (C), through the coil around the armature, and finally back into the commutator. The magnetic field in the armature cause the armature to turn clockwise. Eventually, the graphite brushes come into contact with the other half of the commutator, reversing the direction of the current in the coil to keep the armature rotating.

CHAPTER 7 ELECTRIC CHARGES AND ELECTRIC FIELDS Reflect on Your Learning (Page 316) 1. 2. 3. 4. 5. 6.

The electrocardiograph measures electric potential differences across the surface of the skin. These measurements are compared to what is normally expected when doing similar measurements on other individuals, and the results are then interpreted. The computer towers are encased in metal to shield them against stray electric and magnetic fields. The force of gravity is always attractive, is very weak, depends on the product of the two masses, and is an inverse square law. The electric force can either attract or repel depending on the types of charges, is much stronger, depends on the product of the two charges, and is also an inverse square law. The charge also exerts a force of attraction, but through an electric field. (a) Charges of opposite signs attract (like gravity) and so the electric potential energy also increases. (b) Charges of the same sign repel each other, therefore electric potential energy will decrease as the distance between them increases. (a) The electric field is constant, therefore the charged particle will act like projectile motion (in a gravitational field near the surface of the earth). (b) The charged particle will experience a decreasing force of either attracting or repelling the charge. If the force is attracting then the initial velocity of the particle will decrease. The particle will either stop and reverse its direction, or continue to move forward with an acceleration approaching zero. If the force is repulsive, the particle’s speed will increase and the acceleration will approach zero. Gravity can only attract but all these problems can be easily studied using conservation of energy.

Try This Activity: Charging By Induction (Page 317) (a) Use charging by induction and grounding. Charge the ebonite rod with fur and bring it close to the grounded electroscope. (The electroscope can be grounded by holding your finger on it.) The charge on the electroscope is opposite to the ebonite. (b) You could charge the ebonite rod with silk and show that it repels the electroscope, but that would require the use of extra materials. You cannot just show that they attract each other because that happens even when the electroscope is neutral. However, if the electroscope is neutralized and still attracts the rod, then the charge on the electroscope must have been opposite. You could also show that it attracts the ebonite but repels the fur.

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 419

7.1 ELECTRIC CHARGE AND THE ELECTRICAL STRUCTURE OF MATTER Try This Activity: Charging Objects (Page 323) The smoke particles are attracted to the charged object due to induction.

PRACTICE (Pages 324–325)

Understanding Concepts 1. 2.

3.

4. 5. 6. 7.

If the base layer is positive it attracts the negatively charged beads and the microcapsule will appear dark when viewed from the top. If the base layer is negative it repels the negatively charged beads that float to the top, making the microcapsule appear white when viewed from the top. (a) The charge is the same but opposite in sign. (b) You achieve this transfer of charge by friction. Both objects are neutral at the start, but walking involved some rubbing motion between the carpet and your feet. This resulted in an opposite charge on each object. (c) If the air is not dry then the water vapour and dissolved substances that ionize in the air can act as conductors and reduce the charge on your body. This makes it more difficult to build up and retain any excess charge. (a) Some of the excess electrons from the charged object pass into the electroscope and spread out over the surface of the conducting material. The two leaves become negatively charged and repel each other. (b) The further apart the leaves separate, the greater the charge on the object. (c) No, if the charged object comes close to the electroscope without touching it, the object will cause the electrons in the electroscope to redistribute either towards the leaves or away from them. In either case, the leaves will develop similar charges and repel. Touching the metal casing neutralizes your body so that any excess charge does not damage the memory when you handle the chip. The memory is stored magnetically and the charge could change the magnetization on the disk and cause a loss of data. Heat from pressure rollers is used to melt the toner and fix it to the paper. The paper sticks together because the toner particles are given a negative charge and the paper is given a positive charge. The interaction of these charges causes the papers to stick together. The toner and the paper have to be replenished from time to time. Selenium is a photoconductor. It acts as an insulator in the dark and as a conductor when exposed to light. When light is reflected from the copy through lenses and onto selenium, it loses its charge. Where the copy is dark no light is reflected and the selenium retains its charge. In these places it can attract negatively charged toner particles.

Making Connections 8.

Using warm water will tend to melt the toner particles causing them to stick more readily to your hands. Using cold water will help to remove the particles without additional melting.

Section 7.1 Questions (Page 326)

Understanding Concepts 1. (a) The glass rod will develop a positive charge where it is rubbed and the plastic bag will have a negative charge. Electrons will move from the glass into the plastic and stay in that area. (b) The ebonite rod will develop a negative charge where it is rubbed and the fur will develop a positive charge. Electrons will move from the fur into the ebonite and stay in that area. (c) Both objects will end up with positive charges spread over the surface of the objects. Electrons will move from the neutral rod into the positively charged rod. (d) Both objects will end up with negative charges spread over the surface of the objects. Electrons will move from the negatively charged sphere into the neutral rod.

420 Unit 3 Electric, Gravitational, and Magnetic Fields

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2.

Bring the negatively charged object close to the neutral grounded object. Remove the ground and then the charged object. The object will end up with a positive charge. In this process electrons in the neutral object are repelled by the negatively charged object and flow out of the object into the ground. 3. The charged object causes the electrons to shift in position inside the object, which causes the centre of negative charge to move closer to the positively charged object or further from the negatively charged object. The centre of positive charge does not move. This causes a charge separation within the object. The neutral object is attracted to the charged object because the centre of the opposite charge on the neutral object is closer than the centre of similar charge to the charging object. 4. (a) Water vapour and ions within the air act as conductors causing the charged object to lose charge over time. (b) (i) More water vapour in the air means there are more charge carriers in the air to act as conductors and promote the loss of charge in the object. (ii) X-ray radiation and other types of radiation are more likely to interact with a neutral gas at greater altitudes, changing the gas into a conductor which reduces the charge on the object. 5. (a) Aluminum is a conductor. It is used in the selenium-coated drum to conduct charge away from the selenium that has been exposed to light. (b) Paper is an insulator. Positive charge is spread uniformly over its surface to attract the negatively charged toner particles away from the selenium.

Applying Inquiry Skills 6.

Diagrams and designs may vary but most will involve using charged conductors to attract the charged particles in the air ducts. Maintenance is required to clean these conductors from the particles stuck to the surface. 7. If we use the negatively charged electroscope first and the object is repelled then the object has a negative charge. If it is attracted to the electroscope it could be either neutral or positively charged. If this happened, we could further determine the charge by using the positively charged electroscope. If the object is attracted then the object is neutral, and if it is repelled then the object is positively charged. 8. (a) The paper will be attracted to the charged objects to varying degrees. (b) The paper is an insulator and when the charged object is brought close, it causes an induced charge separation in the object, which is then attracted to the charged object. (c) The charged object can lose its charge over time and loses the ability to hold up the paper. (d) Charge from the sphere is transferred to the paper. The similarly charged objects now repel one another and jump apart.

Making Connections 9. (a) Instead of reflecting light from a copy, the copy is stored digitally and a laser is aimed at the selenium drum to release the toner in places on the copy where it is not needed. (b) The laser beam can be very thin, making the possible detail of the copy greater. 10. (a) The clothes rub up against each other in the dryer and transfer charge in the process. The oppositely charged clothes then attract each other. (b) The fabric softener sheet is used to help transfer charge back and forth between the clothes after the charge develops to help reduce the buildup of charge on the clothes. 11. (a) There is friction between the plane and the air as it moves through the air resulting in a buildup of charge on the plane. (b) There is water vapour and ions in the air that act as conductors to remove the charge from the plane. (c) Charge will concentrate on longer and thinner areas of a conductor. The charge that gathers on these strips is more likely to attract ions which will then reduce the charge on the plane.

7.2 ELECTRIC FORCES: COULOMB’S LAW PRACTICE (Pages 330–331)

Understanding Concepts 1.

d = 10.0 cm F1 = 3.0 ¯ 10–6 N

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Chapter 7 Electric Charges and Electric Fields 421

(a) F2 = ? F ∝ q1q2 F2 ( 2q1 )( 2q2 ) = F1 q1q2 F2 = 4 F1 = 4(3.0 × 10−6 N) F2 = 1.2 × 10−5 N The resulting force would be 1.2 ¯ 10–5 N. (b) F2 = ? F ∝ q1q2 1  q (q ) F2  2 1  2 = F1 q1q2 1 F1 2 1 = (3.0 × 10−6 N) 2 F2 = 1.5 × 10 −6 N F2 =

1 F1 2 The resulting force would be 1.5 ¯ 10–6 N. (c) F2 = ? 1 F∝ 2 r F2 d12 = F1 d 2 2 F2 =

F2 = F1

d12 d22

(10.0 cm ) 2 (30.0 cm ) 2

= (3.0 × 10−6 N)

F2 = 3.3 × 10 −7 N The resulting force would be 3.3 ¯ 10–7 N. 2. F1 = 3.6 ¯ 10–5 N d1 = 0.12 m (a) d2 = 0.24 m F2 = ? 1 F∝ 2 r F2 d12 = F1 d 2 2 F2 = F1

d12 d22

 0.12 m  = (3.6 ×10 N)    0.24 m  F2 = 9.0 × 10 −6 N The force of repulsion is 9.0 ¯ 10–6 N. (b) d2 = 0.30 m F2 = ?

2

−5

422 Unit 3 Electric, Gravitational, and Magnetic Fields

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F2 = F1

d12 d22

 0.12 m  = (3.6 × 10 −5 N)    0.30 m  F2 = 5.8 × 10 −6 N The force of repulsion is 5.8 ¯ 10–6 N. (c) d2 = 0.36 m F2 = ? d2 F1 = F2 1 2 d2

2

2

3.

4.

5.

 0.12 m  = (3.6 × 10 −5 N)    0.36 m  F2 = 4.0 × 10−6 N The force of repulsion is 4.0 ¯ 10–6 N. q1 = 5.0 ¯ 10–8 C q2 = 1.0 ¯ 10–7 C r = 5.0 cm = 5.0 ¯ 10–2 m F=? kq q F = 12 2 r (9.0 ×109 N ⋅ m2 /C2 )(5.0 ×10−8 C )(1.0 ×10−7 C ) = 2 (5.0 ×10−2 m ) F = 1.8 ×10 −2 N The force between the charges is 1.8 ¯ 10–2 N. q1 = 1.5 × 10–8 C q2 = 3.2 × 10–7 C r = 1.5 m F=? kq q F = 12 2 r ×109 N ⋅ m 2 /C 2 )(1.5 ×10 −8 C )(3.2 ×10 −7 C ) 9.0 ( = 2 (1.5 m ) F = 1.9 N The force between the charges is 1.9 N. k = 9.0 ¯ 109 N·m2/C2 r = 4.0 ¯ 10–2 m F = 1.2 ¯ 10–2 N q=? Let the charges be q and 2q in magnitude. kq q F = 12 2 r 9.0 × 109 N ⋅ m 2 /C 2 ) ( q )( 2q ) ( −2 1.2 ×10 N = 2 (4.0 ×10−2 m ) q 2 = (1.07 × 10−22 C )

2

q = 1.0 × 10−11 C Therefore, one charge is 1.0 ¯ 10–11 C, and the other charge is 2(1.0 ¯ 10–11 C) = 2.0 ¯ 10–11 C. It doesn’t matter which charge is positive.

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Chapter 7 Electric Charges and Electric Fields 423

6.

q1 = q2 = 1.1 × 10–7 C F = 4.2 × 10–4 N r=? kq1 q2 r2 kq1q2 r= F

F=

(9.0 ×10 N ⋅ m /C )(1.1×10 C ) 9

=

2

2

−7

2

4.2 ×10 −4 N

= 0.259 m 2

7.

r = 0.51 m The centres of the two charges are 0.51 m apart. Resolving the tension in the x and y directions and using the fact that the object is in equilibrium: T cos 15º = mg T sin 15º = FE FE sin15° = = tan15° mg cos15° FE = mg tan15° = ( 2.0 × 10−3 kg ) (9.8 N/kg )( 0.268 )

FE = 5.25 × 10 −3 N The free body diagram for the sphere on the left:

424 Unit 3 Electric, Gravitational, and Magnetic Fields

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For the distance between the two charges (r): r 2 = sin15° 0.30 m r = 2(0.30 m)(sin15°) r = 0.155 m Then FE =

kq1q2 , but q1 = q2 = q r2 Fd 2 k

q=

(5.25 ×10

=

−2

N ) (0.155 m )

2

9.0 ×109 N ⋅ m 2 /C 2

= 1.40 × 10−14 C 2 q = 1.2 × 10 −7 C The magnitude of the charge on each sphere is 1.2 ¯ 10–7 C.

PRACTICE (Page 334)

Understanding Concepts 8.

qA = –4.0 ¯ 10–6 C qB = –6.0 ¯ 10–6 C qC = +9.0 ¯ 10–6 C rAB = rBC = 10.0 cm F1 , F2 , F3 = ?

For the force exerted on sphere 1 by sphere 2: kq q F12 = 1 2 2 r12

(9.0 ×10 N ⋅ m /C )( 4.0 ×10 C )(6.0 ×10 C ) 9

=

2

−6

2

−6

(0.50 m )

2

F12 = 0.864 N [left]

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Chapter 7 Electric Charges and Electric Fields 425

For the force exerted on sphere 1 by sphere 3: kq q F13 = 1 2 3 r13

(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )(9.0 ×10 C ) 9

=

2

−6

2

−6

(1.0 m )

2

F13 = 0.324 N [right] The total force acting on sphere 1 is: F1 = F12 + F13 = 0.864 N [left] + 0.324 N [right] F1 = 0.54 N [left] For the force exerted on sphere 2 by sphere 3: kq q F23 = 2 2 3 r23

(9.0 ×10 N ⋅ m /C )(6.0 ×10 C )(9.0 ×10 C ) 9

=

2

−6

2

−6

(0.50 m )

2

F23 = 1.944 N [right] The total force acting on sphere 2 is: F2 = F21 + F23 = 0.864 N [right] + 1.944 N [right] F2 = 2.8 N [right] The total force acting on sphere 3 is: F3 = F31 + F32 = 0.324 N [left] + 1.944 N [left]

9.

F3 = 2.3 N [left] The net forces acting on spheres 1, 2, and 3 are 0.54 N [left], 2.8 N [right], and 2.3 N [left]. q1 = q2 = q3 = –4.0 ¯ 10–6 C F1 , F2 , F3 = ?

426 Unit 3 Electric, Gravitational, and Magnetic Fields

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For the magnitude of each repulsive force: kq q F = 12 2 r

(9.0 ×10 N ⋅ m /C )(4.0 × 10 C ) = 9

2

−6

2

2

(0.20 m )

2

F = 3.6 N For sphere 1:

Using components in the x – y plane: F3 x = − F3 cos 60° = −(3.6 N)(0.50) = −1.8 N F3 y = F3 sin 60° = (3.6 N)(0.867) = 3.1 N F2 x = F2 cos 60° = (3.6 N)(0.50) = 1.8 N F2 y = F2 sin 60° = (3.6 N)(0.867) = 3.1 N

∑F ∑F

x

= −1.8 N + 1.8 N = 0

y

= 3.1 N + 3.1 N = 6.2 N

The net force on sphere 1 is: ∑ F1 = 6.2 N [up]

∑F

1

= 6.2 N [outward, 150° away from the side]

The net force acting on sphere 1 is 6.2 N [outward, 150° away from the side]. The net forces on spheres 2 and 3 have the same magnitude as sphere 1, and act along lines away from the triangle at an angle of 150º from the sides.

Section 7.2 Questions (Pages 335–336)

Understanding Concepts 1. (a) The electric force is very strong, is directly proportional to the product of the charges, and is inversely proportional to the distance between them squared. The gravitational force is much weaker and is directly proportional to the product of the masses instead of the charge, but is also an inverse square law. (b) Coulomb’s law: The force between two point charges is inversely proportional to the square of the distance between the charges and directly proportional to the product of the charges. Newton’s law of universal gravitation: The force of gravitational attraction between any two objects is directly proportional to the product of the masses of the objects, and inversely proportional to the square of the distance between their centres.

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 427

Information to be entered into Venn diagram on page 335: Newton’s law differences: Directly proportional to the product of the two masses. The force is relatively weak. Can only attract. Common: Both are inverse square laws. Forces act along the line between the centers of mass or charge. Distance is measured from the centers of the spheres. Coulomb’s law differences: Directly proportional to the product of the two charges. The force is relatively strong. Can attract and repel. 2. (a) F2 = ? 1 F∝ 2 r F2 d12 = F1 d 2 2 F2 = F1

d12 d22

=F

(1) 2 (3 ) 2

F 9 The magnitude of the electric force would be divided by a factor of 9. (b) F2 = ? 1 F∝ 2 r F2 d12 = F1 d 2 2 F2 =

F2 = F1

d12 d22

(1)

2

=F

2

1   2 F2 = 4 F The magnitude of the electric force would be multiplied by a factor of 4. (c) F2 = ? F ∝ q1q2 F2 ( 2q1 )( 2q2 ) = F q1q2 F2 = 4 F The magnitude of the electric force would be multiplied by a factor of 4. (d) The magnitude will not change, but the direction will change. They will now attract instead of repel. (e) F2 = ? We first consider the effect of the neutral sphere: F ∝ q1q2 1  q (q ) F2  2 1  2 = F q1q2 F2 =

1 F 2

428 Unit 3 Electric, Gravitational, and Magnetic Fields

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Now the effect of decreasing the distance: 1 F∝ 2 r F2 d12 = 2 1 F d2 2 1 d2 F2 = F 1 2 2 d2 1 (1) F 2  2 2   3  1  9  =    F  2  4  9 F2 = F 8 2

=

9 . 8

The magnitude of the sphere would increase by a factor of 3.

q1 = 5.0 × 10–6 C q2 = 4.0 × 10–6 C r = 2.0 m F=? F= =

kq1 q2 r2 (9.0 ×109 N ⋅ m2 /C2 )(5.0 ×10−6 C )( 4.0 ×10−6 C )

( 2.0 m )

2

F = 4.5 ×10 −2 N The magnitude of the force is 4.5 ¯ 10–2 N. 4. (a) m1 = m2 = 10.0 kg r = 5.00 × 102 m FG = ? Gm1m2 FG = r2

(6.67 ×10 =

−11

N ⋅ m 2 /kg 2 ) (10.0 kg )

2

(5.00 ×10 m ) 2

2

FG = 2.67 × 10−14 N The force of gravity between the two objects is 2.67 ¯ 10–14 N. (b) q1 = q2 = 1.0 C r = 5.00 × 102 m F=? kq q F = 12 2 r

(9.0 ×10 N ⋅ m /C ) (1.0 C ) = (5.00 ×10 m ) 9

2

2

2

2

2

F = 3.6 ×104 N The electric force between the two objects is 3.6 ¯ 104 N. (c)

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Chapter 7 Electric Charges and Electric Fields 429

(d) FG = –2.67 ¯ 10–14 N FE = 3.6 ¯ 104 N m = 10.0 kg a=? The force of gravity will pull the objects together and the electric force will force them apart. ∑ F = FG + FE = −2.67 × 10−14 N + 3.6 × 104 N

∑ F = 3.6 ×10

4

N

For the acceleration we have: ∑F a= m 3.6 ×10 4 N = 10.0 kg a = 3.6 × 103 m/s 2 The net force acting on each object is 3.6 ¯ 104 N. The acceleration is 3.6 × 103 m/s2 away from the other object. (e) The answers are the same (3.6 ¯ 104 N, 3.6 ¯ 103 m/s2). The force of gravity does not make any significant contribution to the acceleration because it is so small when compared with the electric force. 5. F = 36 N r = 2.0 m q=? kq q F = 12 2 r

(9.0 ×10 N ⋅ m /C ) (q ) 36 N = 9

2

2

2

( 2.0 m )

2

q 2 = 1.6 ×10 −8 C 2

6.

q = 1.3 × 10 −4 C The charge on each sphere is 1.3 ¯ 10–4 C. The charges could either be positive or negative. mA = mB = 0.10 kg q=? 1 The charge on each sphere is now − q . The free body diagram for sphere A is: 2

430 Unit 3 Electric, Gravitational, and Magnetic Fields

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Resolving the tension in the x and y directions and using the fact that the object is in equilibrium: T cos 12º = mg T sin 12º = FE FE sin12° = = tan12° mg cos12° FE = mg tan12° = (1.0 × 10−1 kg ) (9.8 N/kg )( tan12° )

FE = 2.08 × 10 −1 N For the distance between the two charges (r): r = sin12° 2.0 m r = (2.0 m)(sin12°) r = 0.416 m Then FE =

kq1q2 q but q1 = q2 = 2 r2 4 Fr 2 k

q=

4 ( 2.0 × 10−1 N ) (0.416 m )

2

=

7.

9.0 ×109 N ⋅ m 2 /C 2 q = 3.9 ×10 −6 C The initial charge on B is 3.9 ¯ 10–6 C. q1 = +5.0 µC q2 = –6.0 µC q3 = +7.0 µC r12 = 0.80 m r23 = 0.40 m F1 , F2 , F3 = ?

For the force exerted on sphere 1 by sphere 2: kq q F12 = 1 2 2 r12

(9.0 ×10 N ⋅ m /C )(5.0 ×10 C )(6.0 ×10 C ) 9

=

2

−6

2

−6

( 0.80 m )

2

F12 = 0.422 N [right]

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Chapter 7 Electric Charges and Electric Fields 431

For the force exerted on sphere 1 by sphere 3: kq q F13 = 1 2 3 r13

(9.0 ×10 N ⋅ m /C )(5.0 ×10 C )(7.0 ×10 C ) 9

=

2

−6

2

−6

(1.2 m )

2

F13 = 0.219 N [left] The total force on sphere 1 is: F1 = F12 + F13 = 0.422 N [right] + 0.219 N [left] F1 = 0.203 N [right] For the force exerted on sphere 2 by sphere 3: kq q F23 = 2 2 3 r23

(9.0 ×10 N ⋅ m /C )(6.0 ×10 C )(7.0 ×10 C ) 9

=

2

−6

2

−6

(0.40 m )

2

F23 = 2.363 N [right] The total force acting on sphere 2 is: F2 = F21 + F23 = 0.422 + 2.363 F2 = 1.94 N [right] The total force acting on sphere 3 is: F3 = F31 + F32 = 0.219 N [right] + 2.363 N [left]

8.

F3 = 2.14 N [left] The net force acting on sphere 1 is 0.20 N [right]; the net force acting on sphere 2 is 1.94 N [right]; the net force acting on sphere 3 is 2.14 N [left]. Let the subscript S represent the sides, d represent the smaller diagonal, and D represent the larger diagonal. q1 = q2 = q3 = q4 = 1.0 ¯ 10–5 C r = 1.0 m F on all sides = ? We must calculate forces along the sides and the two on the diagonals. Along the sides we have: kq q FS = 12 2 r

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C ) = 9

2

−6

2

2

(1.0 m )

2

FS = 9.0 × 10−3 N The smaller diagonal d can be found using the cosine law: d 2 = a 2 + b 2 − 2ab cos D = (1.0 m)2 + (1.0 m)2 − 2(1.0 m)(1.0 m) cos 45° d = 0.765 m

432 Unit 3 Electric, Gravitational, and Magnetic Fields

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To calculate the force along the smaller diagonal: kq q Fd = 12 2 r

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C ) = 9

2

−6

2

2

(0.765 m )

2

Fd = 1.54 × 10−2 N The larger diagonal D is: D2 = (1.0 m)2 + (1.0 m)2 – 2(1.0 m) (1.0 m)cos135º D = 1.848 m The force along the larger diagonal D is: kq q FD = 12 2 r

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C ) = 9

2

−6

2

2

(1.848 m )

2

FD = 2.64 ×10 −3 N For the sphere in the bottom left corner, the angle between adjacent forces is 22.5º:

Components in the x direction: x = −9.0 ×10−3 N − 2.64 ×10−3 N(cos 22.5°) − 9.0 ×10−3 N(cos 45°) = −1.78 × 10−2 N Components in the y direction: y = −2.64 × 10−3 N(sin 22.5°) − 9.0 × 10−3 N(sin 45°) = −7.37 × 10−3 N Therefore,

∑F ∑F θ=

+ ( Fy )

=

( Fx )

=

( −1.78 ×10

2

2

−2

N ) + ( −7.37 ×10 −3 N ) 2

2

= 1.9 ×10 −2 N

Fy Fx

 −7.37 × 10−3 N  = tan −1   −2  −1.78 × 10 N  θ = 22° The force on the sphere in the bottom left corner is 1.9 ¯ 10–2 N [158° from the side]. From symmetry the force on the sphere in the upper right corner is also 1.9 ¯ 10–2 N [158° from the side].

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Chapter 7 Electric Charges and Electric Fields 433

For the top left charge we have:

Components in the x direction: x = −9.0 ×10−3 N − 1.54 ×10−3 N(cos 67.5°) + 9.0 ×10−3 N(cos 45°) = −3.22 × 10−3 N Components in the y direction: y = 1.54 × 10−3 N(sin 67.5°) + 9.0 × 10−3 N(sin 45°) = 7.79 × 10−3 N Therefore,

∑F ∑F θ=

9.

+ ( Fy )

=

( Fx )

=

( −3.22 ×10

2

2

−3

N ) + (7.79 ×10−2 N ) 2

2

= 8.4 ×10 −2 N

Fy Fx

 7.79 × 10−3 N  = tan −1   −3  3.22 × 10 N  θ = 68° The force on the top left charge is 8.4 ¯ 10–3 N [112° from the side]. From symmetry the force on the sphere in the bottom right corner is also is 8.4 ¯ 10–3 N [112° from the side]. q1 = 1.6 ¯ 10–5 C q2 = 6.4 ¯ 10–5 C x = 2.0 m q3 = 3.0 ¯ 10–6 C x=? If we place the charge anywhere other than on the line joining the two charges, the forces from the two charges on the third charge will not be parallel and will not cancel. Therefore we must place the charge on the line joining the two charges. If the charge is placed to the left of the smaller charge then both forces will point in the same direction. The same is true if we place it to the right of the larger charge. Either case will not produce a net force of 0. The third force must be placed between the two charges q1 and q2, a distance x from q1, as shown.

434 Unit 3 Electric, Gravitational, and Magnetic Fields

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If there is no net force on q3, F31 = F32 kq3 q1 kq3 q2 = 2 2 x ( 2.0 m − x )

kq3 (1.6 ×10 −5 C ) x2

=

kq3 (6.4 × 10 −5 C )

( 2.0 m − x )

2

Omitting units for simplicity, multiplying both sides by 105 and cancelling q3: (1.6)(2 − x) 2 = 6.4 x 2 6.4 − 6.4 x + 1.6 x 2 = 6.4 x 2 (divide by a factor of 1.6) 4 − 4 x + x2 = 4 x2 3x 2 + 4 x − 4 = 0 Using the quadratic formula: x=

−b ± b 2 − 4ac 2a

−4 ± 16 − 4(3)(−4) 6 −4 ± 8 = 6 4 −12 x= , or 6 6 The negative answer is inadmissible and so the third charge is located 0.67 m from the 1.6 × 10–5-C charge. The third charge divides out of the original equation. We do not need to know its value or even the sign. 10. r = 8.0 cm, 2r = 16.0 cm q1 = q2 = 2.5 ¯ 10–6 C k=? =

The distance between the charges doubles meaning each spring must compress 4.0 cm. The force of the spring must be equal to the repulsion from the other charge: Fs = Fe kx =

kq 2 r2

(9.0 ×10 N ⋅ m /C )( 2.5 ×10 C ) = 9

k ( 0.04 m )

2

−6

2

2

(0.16 m )

2

k = 55 N/m The force constant of the springs is 55 N/m.

Applying Inquiry Skills 11. Determine the spring constant if it is not known. Use the technique of charging one sphere and then touching it to the other, which will divide the charge in half. Grounding and charging by contact will help to reduce the charge by factors of one half again. The charges are similar and so they will repel and compress the spring until the force of the spring is equal to the electric force. The compression of the spring is directly related to the product of the charge and inversely related to the square of the distance between the charges.

Making Connections 12. (a) The universal gravitation constant is very small and often the charges on objects are very small. (b) One set of examples are the force of Earth on objects, the force of gravity on the Moon, and its effect on the tides. In all cases one of the objects has a very large mass. Another set of examples are static cling between clothes from a dryer, paper sticking together from a copier, charge stored on the surface of a TV screen. In all cases the charge can be large and/or the distance between the charges is small.

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Chapter 7 Electric Charges and Electric Fields 435

13. (a) The Moon is in circular orbit around Earth. Fc = FE 4π 2 mm ro kq 2 = 2 T2 ro q=

4π 2 mm ro 3 kT 2 4π 2 (7.35 ×10 22 kg )(3.84 ×108 m )

3

=

(9.0 ×10 N ⋅ m /C )(2.36 ×10 s ) 9

2

2

6

2

q = 5.7 × 1013 C The magnitude of the charge required on each to hold the Moon in orbit is 5.7 ¯ 1013 C. (b) The charge would decrease over time since the charge is so large it might be lost to the atmosphere and then into space. The attraction will decrease and the Moon will escape its orbit around Earth.

7.3 ELECTRIC FIELDS PRACTICE (Pages 343–344)

Understanding Concepts 1. (a) q = 2.4 ¯ 10–6 C FE = 3.2 N [left]

FE q 3.2 N [left] = −2.4 × 10 −6 C = −1.3 × 10 6 N/C [left]

ε =

ε = 1.3 × 106 N/C [right] The magnitude of the electric field at that point is 1.3 ¯ 106 N/C [right]. (b) The electric field at that point is independent of the charge placed there. The answer is the same. 2. ε = 12 N/C q = 2.5 ¯ 10–7 C FE = ? FE = qε = ( 2.5 ×10−7 C ) (12 N/C )

3.

FE = 3.0 ×10 −6 N The magnitude of the electric force is 3.0 ¯ 10–6 N. r = 3.0 m q = 5.4 ¯ 10–4 C ε =? F ε= E q kq = 2 r (9.0 ×109 N ⋅ m2 /C2 )(5.4 ×10−4 C ) = 2 (3.0 m )

ε = 5.4 ×105 N/C [right] The magnitude of the electric field is 5.4 ¯ 105 N/C [right]. 436 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

4.

k = 9.0 ¯ 109 N⋅m2/C2 q1 = 50.0 µC = 5.0 ¯ 10–5 C q2 = –10.0 µC = –1.0 ¯ 10–5 C rXY = 0.45 m rYZ = 0.30 m ε =? First we must calculate the electric field exerted on point Z by point X: kq ε ZX = 2 rZX

(9.0 ×10 Nim /C )(5.0 ×10 C ) 9

=

2

−5

2

( 0.75 m )

2

ε ZX = 8.0 × 105 N/C [right] Now we must calculate the electric field exerted on point Z by point Y: kq ε ZY = 2 rZY

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C ) 9

=

2

−5

2

(0.30 m )

2

ε ZY = 1.0 × 106 N/C [left] Therefore,

ε Z = ε ZX + ε ZY = 8.0 × 105 N/C [right] + 1.0 × 106 N/C [left]

5.

ε Z = 2.0 × 105 N/C [left] The electric field at point Z is 2.0 ¯ 105 N/C [left]. qX = +2.0 ¯ 10–8 C qY = +2.0 ¯ 10–8 C ε =? First we must determine the distance between point X and point Z. This will be equal to the distance between point Y and point Z, since X and Y are both 3.0 cm from the centre of the base of the triangle.

Using the Pythagorean theorem: rXZ = rYZ =

(3.0 cm ) + (4.0 cm ) 2

2

rXZ = rYZ = 5.0 cm  4.0  θ = tan −1    3.0  θ = 53°

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Chapter 7 Electric Charges and Electric Fields 437

We can now calculate the electric field exerted on point Z by point X: kq ε ZX = X2 rZX

(9.0 ×10 N ⋅ m /C )( 2.0 ×10 C ) (5.0 ×10 m ) 9

=

2

−8

2

−2

2

ε ZX = 7.2 × 104 N/C [53° U from R] Similarly, ε ZY = 7.2 × 104 N/C [53° U from L]. Components in the x direction: (ε ZX )X = (7.2 ×104 N/C ) cos 53° = 4.3 ×104 N/C

(ε ZY )X = − ( 7.2 × 104 N/C ) cos 53° = −4.3 × 104 N/C

Components in the y direction: (ε ZX )Y = (7.2 ×104 N/C ) sin 53° = 5.75 × 104 N/C

(ε ZY )Y = (7.2 ×104 N/C ) sin 53° = 5.75 × 104 N/C

Therefore,

6. 7.

(ε Znet )X = 0 (ε Znet )Y = 1.15 × 105 N/C

ε Znet = 1.2 × 105 N/C [Up] The electric field at point Z is 1.2 ¯ 105 N/C [up]. Since the electric field is constant between the plates, ε1 = ε 2 = 3.0 × 103 N/C . ε1 = 3.0 ¯ 103 N/C ε2 = ? ε ∝q ε 2 q2 = ε1 q1 q  ε 2 = ε1  2   q1  1 = (3.0 × 103 N/C )   2 3 ε 2 = 1.5 × 10 N/C The electric field strength would now be 1.5 ¯ 103 N/C.

PRACTICE (Page 347)

Understanding Concepts 8. (a) Since the conductor is in electrostatic equilibrium the charges experience no net force. This means the electric field inside the conductor must be zero. (b) A Faraday cage is a container made of a conductor. Any charge on the cage spreads out over the surface and quickly reaches electrostatic equilibrium. Therefore, there is no electric field inside the container. (c) If the field is not perpendicular to the surface of the conductor then the component of the field parallel to the surface of the conductor causes free electrons to move. This is not the case (since the conductor is in electrostatic equilibrium) and so the electric field is perpendicular to the surface.

438 Unit 3 Electric, Gravitational, and Magnetic Fields

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9. (a) No, the electric field from the charge will cause an induced charge separation between the inner and outer surfaces of the sphere. The field lines will start and end on the charge inside the sphere and the inner surface of the sphere. The excess charge on the surface of the sphere will cause an electric field outside the sphere. (b) If the charge is not inside the sphere then the sphere will shield the inside of the sphere. If we leave the charge inside the sphere then we can shield the outside world by placing an equal but opposite charge on the sphere. This charge will be attracted to the inner surface of the sphere and the electric field lines will end there. There will be no external field lines since there is no net charge on the surface of the sphere. 10. A coaxial cable consists of a solid wire in the middle of the cable surrounded by an insulating sleeve, a metallic cylindrical braid, and an outer insulating jacket. Stray electric fields outside the cable stop at the surface of the metallic braid shielding the current inside the central solid wire.

Section 7.3 Questions (Pages 347–348)

Understanding Concepts 1. (a) If the test charge is not small then its field will significantly affect the nature of the field in the area, changing its nature and what we intended to measure. For a smaller charge the effect is minimal, therefore providing a clearer picture of the nature of the field when no test charge is present. (b) If a negative test charge is used then the direction of the field is opposite to the direction of the net force on the charge. We can easily account for this and so it does not have to be positive. 2.

3. (a) Diagram is not to scale. The distance between the charges should be 15 cm and the dotted circle should have a radius of 3 cm.

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Chapter 7 Electric Charges and Electric Fields 439

(b) The electric field is strongest at C. It is the only point that has both electric fields in the same direction. For the other points the distances are larger for at least one of the charges (reducing the force) and the forces are not in the same direction. This results in a smaller net electric field. (c) Since the distance from any point on the circle to the positive charge is 3.0 cm, the field due to the positive charge is the same in magnitude. Therefore we will concentrate on the field from the negative charge. The point on the circle closest to the negative charge is at S and the field is in the same direction so this is the place where the field is strongest. At W the fields are in opposite directions and so it is reasonable to assume the field is weakest here. (See diagram for 3(a).) 4. (a)

5. 6.

7. 8.

(b) The electric field is strongest at D because both electric fields point in the same direction. (c) The electric field is strongest at S because the fields are in the same direction. It is weakest at W where the fields are in opposite directions. Electric field lines indicate the direction of the net force on a small positive test charge. This test charge cannot have two net forces in different directions because then they are not net forces. Therefore, electric field lines cannot cross. (a) The electric field lines indicate the direction of the force on the charge. (b) The electric field lines indicate the direction of the acceleration. (c) The electric field lines are not related to the velocity of the charge. F Magnitude of the electric field is defined as the electric force per unit positive charge, or , therefore they are the same. q The negative charge causes a charge separation between the inner and outer surfaces of the sphere. Electric field lines extend from the inner surface of the sphere (starting from positive charge) and end on the negative charge. Since the outer surface has a negative charge, electric field lines end on the surface of the sphere directed towards the centre of the sphere.

Applying Inquiry Skills 9.

First students must decide on a method to measure or indicate the electric field of the plates. A computer probe or rayon fibers in mineral oil are good choices. Then they must decide what properties they will investigate. Examples include the effect of the amount of charge on the plates, and the distance between the plates. They can then investigate the effect on

440 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

the field between the plates, the field near the end of the plates (curved), and the field outside the plates (zero). They must ensure that they change only one factor at a time. 10. (a) The field lines from the small charge must end on an equal amount of negative charge, which then induces a charge separation on the inner and outer surfaces of the sphere. Since the outside of the sphere now has a net positive charge we get a reading on the electrometer.

(b) The negative charge on the inside of the large sphere is equal to the positive charge on the small sphere. When they come into contact, electrons move into the small sphere, making it neutral.

(c) The positive charge on the outside of the large sphere stays where it is. Students should remember that the excess charge on a conductor spreads out over the outer surface. (d) We cannot shield an electric field from the outside by placing it in a neutral conductor. 11. By definition, a field of force exists in a region of space when an appropriate object (in this case a charge) placed in the field experiences a force. Since the test charge experiences a force then there is a field. 12. (a) The planet experiences a force of attraction towards the star (and vice versa) and so there is a field. In this case the appropriate object is anything with mass (the planet) and it is located in the field of the star. (b) The electron experiences a force of attraction towards the proton (and vice versa) and so there is a field. In this case the appropriate object is anything with charge (the electron) and it is located in the field of the electron. However, students should remember that the field of the electron is similar in magnitude to the field of the proton (unlike the planet and star) and so we can reverse the roles here easily. The mass of the electron is a lot less than the proton and so it is usually described this way. (The electron orbits the proton.) 13. (a) Student answers may include the following concepts: Like gravity this is an action-at-a-distance force, the main reason for introducing fields. Since charges attract and repel when they are not in contact it seems reasonable to describe them with a field because we are familiar with the concept and it has worked well in the past. Both laws are inverse square laws and so it also seems reasonable to assume they both involve fields. Test charges placed anywhere in the area of another charge experience a force. This alone implies a field in the area around the field. (b) It must be an action-at-a-distance force involving an appropriate object that experiences a force anywhere in the field.

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 441

Making Connections 14. One of the devices could be causing an electric field in the area, and if the other device is close enough its operation may be affected if it is not shielded properly. Two possible solutions are to increase the distance between the objects since the field strength obeys an inverse square law, or check the covering of the improperly working device to make sure it is properly in place. 15. Some eels use electric fields to stun fish while most other organisms use electric fields to detect prey. In many cases, the ability to detect these fields is similar to that of the hammerhead.

7.4 ELECTRIC POTENTIAL PRACTICE (Page 354)

Understanding Concepts 1.

r = 25 cm V = –6.4 ¯ 104 V k = 9.0 ¯ 109 N⋅m2/C2 q=? kq r rV q= k (0.25 m ) ( −6.4 × 104 V ) = (9.0 ×109 N ⋅ m 2 /C2 )

V=

2.

q = −1.8 × 10 −6 C The point charge is –1.8 ¯ 10–6 C. W = 4.2 ¯ 10–3 J q = 1.2 ¯ 10–6 C V=? W V= q 4.2 × 10−3 J 1.2 × 10−6 C V = 3.5 × 103 V The potential difference between X and Y is 3.5 ¯ 103 V. r = 5.0 mm = 5.0 ¯ 10–3 m V = 3.0 ¯ 102 V V ε= r 3.0 ×10 2 V = 5.0 × 10−3 m ε = 6.0 ×104 N/C The electric field strength is 6.0 ¯ 104 N/C. r = 1.2 cm = 1.2 ¯ 10–2 m ε = 1.5 ¯ 104 N/C V=? V ε= r V = εr =

3.

4.

= (1.5 × 105 N/C )(1.2 × 10 −2 m )

V = 1.8 × 102 V A potential difference of 1.8 ¯ 102 V would have to be maintained. 442 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

PRACTICE (Page 358)

Understanding Concepts 5. (a) Inside the clouds there is a charge separation, with a negative charge on the bottom. The negatively charged bottom of the cloud repels the electrons in the ground, inducing a positive charge on the surface of the ground. Since the surfaces are oppositely charged there is an attraction between the two. (b) If the bottom of the cloud is negatively charged then the lightning rod will be positively charged because the electrons in the lightning rod are repelled. (c) The electric field near the tip of the lightning rod will be strong enough to ionize the air (because of the small radius of curvature), changing the air from an insulator to a conductor. Lightning will more readily follow this path of ionized air towards the lightning rod striking it instead of the roof of the building. 6. (a) Figure 13 on page 356 shows a positively charged layer of extracellular fluid parallel to a negatively charged layer of intracellular fluid. These two parallel layers are separated by a cell membrane giving the appearance of a set of parallel plates. (b) ∆V = 0.070 V d = 5.0 nm = 5.0 ¯ 10–9 m ε=? If we assume this acts like a set of parallel plates then the electric field is uniform: ∆V ε= d 0.070 V = 5.0 × 10−9 m ε = 1.4 × 107 N/C The magnitude of the electric field is 1.4 ¯ 107 N/C. (c) q = 1.6 ¯ 10–19 C W=? W = Vq = (0.070 V)(1.6 × 10−19 C) W = 1.1× 10 −20 J You would have to do 1.1 ¯ 10–20 J of work on the sodium ion.

Section 7.4 Questions (Pages 358–359)

Understanding Concepts 1.

r = 0.35 m V = 110 V = 1.1 ¯ 102 V k = 9.0 ¯ 109 N·m2/C2 q=? kq r rV q= k (0.35 m ) (1.1× 102 V ) = (9.0 ×109 Nm 2 /C2 )

V=

q = 4.3 × 10 −9 C The electric charge is +4.3 ¯ 10–9 C.

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 443

2.

r1 = 0.16 m r2 = 0.40 m q1 =? q2 Assuming both charges do not equal zero, one charge must be negative and the other positive. (Otherwise they could not add to zero.) kq V = r V1 + V2 = 0 kq1 kq2 + =0 r1 r2 kq1 kq =− 2 r1 r2 q1 r =− 1 q2 r2 =−

0.16 m 0.40 m

q1 = −0.40 q2 q1 is –0.40. q2 3. (a) The electric field is zero at X, the point midway between the two charges. The ratio of

(b) No. Electric potential is a scalar quantity and in this case both potentials are positive therefore the sum of the two potentials is always greater than zero. (c) Electric field is a vector quantity that can equal zero when the two fields from each charge are equal in magnitude but opposite in direction (at X). Electric potential is a scalar quantity, and since the electric potential from each positive charge is positive they cannot add to zero. 4. (a) Electric potential is a scalar measurement and electric field is a vector measurement. The magnitude of the electric field is equal to the change in the potential difference over the distance. (b) Electric potential is the electric potential energy stored per unit charge. Electric potential is a characteristic associated with a single charge while electric potential energy is the energy stored in a system of two or more charges. 5. A trivial example is if there are no charges present then the electric potential is zero and so is the electric field. If there are charges present then there must be at least two charges of the opposite signs otherwise the electric potential cannot be zero. If the charges are equal in magnitude (to simplify the argument) then the electric potential is zero anywhere on the perpendicular bisector of the line joining the two charges, but the electric field can only be zero on the line joining the two charges and it is not zero between the two charges. Therefore, if there are charges present the electric field is not zero. Similar arguments can be given by examining the equations and the vector nature of electric fields. 6. Yes, if the particle has the same sign of charge as the electric potential then the product of the electric potential with the charge will be positive, and the electric potential energy will increase from a lower value to a higher value.

444 Unit 3 Electric, Gravitational, and Magnetic Fields

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7.

V = 120 V ε = 3.0 ¯ 106 N/C r=? V r V r= ε

ε=

120 V 3.0 × 106 N/C r = 4.0 × 10−5 m The smallest possible gap between the plates is 4.0 ¯ 10–5 m. 8. (a) q1 = +4.0 µm = +4.0 ¯ 10–6 C q2 = +2.0 µm = +2.0 ¯ 10–6 C q3 = –4.0 µm = –4.0 ¯ 10–6 C r = 2.0 m ET = ? =

Since electric potential energy is scalar, we simply add all the energy associated with the different combinations of charges. kq q E= 1 2 r (9.0 ×109 N ⋅ m2 /C2 ) ( 4.0 ×10−6 C )(2.0 ×10−6 C ) + (4.0 ×10−6 C )(−4.0 ×10−6 C ) + ( −4.0 ×10−6 C )(2.0 ×10−6 C ) ET = 2.0 m ET = 3.6 ×10−2 J The total electric potential energy is 3.62 ¯ 10–2 J. (b) VT = ? The distance from a charge to the middle of the opposite side is 3 m for triangles with angles 30º, 60º, 90º. (Students may also use the Pythagorean theorem.) Otherwise the distance is 1.0 m. For the bottom side: V=

kq r

 −4.0 ×10 −6 C 2.0 ×10 −6 C 4.0 ×10 −6 C  + + VT = (9.0 × 109 N ⋅ m 2 /C 2 )   1.0 m 3m   1.0 m VT = 2.8 ×103 V For the top left side:  −4.0 ×10 −6 C 4.0 ×10 −6 C 2.0 ×10 −6 C  + + VT = (9.0 × 109 N ⋅ m 2 /C 2 )   1.0 m 3m   1.0 m VT = 1.0 ×10 4 V For the top right side:  4.0 × 10−6 C 2.0 × 10−6 C 4.0 × 10−6 C  + − VT = (9.0 ×109 N ⋅ m 2 /C 2 )   1.0 m 3m   1.0 m VT = 3.3 × 10 4 V The electric potential for the top left side is 1.0 ¯ 104 V, for the top right side is 3.3 ¯ 104 V, and at the midpoint of the bottom side is 2.8 ¯ 103 V. 9. (a) r = 12 cm = 0.12 m V = 85 kV = 8.5 ¯ 104 V q=?

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Chapter 7 Electric Charges and Electric Fields 445

kq r rV q= k (0.12 m ) (8.5 × 104 V ) = (9.0 ×109 N ⋅ m2 /C2 )

V=

q = 1.1× 10 −6 C The charge on the sphere is 1.1 ¯ 10–6 C. (b) ε = ? V ε= r 8.5 ×10 4 V = 0.12 m ε = 7.1× 105 N/C The electric field is 7.1 ¯ 105 N/C. (c) No, the electric field near the surface would have to be at least 3.0 × 106 N/C at the surface to ionize air.

Applying Inquiry Skills 10. Place a charge on a sphere of radius R and measure the potential difference between the sphere and the ground. Measure the potential difference between the charged sphere and another neutral sphere of radius r. Connect the two spheres using a long wire ensuring the spheres are far apart. Measure the potential difference between the two spheres again. Repeat for different amounts of charge on the spheres. 11. The negatively charged bottom of the clouds will induce a positive charge on the lightning rod. Because the radius of curvature of the lightning rod is small it will easily produce large enough electric fields to ionize the air and form positive ions. The positively charged ions will be attracted to the negatively charged plate and the negatively charged sphere reducing the charge on the spheres. The ions will be repelled from the positively charged spheres.

Making Connections 12. (a) The gap is small to maximize the electric field between the conductors and help ionize the gases so that a spark will jump between the conductors. (b) The breakdown value is higher because there are gases around the conductors other than air. The predominant gas is octane and the mixture of these gases and air has a higher breakdown value than air. (c) When a spark jumps from one conductor to the other they become neutral (for an instant) and the potential difference between the two drops to zero and then quickly increases again until the breakdown value is reached. Another spark is created and the process continues. 13. If the bottom of the cloud is negatively charged then the tip of lightning rods will be positively charged. The electric field near the tip of the lightning rod will reach the breakdown value and create positively charged ions in the air. It is believed that these ions might serve to help neutralize the cloud because they are now attracted to the negatively charged cloud. The problem is in the delay (the ions have to reach the cloud before lightning starts) and the shear number of ions required. This implies that if this could work, we would need a high number of lightning rods in the area.

7.5 THE MILLIKAN EXPERIMENT: DETERMINING THE ELEMENTARY CHARGE PRACTICE (Pages 362–363)

Understanding Concepts 1.

q = 8.0 ¯ 10–8 C e = 1.6 ¯ 10–19 C N=?

446 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

q e 8.0 ×10 −8 C = 1.6 ×10 −19 C N = 5.0 × 1011 electrons The number of electrons that must be removed is 5.0 ¯ 1011. N = 1.0 ¯ 108 electrons F=? N=

2.

First we must calculate the charge on each sphere: q = N ( −e) = (1.0 × 108 )(−1.6 × 10−19 C) q = 1.6 × 10−11 C We then calculate the magnitude of the force: kq q F = 12 2 r

(9.0 ×10 N ⋅ m /C )(1.6 ×10 = 9

2

2

(1.0 m )

−11

C)

2

2

3.

F = 2.3 ×10 −12 N The force of electric repulsion is 2.3 ¯ 10–12 N. N = 5.00 ¯ 109 electrons r = 0.500 m ε=? V=? First we must calculate the charge of the object: q = Ne = (5.00 × 109 )(−1.6 × 10−19 C) q = −8.0 × 10−10 C To calculate the electric field intensity: kq ε= 2 r 9.0 ×109 N ⋅ m 2 /C 2 )(8.0 × 10 −10 C ) ( = 2 ( 0.50 m )

ε = 29 N/C To calculate the electric potential: kq V= r 9.0 ×109 N ⋅ m 2 /C 2 )( −8.00 ×10 −10 C ) ( = 0.50 m V = −14 V The magnitude of the electric field is 29 N/C. The electric potential is –14 V. 4. r = 0.050 m FE = 4.5 ¯ 10–15 N (a) q = 6.4 ¯ 10–19 C V=?

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Chapter 7 Electric Charges and Electric Fields 447

First we must calculate the electric field intensity: F ε= E q 4.5 × 10 −15 N 6.4 × 10−19 C ε = 7.0 ×103 N/C =

For the potential difference: V = εr

= (7.0 ×103 N/C ) (0.050 m )

V = 3.5 × 10 2 V The potential difference between the plates is 3.5 ¯ 102 V. (b) m = ? Fg m= g F = E g =

5.

4.5 × 10−15 N 9.8 N/kg

m = 4.6 × 10 −16 kg The mass of the sphere is 4.6 ¯ 10–16 kg. m = 4.95 ¯ 10–15 kg r = 1.0 cm = 1.0 ¯ 10–2 m V = 510 V q=? First we must calculate the force of electric repulsion: FE = Fg = mg = (4.95 × 10−15 kg)(9.8 N/kg) FE = 4.9 × 10−14 N We must also calculate the electric field intensity: V ε= r 510 V = 1.0 ×10 −2 m ε = 5.1× 104 N/C Therefore, to calculate the charge: F ε= E q FE q= ε 4.9 × 10−14 N = 5.1× 10 4 N/C q = 9.6 × 10−19 C

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6.

The charge on the drop is 9.6 ¯ 10–19 C, which can be expressed as a multiple of elementary charge: 9.6 × 10−19 C q= 1.6 × 10−19 C q = 6e Since the upper plate is positive, the charge must be negative due to an excess of electrons. ε = 1.0 ¯ 102 N/C m = 2.0 ¯ 10–15 kg q=? F ε= E q FE = qε = mg q= =

mg ε (2.0 ×10−15 kg ) (9.8 N/kg )

−1.0 × 10 2 N/C q = −2.0 ×10 −16 C The charge on the oil drop is –2.0 ¯ 10–16 C, or a multiple of –1.2 ¯ 103 e.

Section 7.5 Questions (Page 364)

Understanding Concepts 1.

qA = –3q qB = +5q r = 1.5 m FE = 8.1 ¯ 10–2 N (a) N = ? Since the spheres are identical they will share the total charge equally. The charge on each sphere is FE =

( −3q + 5q ) 2

=q.

kq 2 r2

q= =

FE r 2 k

(8.1×10

−2

N ) (1.5 m )

2

9.0 ×109 N ⋅ m 2 /C 2 q = 4.5 × 10 −6 C For this to occur a charge of – 4q must transfer from the negative to the positive charge: − 4 q = N ( − e) N= =

−4 q −e −4 ( 4.5 × 10−6 C )

−1.6 × 10 −19 C N = 1.1× 1014 The number of transferred electrons is 1.1 ¯ 1014. (b) Since the charges are equal, the distance from each charge is equal and the fields are in opposite directions. Therefore, the electric field midway between the two charges is zero.

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Chapter 7 Electric Charges and Electric Fields 449

For the electric potential, r =

1.5 m = 0.75 m, since we are determining the electric potential midway between the two 2

spheres: VT = 2V =2 =2

kq r (9.0 ×109 N ⋅ m2 /C2 )(4.5 ×10−6 C ) 0.75 m 5

VT = 1.1 ×10 V The electric potential is 1.1 ¯ 105 V. (c) FE = ? kq q FE = 12 2 r k (3q )(5q ) = r2 kq 2 = 15 2 r = 15 (8.1× 10−2 N ) FE = 1.2 N The initial electric force is 1.2 N. 2. m = 4.3 ¯ 10–9 kg ε = 9.2 ¯ 102 N/C [up] (a) The charge on the drop must be positive so that the electric force is up to cancel the force of gravity. (b) N = ? FE = Fg qε = mg mg ε mg Ne = ε mg N= εe q=

(4.3 ×10 kg ) (9.8 N/kg ) (9.2 ×10 N/C )(1.6 ×10 N ⋅ m /C ) −9

=

2

−19

2

2

N = 2.9 ×108 The number of extra electrons or protons is 2.9 ¯ 108. 3. m = 4.7 ¯ 10–15 kg (a) q = ? FE = Fg qε = mg ∆V = mg r mgr q= ∆V (4.7 ×10−15 kg ) (9.8 N/kg ) (5.0 ×10−3 m ) = 120 V q = 1.92 × 10−18 C, or 1.9 × 10−18 C The charge on the oil drop is 1.9 ¯ 10–18 C. q

450 Unit 3 Electric, Gravitational, and Magnetic Fields

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(b) N = ? q = Ne q e 1.92 × 10−18 C = 1.6 × 10−19 C N = 12 The number of elementary charges required is 12. (c) The charge must have a deficit of electrons since the top plate is negative. 4. m = 2.0 ¯ 10–5 kg q=? N=

The charges must be similar so that the electric force pushes the objects apart to cancel the force of gravity. FE = Fg kq 2 Gm1m2 = r2 r2 Gm 2 q2 = k q=±

(6.67 ×10

−11

Nm 2 /kg 2 )( 2.0 × 10−5 kg )

2

9.0 ×109 Nm 2 /C 2

q = ±1.7 × 10 −15 C The charge on one object will be +1.7 ¯ 10–15 C, and the charge on the other object will be –1.7 ¯ 10–15 C. 5. m = 5.0 ¯ 10–2 kg NA = 1.0 ¯ 1012 electrons (excess) NB = 4.5 ¯ 1012 electrons (deficit) r = 0.12 m (a) θ = ?

To determine the charge on sphere A: qA = N ( −e ) = (1.0 ×1012 )( −1.6 × 10−19 C) qA = −1.6 ×10 −7 C To determine the charge on sphere B: q B = N ( e) = (4.5 × 1012 )(1.6 ×10−19 C) qB = 7.2 ×10−7 C

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Chapter 7 Electric Charges and Electric Fields 451

We can calculate the electric force: kq q FE = A2 B r ×109 N ⋅ m 2 /C 2 )(1.6 × 10−7 C )(7.2 × 10−7 C ) 9.0 ( = 2 (0.12 m ) FE = 7.2 ×10 −2 N Also, Fg = mg = (5.0 × 10−2 kg)(9.8 N/kg) Fg = 0.49 N From the free body diagram for A: Equating the components: T sin θ = FE T cos θ = Fg Dividing the components: T sin θ FE = T cos θ Fg tan θ =

FE Fg

 7.2 ×10 −2 N  θ = tan −1    0.49 N  θ = 8.4° The angle between the thread and the vertical is 8.4°. (b) T = ? T cos θ = Fg T=

Fg

cos θ 0.49 N = cos 8.4° T = 0.50 N The tension in the thread is 0.50 N.

Applying Inquiry Skills 6. (a) Water evaporates over time and, therefore, while trying to suspend the water droplet the mass is decreasing, which changes the amount of electric force required to suspend the drop. (b) A droplet initially suspended starts to rise as the water evaporates and the mass of the drop decreases. (c) The scientist could not be certain for a variety of reasons. Possible answers include the charge on the drop was not constant, the equipment was not working properly or lacked the ability to suspend the drop for an extended period of time, there is something wrong with the procedure, and the electric field between the plates is not constant. 7. (a) If each charge is a multiple of some elementary charge then each charge can be expressed as an integer times this elementary charge. Dividing any two of the charges will yield a ratio of integers that can be written as a fraction in lowest terms. The numerator and denominator represent the required integers and can be used to find the elementary charge. Other charges should be checked to make sure the others are multiples of this charge. 28125 45 (b) Dividing the charge on oil drop 2 by the charge on oil drop 1 yields 2.8125 or . Therefore the first charge = 10000 16 is 16 times the fundamental charge and the second is 45 times the fundamental charge. Dividing these charges by the number of fundamental charges yields a fundamental charge of 4.0 × 10–20 C. Since this number divides evenly into all the other charges there is no need to make any adjustments.

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(c) No, the fundamental charge could be any positive integer multiple of this charge. We can increase our certainty if we had more experimental data. (d) This charge is actually smaller than the actual fundamental charge and so we must conclude that there is some error in the experiment. If it was a positive integer multiple of the actual fundamental charge then we might conclude that more data is required.

Making Connections 8.

ε = 1.0 ¯ 102 N/C (a) q = ?

ε=

kq rE 2

q=

ε rE 2 k

(1.0 ×10 =

2

N/C )(6.4 ×10 6 m )

2

9.0 ×109 N ⋅ m 2 /C 2 q = 4.5 × 105 C The charge on Earth is 4.5 ¯ 105 C. Since the field lines are directed towards Earth the charge is negative. (b) (i) Both fields can be represented by straight field lines directed towards the centre of the earth. (ii) Gravity always attracts objects while the electric field can either attract or repel objects. (iii) Both fields are inverse square laws meaning they decrease with the square of the distance from the centre of Earth. (c) m = ? FE = Fg eε = mg eε g

m=

(1.6 ×10 =

−19

C )(1.0 × 10 2 N/C )

9.8 N/kg

−18

m = 1.6 × 10 kg The largest mass that could be suspended is 1.6 ¯ 10–18 kg. (d) Both particles have small enough masses but the electric field points towards the centre of Earth meaning Earth has a negative charge. It will attract the proton (which cannot be suspended) and repel the electron. The electron must be at a higher altitude since its mass is so small. 9. The dust particles could have a negative charge and the electric field of the Earth would then suspend them above the surface causing them to float. If this is true a negatively charged ebonite rod will repel them.

7.6 THE MOTION OF CHARGED PARTICLES IN ELECTRIC FIELDS PRACTICE (Page 368)

Understanding Concepts 1.

V = 1.5 ¯ 102 V W = 0.24 J q=?

W = Vq q=

W V

0.24 J 1.5 ×10 2 V q = 1.6 × 10 −3 C The magnitude of the charge is 1.6 ¯ 10–3 C. =

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Chapter 7 Electric Charges and Electric Fields 453

2.

q = 2e m = 6.6 ¯ 10–27 kg V = 2.0 ¯ 103 V v=? (a) Using the equation: W = qV = v= =

1 2 mv 2

2Vq m 2 ( 2.0 × 103 V )(3.2 × 10−19 C ) 6.6 × 10−27 kg

v = 4.4 × 105 m/s The α particle would reach the plate at a velocity of 4.4 ¯ 105 m/s. 1 (b) V = ( 2.0 × 103 V ) = 1.0 × 103 V 2 2Vq v= m =

2 (1.0 × 103 V )(3.2 × 10−19 C ) 6.6 ×10 −27 kg

v = 3.1× 105 m/s The α particle would reach the plate at a velocity of 3.1 ¯ 105 m/s. 3. m = 1.0 × 10–5 kg q = 4.0 × 10–7 C d = 50.0 cm = 0.50 m V = 8.0 × 102 V (a) W = ? W = qV = ( 4.0 × 10−7 C )(8.0 × 102 V )

W = 3.2 ×10 −4 J The work done by the string is 3.2 ¯ 10–4 J. (b) F = ? W F= d 3.2 ×10 −4 J = 0.50 m F = 6.4 ×10 −4 N The average force required is 6.4 ¯ 10–4 N. (c) EK = ? EK = W EK = 3.2 × 10 −4 J The kinetic energy of the pith ball is 3.2 ¯ 10–4 J.

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(d) v = ? 1 2 mv = EK 2 2 EK v= m =

4.

2 (3.2 ×10 −4 J ) 1.0 × 10 −5 kg

v = 8.0 m/s The speed of the pith ball is 8.0 m/s. r = 1.0 ¯ 10–12 m EK = ? v=? kq q EE = 1 2 r

(9.0 ×10 N ⋅ m /C )(1.6 ×10 = 9

2

2

−19

1.0 ×10 −12 m

C)

2

EE = 2.3 × 10 −16 J Etotal = EK + EE = 2.3 × 10 −16 J ′ = EK′ + EE′ Etotal

′ Etotal

(since EK = 0)

1  = 2  mv 2  + 0 2   = mv 2

′ = 2.3 ×10 −16 J But Etotal = Etotal v= =

5.

Etotal m 2.3 × 10 −16 J 9.1× 10 −31 kg

v = 1.6 × 107 m/s The kinetic energy of the each electron is 2.3 ¯ 10–16 J. The speed of each electron is 1.6 ¯ 107 m/s. m = 3.3 ¯ 10–27 kg q = 1.6 ¯ 10–19 C v = 5.0 ¯ 106 m/s V=? 1 W = qV = mv 2 2 2 mv V= 2q

(3.3 ×10 =

−27

kg )(5.0 × 106 m/s )

2

2 (1.6 ×10 −19 C )

V = 2.6 × 105 V The potential difference required is 2.6 ¯ 105 V.

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Chapter 7 Electric Charges and Electric Fields 455

Section 7.6 Questions (Page 371)

Understanding Concepts 1.

V = 1.2 ¯ 103 V d = 0.12 m (a) v = ?

W = qV = v= =

1 2 mv 2

2Vq m 2 (1.2 × 103 V )(1.6 × 10−19 C ) 9.11 ×10 −31 kg

v = 2.1 × 107 m/s The electron will pass through the hole in the positive plate at a speed of 2.1 ¯ 107 m/s. (b) The electron is not pulled back because there is no electric field outside of the parallel plate apparatus. (c) Place a small hole in the negative plate instead. Some source of protons would be required. (A hot filament will not suffice.) (d) v = ? 2Vq v= m =

2 (1.2 × 103 V )(1.6 × 10−19 C ) 1.67 ×10 −27 kg

v = 4.8 × 105 m/s The speed of the emerging proton is 4.8 ¯ 105 m/s. 2. ε = 2.5 ¯ 102 N/C v = 1.2 ¯ 106 m/s (a) r = 2.5 cm = 2.5 ¯ 10–2 m W=? W = qV = qε r

= (1.6 × 10−19 C )( 2.5 × 10 2 N/C )( 2.5 × 10 −2 m ) W = 1.0 × 10−18 J The work done is 1.0 ¯ 10–18 J. (b) v΄ = ? Knowing the relationship between work and energy, W = ∆EK = EK′ − EK , we can calculate the kinetic energy: EK′ = W + EK 1 = W + mv 2 2 = 1.0 × 10 −18 J +

2 1 9.11× 10 −31 kg )(1.2 × 106 m/s ) ( 2

EK′ = 1.7 × 10 −18 J

456 Unit 3 Electric, Gravitational, and Magnetic Fields

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We can then calculate the speed: EK′ = 12 mv′2 2 EK m

v′ =

2 (1.7 ×10 −18 J )

=

9.11× 10−31 kg

v′ = 1.9 × 106 m/s The speed of the electron at this point is 1.9 ¯ 106 m/s. (c) r = ? 1 W = qV = mv 2 2 1 2 qε r = mv 2 1 2 mv r= 2 qε 2 1 (9.11×10−31 kg )(1.2 ×106 m/s ) 2 = (1.6 ×10−19 C )( 2.5 ×102 N/C ) r = 1.6 × 10−2 m The electron will move 1.6 ¯ 10–2 m, or1.6 cm into the field before it comes to rest. 3. v = 3.5 ¯ 106 m/s (a) r = ? ET = ET′ 2 EK = EE′ mv 2 = r= =

ke 2 r2 ke 2 mv 2 k e   mv

9.0 × 109 N ⋅ m 2 /C 2  1.6 × 10−19 C    9.11× 10 −31 kg  3.5 × 106 m/s  r = 4.5 × 10 −6 m The smallest possible distance between the two electrons is 4.5 ¯ 10–6 m. (b) No, this would only occur during a head on collision, which is unlikely to occur for two electrons fired at each other from far apart. 4. m = 6.64 ¯ 10–27 kg q1 = +2e r = 4.7 ¯ 10–15 m q2 = +79e EE = ? kq q EE = 1 2 r =

(9.0 ×10 N ⋅ m /C ) (2 (1.6 ×10 9

=

2

2

4.7 ×10

−19

−15

)(

C ) 79 (1.6 × 10−19 C )

)

m

−12

EE = 7.7 × 10 J The α particle must have an initial energy of 7.7 ¯ 10–12 J.

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Chapter 7 Electric Charges and Electric Fields 457

5.

v = 3.00 ¯ 106 m/s ε = 120 N/C ∆l = 4.0 cm = 4.0 ¯ 10–2 m (a) ∆dy = ? We treat this like a projectile motion problem. The horizontal motion is uniform. The time to cross the plates is: ∆l ∆t = vx 4.0 ×10 −2 m 3.00 × 106 m/s ∆t = 1.3 × 10 −8 s =

The acceleration of the electron is directed up with magnitude: F a= E m eε = m (1.6 ×10−19 C ) (120 N/C ) = 9.11× 10−31 kg a = 2.1× 1013 m/s 2 The initial velocity up is zero, therefore: 1 2 ∆d y = a ( ∆t ) 2 2 1 = ( 2.1 ×1013 m/s 2 )(1.3 ×10 −8 s ) 2 ∆d y = 1.8 × 10 −3 m The vertical deflection of the electron is 1.8 ¯ 10–3 m. (b) v′y = ? v′y = 0 + a∆t = ( 2.1 × 1013 m/s 2 )(1.3 × 10−8 s ) v′y = 2.7 ×105 m/s The vertical component of the final velocity is 2.7 ¯ 105 m/s. (c) θ = ? v′y tan θ = vx  2.7 × 105 m/s  θ = tan −1   6  3.00 × 10 m/s  θ = 5.1° The angle at which the electron emerges is 5.1°.

Making Connections 6. (a) Have the beam of electrons pass through a pair of horizontal parallel plates. The electrons will accelerate towards the positive plate. We can accelerate the electrons up or down by changing the amounts of charge on each plate and changing which plate is negative. (b) Have the beam pass through a pair of vertical parallel plates and deflect them in a similar fashion as explained above. We can increase the deflection from left to right by gradually increasing the potential difference between the plates that will increase the horizontal deflection. We can make the beam jump from right to left by reversing the charge on the plates.

458 Unit 3 Electric, Gravitational, and Magnetic Fields

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(c) The potential difference measured on the surface of the skin of the patient (which is used to indicate potential changes from the heart) is used to control the potential changes on the horizontal plates causing the beam to deflect vertically. The vertical plates change their potential independent of measurements to ensure the beam moves across the screen horizontally at a regular rate. (d) Answers may vary depending on the speed of the electrons, the length of the plates, and the electric field between the plates. For example, in question 5 the deflection is 5.1º, which is close enough. If we double the electric field we will double the acceleration of the electron meaning we will also double the final vertical and so the angle will be about 10º. tan10° This is true even though tangent is not a linear function since = 2 (to two significant digits). tan 5° 7. (a) The objects will accelerate towards each other and as they do so the acceleration and velocity will both increase, but energy and momentum will be conserved. Electric potential energy and gravitational potential energy will be transformed into kinetic energy. (b) The motion will be similar but only gravitational potential energy will be available to convert into kinetic energy.

CHAPTER 7 LAB ACTIVITIES Investigation 7.2.1: Factors Affecting Electric Force between Charges (Pages 372–373)

Questions (i) The force is larger when the distance between the charges is smaller and the charges are larger. The force is directly proportional to the product of the charges, and inversely proportional to the distance between them squared. kq q (ii) FE = 12 2 r

Hypothesis/Prediction (a) The force is directly proportional to the product of the charges and inversely proportional to the distance between them kq q squared: FE = 12 2 . r

Analysis (b)

T cos θ = mg T sin θ = FE F T sin θ = E T cos θ mg FE = mg tan θ (c) Answers may vary. Students will plot a curve (an inverse square).

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Chapter 7 Electric Charges and Electric Fields 459

(d) The electric force is proportional to the inverse of the distance between the charges squared. Graphing produces a straight line. (e) Answers will vary. (f) The electric force is proportional to the product of the two charges. (g) The electric force between two charges is directly proportional to the product of the two charges, and inversely proportional to the distance between them squared.

Evaluation (h) The data clearly shows the predicted results. (i) There is a loss of charge from the spheres to the atmosphere that increases with time. Performing the experiment quickly improves the results. There is some friction between the string and the protractor. Using proper supports for the charges and protractor helps to take accurate measurements. (j) The data at the end of the experiment is not as close to the theoretical prediction as the data at the start. This is due to a loss of charge to the surroundings and a larger percentage error in the smaller measured angles.

Synthesis The forces measured on each charge are equal in magnitude and opposite in direction under all conditions, even when the charges are allowed to move.

Lab Exercise 7.5.1: The Elementary Charge (Page 374) Part 1: An Activity Similar to Millikan’s If each mass measured is a multiple of the mass of one marble then each mass can be expressed as an integer times the mass of one marble. Dividing any two of the measured masses will yield a ratio of integers that can be written as a fraction in lowest terms. The numerator and denominator represent the required integers and can be used to find the mass of a single marble. Other masses should be checked to make sure the others are multiples of this mass. Part 2: Lab Exercise Using Millikan’s Experimental Data

Analysis (a) FE = Fg qε = mg qV = mg r mgr q= V We use this formula to find the charge on each oil drop. Mass of Oil Drop (kg) –15 3.2 × 10 –15 2.4 × 10 –15 1.9 × 10 –15 4.2 × 10 –15 2.8 × 10 –15 2.3 × 10 –15 3.5 × 10 –15 3.7 × 10 –15 2.1 × 10 –15 3.9 × 10 –15 4.3 × 10 –15 2.5 × 10

Electric Potential Difference (V) 140.0 147.0 290.9 214.4 428.8 176.1 214.4 566.6 160.8 597.2 263.4 382.8

460 Unit 3 Electric, Gravitational, and Magnetic Fields

Charge on Oil Drop (C) –18 1.1 × 10 –19 8.0 × 10 –19 3.2 × 10 –19 9.6 × 10 –19 3.0 × 10 –19 6.4 × 10 –19 8.0 × 10 –19 3.2 × 10 –19 6.4 × 10 –19 3.2 × 10 –19 8.0 × 10 –19 3.2 × 10

Copyright © 2003 Nelson

112 7 = , the first charge is 7 elementary charges and the second is 80 5 5 elementary charges. Therefore, the elementary charge is 1.6 ¯ 10–19 C. We check to make sure that all the other charges are multiples of this elementary charge. (b) The method is similar. If each charge is a multiple of some elementary charge then each charge can be expressed as an integer times this elementary charge. Dividing any two of the charges will yield a ratio of integers that can be written as a fraction in lowest terms. The numerator and denominator represent the required integers and can be used to find the elementary charge. Other charges should be checked to make sure the others are multiples of this charge. (c) The individual elementary charges must be constant, the same for all materials (not just oil), and not a multiple of an even more fundamental elementary charge not shown in this data. (d) If just a few measurements were used we might conclude that the elementary charge is larger than the charge we found. All the charges measured might be an even multiple of the elementary charge and we might conclude that the elementary charge is 2e or even larger. If we use the ratio of the first two charges we get

Evaluation (e) To minimize the possibility of finding a smaller elementary charge than what we have calculated the number of values used should be very large. This will help to decrease the chances of only measuring charges that have a common factor in terms of the number of elementary charges on each. If the error in the calculated mass of each charge is too large, or the voltage is too small, then all the elementary charges will be too large.

Activity 7.6.1: The Motion of Charged Particles in Electric Fields (Page 375)

Questions (i) The electric potential energy will decrease and approach zero. The kinetic energy will increase and the sum of the two kinetic energies will approach the initial electric potential energy. (ii) The sum of the electric potential energy and both kinetic energies will always be equal to the initial electric potential energy. (iii) The force between the particles will decrease approaching zero as will the acceleration. The velocity of each particle will increase but approach some constant value.

Analysis (a) (i) The electric potential energy decreases, approaching zero. (ii) The kinetic energy increases and the sum of the two kinetic energies approaches the initial electric potential energy. The kinetic energy of each object is equal. (iii) The electric force decreases approaching zero. The forces are always equal in magnitude but opposite in direction. (iv) The acceleration decreases approaching zero. (v) The velocity increases approaching uniform motion. The velocities are always equal in magnitude but opposite in direction. (b) The sum of the two types of energy is always equal to the initial electric potential energy. The electric potential energy reaches zero and each kinetic energy approaches one half of the initial electric potential energy. (c) The acceleration is not constant meaning our kinematics formulas do not apply. However the total energy (the initial electric potential energy) is constant.

Synthesis (d) Yes, the total energy is always conserved, the forces are always equal in magnitude and opposite in direction, and the total momentum is always zero. As the net force on each object approaches zero the objects approach uniform motion and Newton’s second law can be used to find the instantaneous acceleration.

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Chapter 7 Electric Charges and Electric Fields 461

CHAPTER 7 SUMMARY Make a Summary (Page 376) kq1q2 . Force is a vector and this applies to point charges (not parallel plates r2 or other more complicated charge distributions.) This equation is used to derive many of the other equations in this chapter such as the electric field, electric potential energy, and electric potential either directly or indirectly. For the electric field of a kq point charge we have ε = 21 which is a vector. It can be justified by eliminating the second charge in Coulomb’s law. r The electric potential energy between two point charges is related to the electric force in a similar manner as kq q gravitational potential energy is related to the gravitational force. The equation EE = 1 2 applies to point charges only and r kq1 represents a scalar. The energy per unit charge or electric potential V = is also a scalar for point charges derived from the r electric potential energy. As a result we can write E = qV and for charged parallel plates we can find the electric field between From Coulomb’s law we have the equation FE =

the plates using the equation ε =

V . r

CHAPTER 7 SELF QUIZ (Page 377)

True/False 1. 2. 3. 4. 5. 6.

F The forces are the same in magnitude. F The electric force can also repel and depends on charge not mass, etc. T F The electric field lines indicate the direction of the net electric force on a positive charge. F The charge will stay on the outer surface of the car. T

Multiple Choice 7.

(b)

8.

(e) F = F=

9. 10. 11. 12. 13.

k (2q1 )q2 (3r ) 2 2 kq1 q2 9 r2

 32  (e) ε 2 = ε1  2  6  (e) (b) (b) (b)

CHAPTER 7 REVIEW (Pages 378–379)

Understanding Concepts 1. (a) Charge transmitted to the child’s hair causes the individual strands of hair to repel each other and as a result the hair stands on end. (b) Charge is also transmitted to this child having the same effect on the hair. (c) If the children were grounded the would lose the excess charge and the hair would return to normal. Therefore they are not grounded. 462 Unit 3 Electric, Gravitational, and Magnetic Fields

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2. Answers may vary. Most will include the following: Law Newton’s law of gravitation Coulomb’s law

3.

4.

Factors Affecting Magnitude the product of the masses and the square of the distance between the objects the product of the masses and the square of the distance between the objects

Relative Strength

Direction

weak

always towards the other charge

strong

can repel or attract

The conditions are when we are dealing with small uniformly charged spheres with a small charge, and they are separated by a distance that is large compared to the radius of the sphere. If the charge is uniformly distributed over the surface of the charge then we can treat it as a point charge centered at the centre of the sphere. If the charge or force is large, or the distance is small, the distribution of charge might shift and the distance cannot be measured from the center. Otherwise the objects must be so far apart compared to their own size that they may be considered point charges. F1 = 1.6 ¯ 10–2 N F2 = ? kq q F = 12 2 r  1  1  k  q1   q2   2  2  2 ( 2r ) F2 = kq1q2 F1 r2 2

 1  1   1  F2 = F1       2  2   2  1 = 1.6 × 10 −2 N    16 

5.

F2 = 1.0 × 10 −3 N The magnitude will become 1.0 ¯ 10–3 N. e = 1.602 ¯ 10–19 C F = 4.0 ¯ 10–11 N r=? kq q F = 12 2 r kq1q2 r= F

(9.0 ×10 N ⋅ m /C )(1.6 ×10 9

=

6.

2

2

4.0 ×10 −11 N

−19

C)

2

r = 2.4 ×10 −9 m The distance between the two protons is 2.4 ¯ 10–9 m. r = 5.3 ¯ 10–11 m m1 = 1.67 ¯ 10–27 kg m2 = 9.1 ¯ 10–31 kg

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 463

(a) F = ? F=

kq1 q2 r2

(9.0 ×10 N ⋅ m /C )(1.6 ×10 = (5.3 × 10 m ) 9

2

2

−11

−19

C)

2

2

F = 8.2 ×10 −8 N The electrostatic force is 8.2 ¯ 10–8 N. (b) FG = ? Gm1m2 FG = r2 (6.67 ×10−11 N ⋅ m2 /kg 2 )(1.67 × 10−27 kg )(9.11×10−31 kg ) = 2 (5.3 ×10−11 m ) FG = 3.6 ×10 −47 N The gravitational force is 3.6 ¯ 10–47 N. (c) The electrostatic force is responsible for the electron’s centripetal motion around the proton. (d) v = ? T=? To calculate velocity: mv 2 r Fr v= m

F=

=

(8.2 ×10

−2

N )(5.3 × 10−11 m )

9.11× 10 −31 kg

v = 2.2 ×106 m/s To calculate the period: 2π r v 2π (5.3 ×10 −11 m) = 2.2 × 106 m/s T = 1.5 × 10−16 s The speed of the electron is 2.2 ¯ 106 m/s. The period is 1.5 ¯ 10–16 s. q1 = 4.0 ¯ 10–5 C q2 = –1.8 ¯ 10–5 C r12 = 24 cm = 0.24 m q3 = –2.5 ¯ 10–6 C T=

7.

464 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

(a) r23 = 12 cm = 0.12 m F3 = ? Since the third charge is placed on the side of the negative charge: r13 = r12 + r23 = 0.36 m First, we must calculate the force exerted by the first charge on the third charge: kq q F13 = 1 2 3 r13

(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )( 2.5 ×10 C ) 9

=

2

−5

2

−6

(0.36 m )

2

F13 = 6.9 N F13 = 6.9 N [left] We must then calculate the force exerted by the second charge on the third charge: kq q F23 = 2 2 3 r23

(9.0 ×10 N ⋅ m /C )(1.8 ×10 C )( 2.5 ×10 C ) 9

=

2

−5

2

−6

(0.12m )

2

F23 = 28.1 N F23 = 28.1 N [right] Therefore, for the total force exerted on the third charge: F3 = 28.1 N [right] + 6.9 N [left] F3 = 21 N [right] The force on the third charge is 21 N [right]. (b) r23 = 12 cm = 0.12 m F3 = ? Since the third charge is placed on the side of the positive charge: r13 = r12 – r23 = 0.12 m First we must calculate the force exerted by the first charge on the third charge: kq q F13 = 1 2 3 r13

(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )( 2.5 ×10 C ) = 9

2

−5

2

−6

(0.12 m )

2

F13 = 62.5 N F13 = 62.5 N [right] We then calculate the force exerted by the second charge on the third charge: kq q F23 = 2 2 3 r23

(9.0 ×10 N ⋅ m /C )(1.8 ×10 C )( 2.5 ×10 C ) 9

=

2

−5

2

−6

( 0.12 m )

2

F23 = 3.1 N F23 = 3.1 N [left]

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 465

Therefore, for the total force exerted on the third charge: F3 = 62.5 N [right] + 3.1 N [left] F3 = 59 N [right] The force exerted on the third charge is 59 N [right]. (c) F3 = ? Since the original distance between the two point charges was 0.24 m, F13 = F23 = 0.12 m. First we must calculate the force exerted by the first charge on the third charge: kq q F13 = 1 2 3 r13

(9.0 ×10 N ⋅ m /C )(4.0 ×10 C )( 2.5 ×10 C ) 9

=

2

−5

2

−6

(0.12 m )

2

F13 = 62.5 N F13 = 62.5 N [left] We then calculate the force exerted by the second charge on the third charge: kq q F23 = 2 2 3 r23

(9.0 ×10 N ⋅ m /C )(1.8 ×10 C )( 2.5 ×10 C ) 9

=

2

−5

2

−6

(0.12 m )

2

F23 = 28.1 N F23 = 28.1 N [left] Therefore, the total force on the third charge: F3 = 62.5 N [left] + 28.1 N [left] F3 = 91 N [left] The total force on the third charge is 91 N [left]. 8. The test charge is small so that the effect of the test charge’s electric field on the electric field is minimal. The electric field at any point is the vector sum of all the individual fields present. A ‘small’ test charge would have such a negligible electric field around it that it would not disturb the electric field being measured. 9. It will begin to travel in the direction tangential to the electric field since the electric field is in the direction of the force on the positive test charge. 10. The automobile acts like a Faraday cage. The charge stays on the outside of the car and the electric field inside the car is zero. 11.

466 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

12. q = 1.0 µC = 1.0 ¯ 10–6 C FE = 6.0 ¯ 10–6 N (a) ε = ?

ε=

FE q

6.0 × 10−6 N [right] 1.0 × 10−6 C ε = 6.0 N/C [right] The electric field strength is 6.0 N/C [right]. (b) q = –7.2 ¯ 10–4 C FE = ? =

FE = qε

= ( −7.2 × 10−4 C ) ( 6.0 N/C [right]) = −4.3 × 10 −3 N [right]

FE = 4.3 × 10 −3 N [left] A force of 4.3 ¯ 10–3 N [left] would be exerted on the charge. 13. r = 1.5 m q = 8.0 ¯ 10–3 C ε =? kq ε= 2 r (9.0 ×109 N ⋅ m 2 /C2 )(8.0 ×10−3 C ) = 2 (1.5 m )

ε = 3.2 ×107 N/C [right] The electric field strength is 3.2 ¯ 107 N/C [right]. 14. q1 = 2.0 ¯ 10–5 C q2 = 8.0 ¯ 10–6 C rXZ = 90.0 cm = 0.900 m rYZ = 30.0 cm = 0.300 m ε =? We define the positive direction to the right:

ε = ε X +ε Y kq kq ε = 2X + 2Y rX rY

(9.0 ×10 N ⋅ m /C )(−2.0 ×10 C ) + (9.0 ×10 N ⋅ m /C )(8.0 ×10 C ) = 9

2

−5

2

(0.900 m )

2

9

2

−5

2

(0.300 m )

2

= −2.22 × 105 N/C + 8.00 × 105 N/C

ε = 5.8 × 105 N/C [right] The electric field strength is 5.8 ¯ 105 N/C [right]. 15. m = 3.0 ¯ 10–4 kg r = 10.0 cm = 0.100 m V = 420 V

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 467

(a) ε = ? V r 420 V = 0.10 m ε = 4.2 × 103 N/C The electric field strength is 4.2 ¯ 103 N/C. (b) T = ?

ε=

Drawing a free body diagram for the ball:

Resolving T into horizontal and vertical components: T cos θ = mg T sin θ = FE For θ we have:  0.010 m  θ = sin −1    1.0 m  θ = 0.57° T cos 0.57º = mg T=

(3.0 ×10

−4

)(

kg 9.8 m/s 2

)

cos 0.57° T = 2.9 × 10 N The tension in the thread is 2.9 ¯ 10–3 N. (c) FE = ? T sin θ = FE −3

FE = (2.9 ×10 −3 N)(sin 0.57º ) FE = 2.9 × 10−5 N The magnitude of the electric force is 2.9 ¯ 10–5 N. (d) q = ? F q= E ε 2.9 × 10−5 N = 4.2 × 103 N/C q = 6.9 × 10 −9 C The charge on the ball is 6.9 ¯ 10–9 C.

468 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

16. No net work is required to move a charged particle between two points of the same electric potential. Although there could be paths of zero force, there could also be paths over which the force changes over a great range of values, as long as the net wok is zero. 17. No work is required to move a charged particle through an electric field if it always moves perpendicular to electric field lines. The electric potential remains constant along such a path, by definition. 18. q = 1.2 ¯ 10–3 C w = 30.0 cm l = 40.0 cm FE = ? ε=? V=?

For the length of the diagonal: d = ( w)2 + (l )2 =

(30.0 cm ) + ( 40.0 cm ) 2

2

d = 50.0 cm For the angle of the diagonal: 30.0 cm 40.0 cm θ = 37°

sin θ =

Each charge experiences three forces: one along each side, and one along the diagonal. For the force on q4: kq q F1 = 1 2 4 r14

(9.0 ×10 N ⋅ m /C )(1.2 ×10 C ) = 9

2

−3

2

2

(0.40 m )

2

F1 = 8.1 × 10 4 N F2 =

kq2 q4 r24 2

(9.0 ×10 N ⋅ m /C )(1.2 ×10 C ) = 9

2

−3

2

2

(0.30 m )

2

F2 = 1.44 × 105 N

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 469

F3 =

kq3 q4 r34 2

(9.0 ×10 N ⋅ m /C )(1.2 ×10 C ) = 9

2

2

−3

2

(0.50 m )

2

F3 = 5.2 ×10 4 N Therefore, Fnet = F1 + F2 + F3 . Taking components in the x – y plane: x F1 –8.1 × 104 N 0 F2 F3 (–5.2 × 104 N)(cos 37º) = –4.2 × 104 N –1.23 × 105 N Fnet Fnet = =

2

Fx + Fy

y 0 1.44 × 105 N (5.2 × 104 N)(sin 37º) = 3.1 × 104 N 1.75 × 105 N

2

(−1.23 ×10 N ) + (1.75 ×10 N ) 5

2

5

2

Fnet = 2.1× 105 N  Fy   φ = tan −1   Fx     1.75 × 105 N  = tan −1   5  1.23 × 10 N 

φ = 55° Therefore, Fnet = 2.1× 105 N [55° up from the left] and the force is symmetrically the same at each of the other corners. F = 0 . This is because the forces due to each of the four charges are equal and q opposite, in pairs, so that the net force is 0.

For the electric field at the centre, ε =

For the electric potential: V= =

kq r (9.0 ×109 N ⋅ m2 /C2 )(1.2 ×10−3 C ) 0.25 m

V = 4.32 × 107 V Therefore, for each charge: Vtotal = 4 ( 4.32 ×107 V ) Vtotal = 1.7 ×108 V The electric force on each charge is 2.1 ¯ 105 N [55° up from the left]. The electric field in the centre is zero. The electric potential at the centre is 1.7 ¯ 108 V.

470 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

19. q = 4.5 ¯ 10–4 C r = 0.50 m V=? V= =

kq r (9.0 ×109 N ⋅ m2 /C2 )( 4.5 × 10−4 C ) 0.50 m 6

V = 8.1× 10 V The electric potential is 8.1 ¯ 106 V. 20. q1 = 1.0 ¯ 10–6 C q2 = 3.2 ¯ 10–3 C r1 = 1.0 ¯ 102 cm r2 = 40.0 cm = 0.400 m W=? First we must calculate the electric potential before the charge moved: kq V2 = r2

(9.0 ×10 N ⋅ m /C )(3.2 ×10 C ) 9

=

2

2

−3

0.40 m 7

V2 = 7.2 ×10 V Now we must calculate the electric potential after the charge moved kq V1 = r1

(9.0 ×10 N ⋅ m /C )(3.2 ×10 C ) 9

=

2

2

−3

1.0 m 7

V1 = 2.9 ×10 V To calculate the work: W = q∆V = q (V2 − V1 )

= (1.0 × 10−6 C )(7.2 × 107 V − 2.9 × 107 V ) W = 43 J The amount of work required was 43 J. 21. V = 2.5 ¯ 104 V ∆EK = ? ∆E K = q ∆V

= (1.6 ×10 −19 C )( 2.5 × 10 4 V )

∆EK = 4.0 × 10 −15 J The amount of kinetic energy gained is 4.0 ¯ 10–15 J. 22. r = 1.0 ¯ 10–15 m W=? kq W= r

(9.0 ×10 N ⋅ m /C )(1.6 ×10 = 9

2

2

1.0 × 10−15 m

−19

C)

2

W = 2.3 × 10 −13 J The amount of work done is 2.3 ¯ 10–13 J.

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 471

23. r = 2.0 cm = 2.0 ¯ 10–2 m V = 450 V ε=?

ε=

V r

450 V 2.0 ×10 −2 m ε = 2.3 × 104 N/C The electric field between the two plates is 2.3 ¯ 104 N/C. 24. r = 8.0 cm = 8.0 ¯ 10–2 m ε = 2.5 ¯ 103 N/C V=? V = εr =

= ( 2.5 × 103 N/C )(8.0 × 10−2 m )

V = 2.0 × 102 V The potential difference would be 2.0 ¯ 102 V. 25. The Millikan experiment would not work in a vacuum. The drop could be balanced but when the balancing voltage was removed, it would accelerate under gravity, allowing no measurement of its terminal velocity and hence no calculation of its mass. Also, the oil drops would move very quickly, making them difficult to balance, and they would evaporate. 26. r = ? Fe = Fg kq1q2 = mg r2 r=

ke 2 mg

(9.0 ×10 N ⋅ m /C )(1.6 ×10 C ) (9.11×10 kg ) (9.8 N/kg ) 9

=

2

2

−19

2

−31

r = 5.1 m The other electron would have to be 5.1 m vertically above the first electron. 27. m = 2.6 ¯ 10–15 kg r = 0.50 cm = 5.0 ¯ 10–3 m V = 270 V q=? N=? Since q =

FE , for the electric potential energy: ε FE = Fg FE = mg

= ( 2.6 ×10 −15 kg ) (9.8 N/kg )

FE = 2.55 × 10−14 N For the electric field:

ε=

V r

270 V 5.0 × 10−3 m ε = 5.4 × 104 N/C =

472 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

We can now calculate the charge: F q= E ε 2.55 × 10 −14 N = 5.4 ×10 4 N/C q = 4.7 × 10 −19 C To find the number of electrons: q N= e 4.7 × 10 −19 C = 1.6 × 10−19 C N =3 The charge on the oil droplet is 4.7 ¯ 10–19 C. We cannot tell if there is an excess or a deficit unless we know which plate is positive. 28. m = 0.10 g = 1.0 ¯ 10–4 kg q = 5.0 ¯ 10–6 C r = 25 cm = 2.5 ¯ 10–1 m V=? To calculate the electric potential: FE = Fg FE = mg

= (1.0 ×10 −4 kg ) (9.8 N/kg )

FE = 9.8 × 10−4 N We now calculate the electric field: F ε= E q 9.8 × 10−4 N 5.0 ×10 −6 C ε = 1.96 × 102 N/C =

Using this value, we can calculate the potential difference: V = εr = (1.96 × 10 2 N/C ) (0.25 m )

V = 49 V The potential difference required is 49 V. 29. r = 0.40 m N = 1.0 ¯ 1012 electrons V=? ε=? First we must calculate the charge: q = N ( −e )

= (1× 1012 )( −1.6 × 10 −19 C )

q = −1.6 × 10−7 C

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 473

To calculate the electric potential: kq V= r 9.0 ×109 N ⋅ m 2 /C 2 )( −1.6 ×10 −7 C ) ( = 0.40m V = −3.6 ×103 V To calculate the electric field: kq ε= 2 r 9.0 × 109 N ⋅ m 2 /C 2 )(1.6 × 10−7 C ) ( = 2 (0.40 m )

ε = 9.0 ×103 N/C The electric potential is –3.6 ¯ 103 V. The electric field is 9.0 ¯ 103 N/C. 30. V = 5.0 ¯ 102 V v=? Using the relationships: W = qV = ∆EK =

1 2 mv 2

Therefore, v= =

2qV m 2 (1.6 × 10−19 C )(5.0 × 102 V ) 9.11× 10−31 kg

v = 1.3 × 107 m/s The speed of the electron is 1.3 ¯ 107 m/s. 31. EK = 1.9 ¯ 10–15 J q = +2e V=? ∆EK = qV V =

∆E K q

1.9 ×10 −15 J 3.2 ×10 −19 C V = 5.9 ×103 V The potential difference would be 5.9 ¯ 103 V. 32. vi = 5.0 ¯ 106 m/s vf = 1.0 ¯ 106 m/s V=? =

Using the relationships: W = qV = ∆EK = EK′ − EK We must first calculate the work required: 1 1 W = mvf 2 − mvi 2 2 2 2 2 1 = (9.11× 10 −19 kg ) (1.0 × 106 m/s ) − (5.0 × 106 m/s ) 2 W = −1.09 × 10 −17 J

(

474 Unit 3 Electric, Gravitational, and Magnetic Fields

) Copyright © 2003 Nelson

We now calculate the potential difference: ∆E K V= q 1.09 × 10−17 J 1.6 × 10 −19 C V = 68 V The potential difference is 68 V. 33. rXW = rWY = rYZ = 4.0 cm V1 = 3.0 ¯ 102 V V2 = 5.0 ¯ 102 V (a) v = ? 1 2 mv = qV 2 2qV v= m =

=

2 (1.6 × 10−19 C )(3.0 × 102 V ) 9.11× 10−31 kg

v = 1.0 × 107 m/s The speed of the electron at hole W is 1.0 ¯ 107 m/s. (b) Since plates W and Y are connected together, there is no field between them, so the charge drifts with a constant speed through hole Y. Since 3.0 ¯ 102 V was used to accelerate the electron then the same amount is required to stop it. The distance from the plate Y is: 3.0 × 102 V d = 4.0 cm 5.0 × 102 V d = 2.4 cm Therefore the electron is 4.0 cm – 2.4 cm = 1.6 cm from plate Z when it stops. (c) The path is reversible, so that the electron arrives back at X with v = 0 m/s. Since this 3.0 × 102 V was initially used to accelerate the electron then it will also stop it. 34. v = 3.0 ¯ 106 m/s r=? First we must calculate the kinetic energy: 1  EK = 2  mv 2  2  = (6.6 × 10−27 kg )(3.0 × 106 m/s )

2

EK = 5.94 × 10−14 J At minimum separation: ∆EE = −∆EK

= − (0 J − 5.94 × 10 −14 J )

∆EE = 5.94 × 10−14 J Therefore, kq1 q2 r kq q r= 1 2 EE

EE =

(9.0 ×10 N ⋅ m /C )(3.2 ×10 = 9

2

2

5.94 × 10−14 J r = 1.6 ×10 −14 m The minimum separation is 1.6 ¯ 10–14 m. Copyright © 2003 Nelson

−19

C)

2

Chapter 7 Electric Charges and Electric Fields 475

35. ∆dx = 10 cm = 0.10 m vx = 8.0 ¯ 107 m/s V = 6.0 ¯ 102 V (a) ∆dy = ? For the time the electron takes to move through the apparatus: ∆d x ∆t = vx 0.10 m 8.0 × 107 m/s ∆t = 1.25 × 10 −9 s =

For its vertical acceleration: F a= E m qε a= m

where ε =

V r

Therefore, qV mr (1.6 ×10−19 C )(6.0 ×102 V )

a= =

(9.11×10

−31

kg )( 2.0 × 10−2 m )

a = 5.27 × 1015 m/s 2 For its vertical deflection: 1 a ∆t 2 2 2 1 = (5.27 × 1015 m/s 2 )(1.25 × 10−9 s ) 2 ∆d y = 4.1× 10 −3 m ∆d y =

The vertical deflection of the electron is 4.1 ¯ 10–3 m, or 0.41 cm. (b) v2 = ? For its vertical velocity: v2 y = a ∆ t = (5.27 × 1015 m/s 2 )(1.25 × 10−9 s ) v2 y = 6.6 × 106 m/s To calculate the velocity when the electron leaves the plate apparatus: v2 = v x 2 + v2 y 2 =

(8.0 ×10 m/s ) + (6.6 ×10 m/s ) 7

2

6

2

v2 = 8.0 × 107 m/s To find the angle at which the electron leaves:  6.6 × 106 m/s  θ = tan −1   7  8.0 × 10 m/s 

θ = 4.7° Therefore the electron leaves the plate apparatus at a velocity of 8.0 ¯ 107 m/s [4.7° up from the right].

476 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

Applying Inquiry Skills 36. Students will find that the television screen is not charged if it has been off for a long time. It develops a charge after turning it on and it retains this charge for a while after it has been turned off. 37. Students should investigate the nature of the field near the conductor noticing that it is perpendicular to the surface and zero inside the conductor. This works well if a large metal ring is available. 38. (a) Earth and the plates are negatively charged since the field lines are directed towards them. Beneath the plates the surface of Earth is neutral. (b) The top plate will shield the lower plate from the field lines. The lower plate will lose its charge since the electrons will move into Earth. The ammeter will measure a current. (c) The lower plate will no longer be shielded from the electric field lines. Electrons will flow back into the lower plate and the ammeter will measure a current in the opposite direction. (d) The ammeter will measure an AC current since the charge will repeatedly flow out of and back into the lower plate due to the periodic shielding effect of the upper plate. (e) If the electric field is larger the charge flowing into and out of the lower plate is larger meaning we could calibrate the ammeter to indicate the magnitude of the electric field present. 39. (a) Electric field lines are always perpendicular to the surface of the conductor. The field is strongest at the surface of the conductor closest to the charged plate and very weak on the opposite side. (b) The electric field inside the conductor is zero. The conductor shields the inside from the external charge 40. Electric potential differences across the fluorescent light bulbs cause them to glow. Placing charges on the bulbs causes the necessary potential differences which make the light bulbs glow, as well as bringing a charged object close. The closer the charge and the larger the charge the greater the electric potential difference across the bulb and the brighter the glow. The electric potential difference from the ground to a rural power line is high enough to cause a light bulb to glow as well, as demonstrated in Figure 8 on page 381.

Making Connections 41. Negatively charged dust particles will be attracted to the positively charged film. Introducing positively charged particles into the air will help reduce the number of negatively charged dust particles because these particles will attach themselves to the dust, making them electrically neutral before they reach the film. 42. Pressure on these crystals cause a charge separation and induces an electric potential difference. This electric potential difference is used to help wrist watches keep time since the vibrating crystal has a regular frequency and it loses very little energy. 43. The arc starts when the electric field between the plates is at the breakdown value. This causes the air to ionize, changing it from an insulator to a conductor. The largest electric field will occur between the plates when the distance between the V plates is smallest since ε = . Once the air is ionized in an area it will promote charge to flow from one plate to the other r in that area, in turn causing the spark to climb up the plates.

Extension 44. Gauss’s law relates the electric field through a surface area surrounding a charge (or flux) to the amount of charge inside the surface area. It is used to determine the nature of electric fields near charge distributions or around wires etc. It gives a clearer picture of the electric field produced by charge distributions rather than restricting ourselves to just point charges. 45. v = 2.4 ¯ 106 m/s [45° up from the horizontal] r = 2.5 mm = 2.5 ¯ 10–3 m V = 1.0 ¯ 102 V (a) Approach this like a projectile motion problem except use the electric force instead of gravity. For the vertical acceleration: Fnet = FE ma = qε a= =

qV mr (1.6 ×10−19 C )(1.0 × 102 V )

(9.11×10

−31

kg )( 2.5 × 10−3 m )

a = 7.0 ×1015 m/s 2 [down]

Copyright © 2003 Nelson

Chapter 7 Electric Charges and Electric Fields 477

For the maximum height above the bottom plate the velocity in the y direction is zero: v y′ = v y + a∆t 0 = ( 2.4 ×106 m/s ) sin 45° + ( −7.0 × 1015 m/s 2 ) ∆t ∆t = 2.4 × 10 −10 s Therefore, ∆d y = v y ∆t + 12 a∆t 2 = ( 2.4 ×106 m/s (sin 45° ))( 2.4 × 10−10 s ) +

2 1 −7.0 × 1015 m/s 2 )( 2.4 × 10−10 s ) ( 2

∆d y = 2.1× 10 −4 m If the electron enters in the middle of the plate then it comes 1.25 × 10-3 m – 2.1 × 10-4 m = 1.0 × 10-3 m or about 1.0 mm to the top plate. (b) We must first determine how much time it takes for the electron to reach the bottom plate: 1 ∆d y = v y ∆t + a ∆t 2 2 1 −3 −1.25 × 10 m = ( 2.4 × 106 m/s (sin 45° )) ∆t + ( −7.0 × 1015 m/s 2 ) ∆t 2 2 15 2 6 −3 3.5 × 10 ∆t − 1.7 × 10 ∆t − 1.25 × 10 = 0 Using the quadratic formula: ∆t =

− ( −1.7 × 106 ) ±

( −1.7 ×10 ) − 4 (3.5 ×10 )(−1.25 ×10 ) 2 (3.5 ×10 ) 6 2

15

−3

15

∆t = 8.9 × 10 −10 s To find where it strikes the bottom plate: ∆d x = v x ∆t

= ( 2.4 ×106 m/s ) (cos 45° ) (8.9 ×10−10 s )

∆d x = 1.5 × 10 −3 m The electron strikes the bottom plate at a distance of 1.5 ¯ 10–3 m.

478 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

CHAPTER 8 MAGNETIC FIELDS AND ELECTROMAGNETISM Reflect on Your Learning (Page 382) 1. (a) (b) (c) (d)

The magnetic field forms concentric circles around the conductor. The conductor is perpendicular to these circles. The magnetic field strength is inversely proportional to the distance from the conductor. The current in the conductor affects the magnitude of the magnetic field. The number of turns per unit length, the current in the conductor, and the material in the core all affect the magnitude of the magnetic field. 2. (a) Yes, the charge must be moving with respect to the magnetic field. It will not experience a force when moving parallel to the field. (b) The angle between the velocity and the field, the magnitude and type of charge, and the magnitude and direction of the magnetic field affect the magnitude and direction of the force. (c) The direction will change but the magnitude will not change. The particle will undergo uniform circular motion (not parabolic) until it escapes the field. (d) It will not look the same because the force is always perpendicular to the velocity of the charge. The particle will move with uniform circular motion rather than parabolic. 3. Yes, according to Ampère’s law the total current within a circle outside the cable is zero, and therefore the magnetic field outside the cable is zero. 4. Answers may vary. Possible examples are electric relays, bells, motors, generators, televisions, photocopiers, etc.

Try This Activity: Magnetic Fields (Page 383) (a) The beam is deflected. (b) The beam is deflected in the opposite direction. (c) The presence of a magnetic field and moving charges cause a force on the charges. Light and water particles do not behave in this manner.

8.1 NATURAL MAGNETISM AND ELECTROMAGNETISM Section 8.1 Questions (Page 391)

Understanding Concepts 1.

Iron has domains that act as tiny magnets. These domains can be lined up under certain circumstances to form magnets. Copper does not contain these domains. 2. (a) The iron filings will align along magnetic field lines. If they are carefully removed, the pattern formed by the iron filings will remain because they will still be partially magnetized. (b) Shaking the tube will decrease the magnetic field strength of the iron filings and they will rearrange randomly. (c) The individual iron filings behave similar to domains that line up with external magnetic fields and randomly change when the field is removed and the magnet is struck. Keep in mind that the domains do not physically rotate or move like the iron filings do. 3. 1.

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Chapter 8 Magnetic Fields and Electromagnetism 479

2.

3.

4.

5.

480 Unit 3 Electric, Gravitational, and Magnetic Fields

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6.

4. Field of Earth Magnitude Direction Acts on

gravitational

electric

magnetic

9.8 N/kg

100 N/C

10

toward centre

toward centre

toward north pole

mass

charge

charge

–12

T

5. (a) If we consider a short segment of the loop of wire then we can imagine that it produces a magnetic field similar to a straight conductor. The fields from each segment of the loop combine to produce the magnetic field around the entire loop. The field is stronger inside the loop and weaker outside the loop, and appears similar to the field produced by a bar magnet. The right-hand rule for conductors can be used on an individual segment to predict the direction of the magnetic field of the loop. (b) The field around a long coil of wire is the sum of all the fields of all its loops. As the number of loops per unit length increases, the strength of the field increases and becomes more uniform. The field of such a coil closely resembles the field of a bar magnet.

Applying Inquiry Skills 6. (a) Think of the disk as a wheel that can be turned by applying a force anywhere along its surface at some distance r from the centre of the wheel, where the current is. The further from the current the force is applied, the easier it is to turn the wheel. In fact the turning action (torque) is proportional to r. When there is a steady current in the wire the magnetic field will exert a force on the magnets placed symmetrically about the wire. If the field was uniform in strength then the field closer to the N-poles would cause a greater turning action than the field closer to the S-poles, causing the disk to turn clockwise. Since the disk does not turn, the field must decrease as the distance from the wire increases (according 1 to ) and have circular symmetry because the total turning action is zero. r (b) There is a way of determining if the disk turns at all other than just observing. Placing a protractor beneath the disk on a table and marking a point of reference on the disk will work. The current should be systematically increased and the angle of deviation measured. 7. When the switch is closed there is a current in the circuit and, of course, in the two solenoids with metal cores. This creates a strong magnetic field that attracts the arm of the hammer, swings it towards the bell, and causes the bell to ring. This breaks the circuit at the contact point on the arm, the current stops, and the magnetic field decreases rapidly. The spring then pulls the arm back and completes the circuit again. The process continues to repeat until the person stops ringing the doorbell.

Making Connections 8.

The crystals act as a magnetic compass for the bacterium to help it sense direction and navigate. It uses the magnetotactic crystals to detect the magnetic field of Earth to sense changes in direction. You could test this by carefully monitoring the motion of the bacteria and then placing a magnet in the area to see if if it can influence its motion in any way.

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Chapter 8 Magnetic Fields and Electromagnetism 481

8.2 MAGNETIC FORCE ON MOVING CHARGES Try This Activity: The Vector or Cross Product (Page 394)

G G (a) Since A = qv , we calculate the magnetic field: G G FM = C G G = A× B G G = qv × B G FM = q ( vB sin θ ) This is the magnitude of the magnetic force. (b) The direction of the magnetic force is perpendicular to both the velocity and the magnetic force according to the righthand rule for a cross product. We start with our fingers in the direction of the velocity and turn them towards the magnetic field with our thumb stretched in the direction of the force. This is the right-hand rule for a charge in a magnetic field as outlined in this section.

PRACTICE (Page 396)

Understanding Concepts 1.

2.

The MHD propulsion system is composed of two large electrodes with a large potential difference across them. This causes a current in the seawater due to the presence of charged ions. Large superconducting magnets create a magnetic field perpendicular to the current. According to the right-hand rule, a force will be exerted on the ions towards the rear of the ship. According to Newton’s third law, a force will be exerted on the ship towards the front. v = 8.6 ¯ 104 m/s B = 1.2 T m = 1.67 ¯ 10–27 kg G F= ? F = qvB sin θ

(

)(

)

= 1.6 ×10−19 C 8.6 × 104 m/s (1.2 T )(sin 90° )

3.

G F = 1.7 × 10 −14 N [E] The magnetic force is 1.7 ¯ 10–14 N [E] where the direction is given by the right-hand rule. v = 2.0 ¯ 106 m/s F = 5.1 ¯ 10–14 N G B= ? F B= qv sin θ =

4.

(1.6 ×10

5.1× 10 −14 N

−19

C )( 2.0 × 106 m / s ) (sin 90° )

G B = 0.16 T [horizontal, toward observer] The magnetic field is 0.16 T [horizontal, toward observer]. q = 3.2 ¯ 10–19 C m = 6.7 ¯ 10–27 kg v = 1.5 ¯ 107 m/s B = 2.4 T r=?

482 Unit 3 Electric, Gravitational, and Magnetic Fields

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r= =

5.

mv Bq

(6.7 × 10

−27

kg )(1.5 × 107 m / s )

( 2.4 T ) (3.2 × 10−19 C )

r = 0.13 m The radius of the path of the α particle is 0.13 m. r = 8.0 cm = 8.0 ¯ 10–2 m B = 1.5 T m = 1.67 ¯ 10–27 kg v=? V=? To calculate speed: v= =

rBq m 8.0 ( ×10−2 m ) (1.5 T ) (1.6 ×10−19 C ) 1.67 × 10−27 kg

v = 1.14 × 107 m/s, or 1.1 ×10 7 m/s To calculate voltage: 1 2 mv 2 mv 2 V = 2q

qV =

(1.67 ×10 =

−27

kg )(1.14 × 107 m / s )

2

2 (1.6 × 10−19 C )

6.

V = 6.9 ×105 V The speed of the proton is 1.1 ¯ 107 m/s, and the voltage required is 6.9 ¯ 105 V. v = 2.0 ¯ 102 m/s q = 1.0 ¯ 102 C B = 5.0 ¯ 10–5 T F=? F = qvB sin θ

(

)(

)(

)

= 1.0 ×102 C 2.0 ×102 m/s 5.0 ×10 −5 T (sin 90° ) F = 1.0 N The maximum magnitude of the magnetic force is 1.0 N.

Explore an Issue: Government Spending on Developing New Technologies Students can address the many advantages of these technologies along the lines of improving communications, monitoring the environment, and the many economic improvements. Some may claim that this should be funded solely by the private sector since private companies often benefit from this research.

Section 8.2 Questions (Pages 402–403)

Understanding Concepts 1. (a) The force is down towards the bottom of the page. (b) The velocity is up towards the top of the page.

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Chapter 8 Magnetic Fields and Electromagnetism 483

2.

q = 25 nC = 2.5 ¯ 10–9 C B = 5.0 ¯ 10–5 T [N] v = 12 m/s [W] G F= ? F = qvB sin θ

(

3.

)

(

)

= 2.5 ×10−9 C (12 m/s ) 5.0 × 10−5 T (sin 90° ) G F = 1.5 ×10−12 N [up] The resulting magnetic force is 1.5 ¯ 10–12 N [up], where the direction is given by the right-hand rule assuming the charge on the rod is negative. V = 10.0 kV = 1.00 ¯ 104 V B = 0.40 T r=? First we must calculate the speed of the electron: 1 qV = mv 2 2 2qV v= m =

2 (1.6 × 10 −19 C )(1.00 ×10 4 V ) 9.11× 10−31 kg

v = 5.927 × 107 m/s To calculate the radius: r=

mv Bq

(9.11×10 =

−31

kg )(5.927 × 107 m / s )

(0.40 T ) (1.6 × 10−19 C )

r = 8.4 × 10−4 m The radius of the circular path followed by the electron is 8.4 ¯ 10–4 m. 4. (a) A magnetic field will not cause the electron to move since the magnetic force is directly proportional to the velocity and the velocity is zero (therefore, the force is zero). (b) An electric field will cause the electron to move since the magnitude of the electric force on the charge has nothing to do with the velocity of the charge. 5. The magnetic force is directed towards the centre of the circle while the electric force is directed parallel to the magnetic field and therefore perpendicular to the magnetic force. The electric force will not change direction but the magnetic force will continue to rotate and so the particle will follow a coiled (helix-like) path around the field lines. 6. The direction of the palm and the fingers of the left hand point in the same direction, and the thumb points in the direction of the motion of the negative charges. Therefore, the left-hand rule for negative charges is equivalent to the right-hand rule for positive charges. 7. If the charged particle is moving parallel to the magnetic field lines then the magnitude of the magnetic force is zero since sin 0º = 0 and sin 180º = 0. Therefore a magnetic field could be present with no change in velocity since the acceleration is zero. 8. The kinetic energy will not change since the particle undergoes uniform circular motion and therefore the speed of the particle does not change. Another way to look at it is the force is always perpendicular to the velocity, which means it will do no work and not cause any change in energy. 9. Magnetic field lines indicate the direction of the total force on a charge at any point in space. If the field lines crossed, the total force would have two directions simultaneously or two total forces. Either explanation is not possible implying that magnetic fields can never cross. 10. Magnetic fields cause the particle to move in circular motion (or a helix-shaped path) and the force is proportional to the speed of the particle and the sine of the angle between the velocity and the field. Changing the speed or the direction of the velocity will result in a change in the force and the resulting motion. The electric force from a particular field depends only on the amount of charge and the particles follow parabolic paths. If both fields are present, all of the above factors will affect the motion of the particle and the resulting motion will be more complicated. 484 Unit 3 Electric, Gravitational, and Magnetic Fields

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11. (a) Initially the proton would be deflected east according to the right-hand rule. (b) Initially the electron would be deflected west according to the right-hand rule. (c) The neutron will not be deflected at all since it has no charge. 12. (a) A field of force exists in a region of space when an appropriate object placed at any point in the field experiences a force. (b)

Appropriate Particle Factors Affecting Magnitude of Force Relative Strength

Gravitational any particle with mass

Electric charged particle

Magnetic moving charged particle

magnitude of field, mass of particle

magnitude of field, charge on particle

magnitude of field, charge on particle, speed of particle, angle between velocity and field

very weak

strong

strong

Note the electric and magnetic forces are about 1036 times stronger than the gravitational force. (c) Field theory is considered a general scientific model for several reasons: It can be used to explain these three types of forces and how the force can be transmitted over distances; there are more similarities than differences between the three types of fields; all forces encountered up to this point by students and scientists at the time of its development are explained by fields, etc.

Applying Inquiry Skills 13. Place the magnet near the tube at various distances up to touching the screen noting the effect on the picture. Note any deviations from a normal picture (using another identical tube), and remember that any dark areas indicate a magnetic mirror. Note the size and shape of the dark area.

Making Connections 14. Both use fuel ejected out the back of the units to exert a force forward on the vehicle however MHD uses electric and magnetic fields to accelerate charged particles in seawater and so the fuel will never run out as long as the electric and magnetic field is present. The jet engine must have a fuel source carried along with it and involves internal combustion. 15. (a) The beam of electrons in the tube that are used to form the picture are deflected by the external magnetic field of the magnet. (b) Where the field is strongest the component of the electrons’ velocity that is parallel to the magnetic field is reduced and even reversed because the field is becoming significantly stronger in that area as it approaches the screen. 16. Although presently impractical, forming a magnetic mirror near the cells could help to protect them. Increasing the magnetic field strength as the distance from the cells decreases causes incoming charged particles to spiral around the field lines with the component of the particles velocity parallel to the field lines decreasing or even reversing direction. 17. (a) Both Thomson and Millikan used similar principles. Millikan used the electric force to balance the force of gravity on an oil drop to determine the charge on an electron. Thomson balanced the electric and magnetic forces on electrons to determine the charge to mass ratio of the electron. (b) Millikan used parallel plates and Thomson used cathode ray tubes involving parallel plates and solenoids. (c) Using this technology revealed that what the charge was and that it was quantized. They also revealed that each electron has the same mass and determined the value for it. 18. Electric and magnetic fields are involved in the design and construction of the satellite while an understanding of the force of gravity and how it changes with height is required to launch the satellite. This technology is used in telecommunications, researching weather patterns, tracking animal migrations, studying changes in Earth’s atmosphere and surface, collecting astronomical data, etc.

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Chapter 8 Magnetic Fields and Electromagnetism 485

8.3 MAGNETIC FORCE ON A CONDUCTOR PRACTICE (Page 405)

Understanding Concepts 1.

l = 25 cm = 0.25 m B = 0.20 T I = 15 A θ = 90° F=?

F = IlB sin θ = (15 A )(0.25 m )(0.20 T )(sin 90° )

2.

3.

4.

F = 0.75 N The force exerted on the wire is 0.75 N. B = 0.033 T I = 20.0 A F = 0.1 N θ = 90° l=? F l= BI sin θ 0.10 N = 0.033 T ( )( 20.0 A )(sin 90° ) l = 0.15 m The length of conductor that experiences this force is 0.15 m. l = 1.0 m F = 6.0 ¯ 10–5 N I = 1.5 A θ = 90° B=? F B= Il sin θ 6.0 ×10 −5 N = (1.5 A )(1.0 m )(sin 90° ) B = 4.0 × 10−5 T The magnitude of Earth’s magnetic field is 4.0 ¯ 10–5 T. l = 50.0 m I = 2.0 ¯ 102 A B = 5.0 ¯ 10–5 T θ = 45° F=? F = IlB sin θ

(

)(

)

= 5.0 ×10 −5 T 2.0 × 102 A (50.0 m )(sin 45° ) F = 0.35 N The magnitude of the magnetic force on the wire is 0.35 N.

486 Unit 3 Electric, Gravitational, and Magnetic Fields

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Section 8.3 Questions (Page 407)

Understanding Concepts 1.

l = 1.8 m F = 1.8 N B = 1.5 T (a) Since the force is a maximum then the angle must be 90º so that sin θ is a maximum. (b) I = ? F I= Bl sin θ 1.8 N = (1.5 T )(1.8 m )(sin 90° )

I = 0.67 A The current in the wire is 0.67 A. (c) If the wire is parallel to the magnetic field then sin 0º = 0 and the force is 0 N. 2. l = 2.0 m I = 2.5 A [E] B = 5.0 ¯ 10–5 T [N, horizontal] F=? F = IlB sin θ

(

)

= ( 2.5 A )( 2.0 m ) 5.0 ×10−5 T (sin 90° )

3.

F = 2.5 ×10−4 N The force on the conductor is 2.5 ¯ 10–4 N. The direction of the force is up according to the right-hand rule. l = 1.2 m I = 3.0 A θ = 45° B = 0.40 T F=? F = IlB sin θ = (3.0 A )(1.2 m )(0.40 T )(sin 45° )

F = 1.0 N The force on the wire is 1.0 N. 4. (a) When the circuit is complete, a current will be directed from left to right in the horizontal conductor. Since the field lines are directed up, the magnetic force will act on the conductor into the page. If the force is strong enough the conductor will move towards the magnet. (b) (i) Increasing the current will increase the magnetic force causing the conductor to move more. (ii) If the magnet is inverted the field lines will be directed down and the conductor will move away from the magnet. (iii) If a stronger magnet is used the magnetic force will increase causing the conductor to move more. (c) One can agree with this statement. With the present setup the conductor (and current) is perpendicular to the external magnetic field and therefore sin θ is a maximum. If the bar is not horizontal the angle between the current and the magnetic field will not be 90º and the magnetic force will decrease because sin θ will decrease. 5. (a) Electromagnets mounted on the train and track guideway are set up so they attract each other to pull the train forward and repel each other to push it forward. The magnets on the guideway are rapidly changing by changing the current to ensure that they are always forcing the train forward. (b) To slow the train down, the magnets on the guideway are set up so that they repel the train and push it backward or attract it backward and the current is changed to ensure the magnets keep doing this until it stops.

Applying Inquiry Skills 6. (a) The student will not be able to measure any force on the wire since the angle between the wire and the magnetic field is zero and so the force will always be zero no matter how large the current is. (b) The wire should be turned 90º in the horizontal so the current is directed in the east-west direction. This will maximize the force on the conductor since the current is perpendicular to the field. (c) Using a large U-shaped piece of tinfoil works better since it is light is an excellent conductor (aluminum).

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Chapter 8 Magnetic Fields and Electromagnetism 487

Making Connections 7.

Mercury (a liquid metal) is an excellent conductor and will allow charge to flow through it readily. On the left side, the wire is fixed and will have a circular magnetic field around it and therefore less drag than the end in the mercury. This will cause the top of the magnet to rotate. The reverse is also true when the magnet is fixed and the bottom of the wire is not fixed. Due to Newton’s third law, if the wire exerts a force on the magnet then the magnet exerts an equal and opposite force on the wire.

8.4 AMPÈRE’S LAW PRACTICE (Pages 409–410)

Understanding Concepts 1.

r = 3.5 cm = 3.5 ¯ 10–2 m I = 1.8 A B=?

I 2π r (4π ×10−7 T ⋅ m/A ) (1.8 A )

B = µ0 =

2.

B = 1.0 × 10−5 T The magnetic field strength is 1.0 ¯ 10–5 T. l = 10.0 cm = 0.10 m B = 2.4 ¯ 10–5 T I=? 2π rB I= µ0 =

3.

The current is 12 A. I = 2.4 A B = 8.0 ¯ 10–5 T r=?

( 2π )(0.10 m ) (2.4 × 10−5 T )

(4π ×10

−7

T ⋅ m/A )

I = 12 A

r= =

4.

( 2π ) (3.5 ×10−2 m )

µ0 I 2π B (4π ×10−7 T ⋅ m/A ) (2.4 A )

( 2π ) (8.0 ×10−5 T )

r = 6.0 ×10 −3 m The distance of the magnetic field from the straight conductor is 6.0 ¯ 10–3 m. r = 0.5 m I1 = 10.0 A I2 = 20.0 A B=? To calculate the magnetic field for the first current: I B1 = µ 0 1 2π r π 4 ( ×10−7 T ⋅ m/A ) (10.0 A ) = ( 2π )(0.50 m ) B1 = 4.0 × 10−6 T

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Similarly, B2 = 8.0 × 10−6 T . (a) If I1 and I2 are in opposite directions, B1 and B2 are in the same direction, so that B = B1 + B2 = 1.2 × 10–5 T. (b) If I1 and I2 are in the same direction, B1 and B2 are in the opposite directions, so that B = B1 – B2 = 4.0 × 10–6 T. 5. d = 10.0 cm = 0.10 m I = 5.0 A (a) B = ?  I  B = µ0    2π r   5.0 A  = µ 0    2π (0 m )  B=0T The magnetic field strength at the centre of the rod is 0 T. (b) B = ? First we must calculate the amount of current through a circle of radius 2.5 cm from the center. If we make the assumption that the current is uniformly distributed throughout the rod then the amount of current through this circle is equal to the area of the circle divided by the cross-sectional area of the rod all multiplied by the total current in the rod.

π ( 2.5 cm )

2

I=

π (5.0 cm )

2

(5.0 A )

I = 1.25 A To calculate the magnetic field:  I  B = µ0    2π r  =

(4π ×10

−7

T ⋅ m/A ) (1.25 A )

( 2π ) ( 2.5 × 10−2 m )

B = 1.0 × 10−5 T The magnetic field 2.5 cm from the centre is 1.0 ¯ 10–5 T. (c) B = ?  I  B = µ0    2π r  =

(4π ×10

−7

T ⋅ m/A ) (5.0 A )

( 2π ) (5.0 × 10−2 m )

B = 2.0 × 10−5 T The magnetic field at 5.0 cm from the centre is 2.0 ¯ 10–5 T. (d) B = ? Note that the circle is outside the rod and so we just use the total current in the rod.  I  B = µ0    2π r  =

(4π ×10

−7

T ⋅ m/A ) (5.0 A )

( 2π ) ( 7.5 × 10−2 m )

B = 1.3 × 10 −5 T The magnetic field at 7.5 cm from the centre is 1.3 ¯ 10–5 T.

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Chapter 8 Magnetic Fields and Electromagnetism 489

PRACTICE (Page 411)

Understanding Concepts 6.

I = 12 A l = 15 cm = 0.15 m B = 5.0 ¯ 10–2 T N=? NI L BL N= µ0 I B = µ0

(5.0 ×10 T ) (0.15 m ) = (4π ×10 T ⋅ m/A ) (12 A ) −2

−7

7.

8.

N = 5.0 ×10 2 The number of turns that would have to be wound on the coil is 5.0 ¯ 102. l = 10.0 cm = 0.10 m N = 420 I = 6.0 A B=? NI B = µ0 L π 4 ( ×10−7 T ⋅ m/A ) ( 420 )(6.0 A ) = (0.10 m ) B = 3.2 ×10 −2 T The magnetic field strength is 3.2 ¯ 10–2 T. l = 8.0 cm = 0.80 m N = 400 B= BL I= µ0 N

(1.4 ×10 T )(8.0 ×10 m ) ( 4π ×10 T ⋅ m/A ) ( 400 ) −2

=

−2

−7

I = 2.2 A The current in the coil is 2.2 A.

PRACTICE (Page 413)

Understanding Concepts 9.

d = 1.0 cm = 0.10 m I1 = I2 = 8.0 A F =? l

490 Unit 3 Electric, Gravitational, and Magnetic Fields

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F µ0 I1 I 2 = l 2π d 4π × 10−7 T ⋅ m/A ) (8.0 A )(8.0 A ) ( = ( 2π ) (1.0 × 10−2 m ) F = 1.3 × 10 −3 N/m l The magnitude of the force per unit length is 1.3 ¯ 10–3 N/m. 10. l = 5.0 m d = 12 cm = 0.12 m F = 2.0 ¯ 10–2 N F µ 0 I1 I 2 = l 2π d π dF 2 I2 = µ0 l =

( 2π )( 0.12 m ) ( 2.0 ×10−2 N ) (4π ×10−7 T ⋅ m/A ) (5.0 m )

I = 49 A The maximum possible current in each conductor is 49 A. 11. I1 = 5.0 A I2 = 10.0 A F = 3.6 × 10−4 N/m l d=? F µ0 I1 I 2 = l 2π d µ 0 I1 I 2 d= F 2π   l 

( 4π ×10 =

−7

T ⋅ m/A ) (5.0 A )(10.0 A )

( 2π ) (3.6 × 10−4 N/m )

d = 2.8 ×10 −2 m The wires should be 2.8 ¯ 10–2 m apart. 12. I1 = I2 = 1.0 A d = 2.0 mm = 2.0 ¯ 10–3 m F =? l F µ0 I1 I 2 = l 2π d 4π × 10−7 T ⋅ m/A ) (1.0 A )(1.0 A ) ( = ( 2π ) ( 2.0 × 10−3 m ) F = 1.0 ×10 −4 N/m l The magnitude of the force per metre is 1.0 ¯ 10–4 N/m.

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Chapter 8 Magnetic Fields and Electromagnetism 491

Section 8.4 Questions (Page 414)

Understanding Concepts 1.

r = 0.50 m I = 8.0 A B=? I 2π r (4π ×10−7 T ⋅ m/A ) (8.0 A )

B = µ0 =

2.

3.

4.

( 2π )( 0.5 m )

B = 3.2 ×10 −6 T The magnitude of the magnetic field is 3.2 ¯ 10–6 T. I = 2.0 ¯ 104 A r = 2.5 m B=? I B = µ0 2π r π ×10 −7 T ⋅ m/A )( 2.0 ×104 A ) 4 ( = ( 2π )( 2.5 m ) B = 1.6 × 10−3 T The magnitude of the magnetic field is 1.6 ¯ 10–3 T. L = 25 cm = 0.25 m N = 250 I = 1.2 A B=? NI B = µ0 L (4π ×10−7 T ⋅ m/A ) (250 )(1.2 A ) = (0.25 m ) B = 1.5 × 10−3 T The magnitude of the magnetic field is 1.5 ¯ 10–3 T. l1 = l2 = 3.0 m d = 5.0 mm = 5.0 ¯ 10–3 m I1 = I2 = 2.2 A µ IIl F= 0 1 2 2π d 4 ( π ×10−7 T ⋅ m/A ) ( 2.2 A )( 2.2 A )(3.0 m ) = ( 2π ) (5.0 × 10−3 m )

F = 5.8 × 10−4 N The magnetic force acting on each conductor is 5.8 ¯ 10–4 N. 5. Electric charge will flow through the coil and the mercury causing a magnetic field in the coil as a result of the field of each individual loop of wire. The field from each loop resembles that of a bar magnet resulting in an attractive force between the loops. If the current is large enough the coil will compress and might even lift out of the mercury. The current and the magnetic field will drop to zero in this case and the end of the coil will fall back into the mercury repeating the process over and over again. If an iron rod is suspended in the core of the coil the field will become much stronger and the possibility of the coil contracting will be far more likely. 6. (a) The magnitude of the magnetic field between the two conductors is given by I according to Ampère’s law. The field is circular in nature and decreases in strength further from the centre B = µ0 2π r of the wire. 492 Unit 3 Electric, Gravitational, and Magnetic Fields

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(b) Outside of the cable the magnetic field is zero according to Ampère’s law. The total current within a circular path outside the cable is zero (equal currents in opposite directions) and so there is no magnetic field. 7. The current in each of two long straight, parallel conductors is 1 A when they are 1 m apart in a vacuum and when the magnetic force between them is 2 × 10–7 N per metre of length.

Applying Inquiry Skills 8.

Using magnetic probes, measure the magnetic field strength at several different places within the centre of the solenoid of different lengths until the readings are fairly constant. Note that if there are not enough turns per unit length then no matter how long the solenoid the field will not be uniform. 9. l = 1.0 m m = 6.0 ¯ 10–2 kg θ = 12° (a) I = ?

First we must calculate the force: T cos 6º = mg F = T sin 6º  mg  sin 6° F = D   cos 6  = mg tan 6° F = 6.2 ×10−2 N To find the distance between the wires: d = 2(0.80 m ¯ sin 6º) = 1.6 m ¯ sin 6º = 0.17 m To calculate the current: F µ 0 I1 I 2 = l 2π d 2π dF 2 I = µ0 l I=

( 2π )( 0.17 m ) (6.2 ×10−2 N ) (4π ×10−7 T ⋅ m/A ) (1.0 m )

I = 230 A The current in each wire is 230 A. (b) To make this procedure work something must be done to decrease the amount of current required, otherwise the wires would melt before the angle could be measured. Lighter wires could be used and angles smaller than 12° might be measured although the latter may be difficult. Tinfoil could be used as the conductor since it is very light. Assuming this works, students could conduct controlled experiments by changing one variable at a time including the currents in the wires and the length of the wires.

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Chapter 8 Magnetic Fields and Electromagnetism 493

Making Connections 10. Sound passing into the microphone is converted into an electric current with the same frequency as the sound and magnitude related to the amplitude of the sound. This electric current is in the coil wrapped around the recording head that produces a magnetic field near the recording magnetic tape. The head causes the magnetic tape to become magnetized in strips as shown. These strips of magnetic areas on the tape consist of various degrees of thickness and strengths depending on the size of the current.

8.5 ELECTROMAGNETIC INDUCTION PRACTICE (Page 418)

Understanding Concepts 1.

Two magnets are attached to opposite sides of the drive shaft. The speed of rotation of the shaft is an indication of the speed of the vehicle. A coil near the drive shaft is used to detect the changes in the magnetic field due to the rotating magnets. The coil is connected to a microprocessor used to control the speed of the vehicle. 2. If we examine Figure 7, we see that this situation is not possible since it does not obey the law of conservation of energy. The S-pole of the coil will attract the N-pole of the magnet causing it to speed up. This will in turn cause more current in the coil making the magnetic field of the coil stronger causing even more attraction. The total energy increases since kinetic, electric, and magnetic energies all increase and so this is not possible meaning the S-pole of the coil must be an N-pole meaning it must resist the motion of the magnet. In this way the kinetic energy will decrease when the electric and magnetic energies increase with the total energy remaining constant. 3. (a) The metal ring with the break in it will not allow a current to flow around the circumference and so no force will be experienced by the magnet. The ring is very thin which restricts any eddy currents.

(b) If the rings are replaced by long cylinders there will be enough room for significant eddy currents which will result in a force in both situations. However, the force will be smaller for the one with the gap since the circular current around the cylinder is still not possible.

Section 8.5 Questions (Pages 419–420)

Understanding Concepts 1. (a) When the switch is closed, the suddenly increasing magnetic field of the coil will induce a current in the copper ring and cause the ring to jump up momentarily. (b) The steady current will have no effect on the copper ring. (c) When the switch is opened the ring jumps again because the magnetic field is changing. (d) The ring will still jump in the same manner but the induced current in the ring will be in the opposite direction around the ring. 2. The light bulb is connected to a coil. The base is also an electromagnet attached to a metal cylinder. Since the coil is not moving (in fact one could not move it fast enough) the current in the electromagnet must be AC. This causes the magnetic field to vary in strength and direction, rapidly inducing a current in the coil and causing the light bulb to glow. 3. The current associated with a bolt of lightning can be high enough to cause a significant magnetic field for a short period of time that can induce electric currents in nearby conductors. 4. The AC current will create a rapidly changing magnetic field in its core that will induce a current in the ring. The current will cause the ring to heat up due to the internal resistance of the ring. 494 Unit 3 Electric, Gravitational, and Magnetic Fields

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Applying Inquiry Skills 5.

6.

Factors to be considered include the strength of the magnetic field, the angle between the coil and the field, and the rate of change of area enclosed inside the coil. Each of these factors should be tested separately. For example, students could compress the coils slowly and then repeat the task by compressing them faster each time. Alternatively, students could compress the coils at a moderate rate using different field strengths for each trial. The deflection of the compass needle can be used as a measure of the size of the current induced in the coil by placing it at the opposite end of the coil at the opposite end from the bar magnet. Diagrams will vary based on suggestions made by the students.

Making Connections 7.

Powerful superconducting magnets on the train induce a current in coils placed on the guideway. According to Lenz’s law the magnetic field produced by the induced current in these coils will resist the inducing action by repelling the superconducting magnets causing the train to levitate. 8. Electromagnets operating on an AC current induce eddy currents in metal pots placed on the surface of the stove. Internal resistance of the metal pot causes the metal to heat up. 9. (a) If the circuit is operating normally, the current in to the device is the same as the current out of the device. According to Ampère’s law the external magnetic field around the wires is zero and so no current will be induced in the sensing coil. (b) In this case, the current in one wire inside the sensing coil will be larger than the other and so the field around the wires will increase from zero and induce a current in the sensing coil. The induced current will trip the sensitive circuit breaker and interrupt the circuit, which then stops the current. (c) Only devices with large enough currents to pose a danger to people should be plugged into ground-fault interrupters. 10. The lights on the side streets are set to red unless a car pulls up to the intersection. The metal in the car causes a change in the magnetic field in the area inducing a small current in the coils detected by the microprocessor. Once the microprocessor detects this, a signal is sent to the light to change to green. 11. The ride uses Lenz’s law by using the induced current caused by a powerful electromagnet to resist the falling action of the ride and slow it down. 12. (a) The vibrating magnetized guitar string induces a current in the coil in the pickup with the same frequency and magnitude related to the amplitude of the vibrating string. (b) The pickup should be placed at the antinode since a changing magnetic field is required to produce an induced current.

CHAPTER 8 LAB ACTIVITIES Investigation 8.2.1: Magnetic Force on a Moving Charge (Page 421)

Question The magnitude and type of charge, the magnitude of the magnetic field, the speed of the charge, and the angle between the magnetic field and the velocity will affect the magnitude and direction of the magnetic force on a moving charge.

Hypothesis/Prediction (a) (i) The magnitude of the magnetic force is directly proportional to the magnitude of the magnetic field. (ii) The magnitude of the magnetic force is directly proportional to the speed of the charge. (iii) The magnitude of the magnetic force is directly proportional to the sine of the angle between the velocity and the magnetic field. (iv) The magnitude of the magnetic force is directly proportional to the magnitude of the charge. (v) Reversing the sign of the charge will reverse the direction of the magnetic force on the charge.

Analysis (b) The charged particles move along circular paths (arcs and even complete circles if they start in the magnetic field) with uniform circular motion. The larger the radius of curvature the smaller the magnetic force on the particle according to the  mv 2  formula for circular motion  F = . r   (c) The magnetic field strength is directly proportional to the magnetic force. When the magnetic field strength is increased the radius of curvature of the paths of the particles decreases indicating a larger force.

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Chapter 8 Magnetic Fields and Electromagnetism 495

(d) The speed of the charge is directly related to the magnetic force. When the charge moves into a magnetic field it undergoes uniform circular motion indicating that the force is perpendicular to the velocity and the magnetic field. If the speed is higher the radius of curvature is smaller indicating a larger force. (e) The force is a maximum when the angle between the magnetic field and the velocity is 90º and it decreases to zero when the angle is 0º or 180º. (f) The amount of charge on the particle is directly proportional to the magnetic force. Increasing the charge on the particles causes a smaller radius of curvature indicating a greater force on the charge. (g) The particle curves in the opposite direction indicating that the direction of the force is reversed.

Evaluation (h) Answers may vary. Students might suggest a need for more versatile simulation programs, better sources of uniform magnetic fields, more versatile cathode ray tubes, or even other sources of charged particles.

Synthesis (i) Charged particles undergo uniform circular motion in magnetic fields. This type of motion has been observed in planetary mechanics. The speed of the particle does not change when the particle is in the field.

Investigation 8.3.1: Force on a Conductor in a Magnetic Field (Pages 422–423)

Question The length of the conductor in the field, the angle between the conductor and the magnetic field, the current in the conductor, and the magnitude of the magnetic field all affect the magnitude of the force on a conductor in a magnetic field.

Hypothesis/Prediction (a) (i) The magnetic force on a conductor is directly proportional to the strength of the magnetic field. Reversing the direction of the magnetic field will reverse the direction of the magnetic force on the conductor. (ii) The magnetic force on a conductor is directly proportional to current in the conductor. Reversing the direction of the current will reverse the direction of the magnetic force on the conductor.

Analysis (b) Answers will vary depending on the dimensions of the coils. For a typical coil constructed in class of length 10.0 cm (0.10 m), 20 turns, and a current of 1.00 A:  NI  B = µ0    L   ( 20 )(1.00 A )  = 4π × 10−7 T ⋅ m/A    0.10 m 

B = 2.51× 10−4 T The magnetic field strength is 2.51 ¯ 10–4 T. (c) Answers will vary. Students will use the equation found in part (b). (d) The magnetic force is directly proportional to the current in the wire.

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(e) The magnetic force is directly proportional to the magnetic field strength.

(f) The magnetic force is directly proportional to the product of the current and the magnetic field strength. The magnetic force also depends on the length of the conductor in the field and the angle between the current and the magnetic field.

Activity 8.4.1: Magnetic Fields near Conductors and Coils (Page 424)

Question The magnetic field around a long straight conductor is circular decreasing with the distance from the conductor. The magnetic field inside a coil is uniform and appears like the field of a bar magnet outside the coil.

Experimental Design (a) Students should investigate each factor listed, one at a time, keeping all other factors constant.

Analysis (b) The magnitude of the magnetic field is inversely proportional to the distance from the conductor. (c) The magnitude of the magnetic field is directly proportional to the current in the conductor. (d) The magnitude of the magnetic field inside the conductor is directly proportional to the current in the coil and to the number of turns per unit length. (e) (i) The field is circular in nature and decreases with the distance from the conductor. (ii) The magnetic field inside a coil is uniform and appears like the field of a bar magnet outside the coil.

Evaluation (f) Answers may vary. Students can include sources of experimental error, ways to improve their design, any different materials they would have used, etc.

CHAPTER 8 SUMMARY Make a Summary (Page 426)

Right-Hand Rule Motion Reversing Motion of Charge or Sign of Charges θ

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Charged Particle thumb in direction of velocity of charge, otherwise the same

reverses direction of force

Conductor thumb in direction of electric current; note the flow of individual charges creates the electric current usually none; wire might be deflected to one side reverses direction of force

maximum when perpendicular minimum when parallel (sin θ)

maximum when perpendicular minimum when parallel (sin θ)

uniform circular

Chapter 8 Magnetic Fields and Electromagnetism 497

CHAPTER 8 SELF QUIZ (Page 427)

True/False 1. 2. 3. 4. 5.

F The speed does not change. The velocity changes since the charge will move along a circular path. T T T F There is no magnetic field outside a coaxial cable according to Ampère’s law since the total current enclosed by a circular path outside the cable is zero. T F It is not possible to shield against gravitational fields.

6. 7.

Multiple Choice 8. 9. 10. 11. 12.

(b) (b) (b) (a) (d)

13. (a) B = µ0

I 2π r

I   Therefore, 2 B = 2  µ 0   2π r  3I = µ0 2π r2 r2 =

3r 2

CHAPTER 8 REVIEW (Pages 428–429)

Understanding Concepts 1.

2.

3. 4.

5.

Earth’s magnetic field deflects charged particles from outer space away from Earth, according to the right-hand rule. Charged particles approaching along the polar axis may reach Earth without deflection since they are moving essentially parallel to the magnetic field lines. 1 The electric current in home wiring is AC. Its direction reverses every th of a second, as does the polarity of the 120 magnetic field it creates. No magnetic compass is capable of moving fast enough to respond to such a rapidly changing field. Using the right-hand rule, the positively charged particles will be deflected towards the east since they approach from above. If they pass through the field and leave from below, they will be deflected west. (a) When the strength of the magnetic field increases, r decreases and the charged particle moves in a smaller circle. (b) The circular path will become a spiral path parallel to the magnetic field. (c) When both fields are switched off simultaneously, the charged particle will move off at a constant speed tangent to the spiral at the point where the fields are removed. (a) The doubly ionized ion is accelerated by the potential difference to a velocity that is 2 times that of the singly ionized ion. v2 =

2 ( 2q )V

= 2

m 2qV m

v2 = 2v1

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(b) The doubly ionized ion moves in a circle whose radius is r2 = r2 = 6.

1 2

that of the singly ionized ion.

2Vm qV 2 r1 2

l = 15 cm = 0.15 m I = 12 A F = 0.40 N θ = 90° B=? F = lIB sin θ B= =

7.

F lI sin θ

(0.40 N ) (0.15 m )(12 A ) sin 90°

B = 0.22 T The magnitude of the magnetic field is 0.22 T. l = 45 cm = 0.45 m m = 15 g = 1.5 ¯ 10–2 kg θ = 90° B = 0.20 T I=? First we must calculate the force: F = mg

(

)

= 1.5 ×10−2 kg (9.8 N/kg ) F = 0.147 N To calculate the current: F = IlB sin θ I= =

8.

F lB sin θ

(0.147 N ) (0.45 m )(0.20 T ) sin 90°

I = 1.6 A The current in the conductor is 1.6 A. v = 3.2 ¯ 106 m/s B = 1.2 T [S] G F =? F = qvB sin θ

(

)(

)

= 1.6 × 10−19 C 3.2 × 106 m/s (1.2 T )(sin 90° )

9.

G F = 6.1× 10 −13 N [vertically down] The force s 6.1 ¯ 10–13 N [vertically down] according to the right-hand rule. q = +3.2 ¯ 10–19 C m = 6.7 ¯ 10–27 kg V = 1.2 ¯ 103 V B = 0.25 T θ = 90° F=?

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Chapter 8 Magnetic Fields and Electromagnetism 499

First we must calculate the speed of the α particle: 1 qV = mv 2 2 2qV v= m =

2 (3.2 × 10 −19 C )(1.2 × 103 V ) 6.7 ×10 −27 kg

v = 3.39 × 105 m/s To calculate the force: F = qvB sin θ

(

)(

)

= 3.2 × 10−19 C 3.39 ×105 m/s (0.25 T )(sin 90° ) F = 2.7 × 10 −14 N The magnetic force is 2.7 ¯ 10–14 N. 10. r = 25 cm = 0.25 m [E] I = 12 A G B=?  I  B = µ0    2π r   12 A  = 4π × 10−7 T ⋅ m/A    2π (0.25 m) 

JG B = 9.6 ×10 −6 T [horizontally, south] The magnetic field is 9.6 ¯ 10–6 T [horizontally, south] according to the right-hand rule. 11. I = 15 A [vertically upward] l = 0.10 m v = 5.0 ¯ 106 m/s G F =? First we must calculate the magnitude of the magnetic field:  I  B = µ0    2π r  =

(4π ×10

−7

T ⋅ m/A ) (15 A )

( 2π )( 0.10 m )

B = 3.0 ×10 −5 T To calculate the force on the electron moving in this field: F = qvB sin θ

= (1.6 × 10−19 C )(5.0 × 106 m/s )(3.0 × 10 −5 T ) (sin 90° )

G F = 2.4 ×10−17 N [horizontally, toward the wire] The force on the electron is 2.4 ¯ 10–17 N [horizontally, toward the wire] according to the right-hand rule. 12. L = 0.10 m N = 1200 I = 1. A B=?  NI  B = µ0    L   1200 (1.0 A )  = ( 4π × 10−7 T ⋅ m/A )    0.10 m  B = 1.5 × 10−2 T The magnetic field strength is 1.5 ¯ 10–2 T.

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13. L = 15 cm = 0.15 m N = 100 I = 20.0 A m = 1.8 ¯ 10–2 g = 1.8 ¯ 10–5 kg IXY = ? First we must calculate the magnetic field on the solenoid:  NI  B = µ0    L 

 100 ( 20.0 A )  = ( 4π × 10−7 T ⋅ m/A )    0.15 m  B = 1.68 × 10 −2 T

To balance, the force on XY must be: F = mg

(

)

= 1.8 ×10−5 kg (9.8 N/kg ) −4

F = 1.76 × 10 N Therefore, the current in the conductor XY is: F = IlB sin θ I=

F lB sin θ

(1.76 ×10 N ) (0.15 m ) (1.68 × 10 T ) (sin 90° ) −4

=

−2

I = 0.70 A The current that must flow through the conductor is 0.70 A. (There is no force on WX and YZ.) 14. l = 45 m d = 0.10 m I1 = I2 = 1.0 ¯ 102 A F=? µ IIl F= 0 1 2 2π d ( 4π ×10−7 T ⋅ m/A )(1.0 ×102 A )(1.0 ×102 A ) ( 45 m ) = 2π (0.10 m ) F = 0.90 N The magnetic force between the two parallel wires is 0.90 N. 15. Let the subscript 1 represent the top wire, and subscript 2 represent the bottom wire. D = 150 g/m = 0.150 kg/m I1 = 40.0 A r = 4.0 cm = 4.0 ¯ 10–2 m I2 = ? To balance, the upward magnetic force on a one metre length of wire must be: Fg = mg = (0.150 kg/m )(9.8 N/kg ) Fg = 1.47 N/m

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Chapter 8 Magnetic Fields and Electromagnetism 501

To calculate the magnetic field strength at its location: F = I1lB sin θ B= =

F I1l sin θ

(1.47 N )

( 40.0 A )(1.0 m ) sin 90°

B = 3.68 ×10 −2 T To calculate the current required in the bottom wire to produce this magnetic field:  I  B = µ0  2   2π r  2π rB I2 = µ0 =

( 2π ) ( 4.0 ×10−2 m )(3.68 ×10−2 T )

4π × 10−7 T ⋅ m/A I 2 = 7.4 × 103 A The current required in the bottom wire is 7.4 ¯ 103 A. 16. v = 6.0 ¯ 106 m/s r = 1.8 cm = 0.18 m B = 3.2 ¯ 10–2 T q =? m q v = m Br 6.0 ×106 m/s ) ( = (3.2 ×10−2 T )(1.8 ×10−2 m )

q = 1.0 × 1010 C/kg m The charge-to-mass ratio is 1.0 ¯ 1010 C/kg. 17. m = 1.67 ¯ 10–27 kg B = 1.8 T r = 3.0 cm = 3.0 ¯ 10–2 m v=? qBr v= m 1.6 × 10−19 C ) (1.8 T ) (3.0 × 10−2 m ) ( = 1.67 × 10−27 kg v = 5.2 ×106 m/s The speed of the proton is 5.2 ¯ 106 m/s. 18. q = +e = 1.6 ¯ 10–19 C v = 1.9 ¯ 104 m/s B = 1.0 ¯ 10–3 T r = 0.40 m m=? qBr m= v 1.6 ×10−19 C )(1.0 ×10−3 T ) (0.40 m ) ( = 1.9 × 104 m/s −27 m = 3.4 ×10 kg The mass of the ion is 3.4 ¯ 10–27 kg. 502 Unit 3 Electric, Gravitational, and Magnetic Fields

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19. m = 3.9 ¯ 10–25 kg V = 1.0 ¯ 105 V r = 0.40 m (a) v = ? 1 2 mv 2 2qV v= m

qV =

=

2 (1.6 × 10 −19 C )(1.0 ×105 V ) 3.9 × 10 −25 kg

v = 2.9 ×105 m/s The ionized U atom’s maximum speed is 2.9 ¯ 105 m/s. (b) θ = 90° B = 0.10 T r=? mv 2 r mv r= qB

qvB =

(3.9 ×10 kg )(2.9 ×10 m/s ) (1.6 ×10 C ) (0.10 T ) −25

=

5

−19

r = 7.1 m The radius of the path would be 7.1 m. 20. (a) According to Lenz’s law the current is directed along ZYX to resist the motion of the loop. (b) The current will switch direction (XYZ) to resist the motion in the opposite direction. (c) The loop will swing back and forth with decreasing amplitude until it comes to rest. The amplitude will decrease due to Lenz’s law because the current induced in the loop will cause a magnetic field that will resist the motion of the loop.

Applying Inquiry Skills 21. Test each of the following factors one at a time: the current in the wire, the strength of the magnetic field, the length of the conductor in the field, and the angle between the current and the field 22. (a) Measure the field when there is no current. Vary the current and measure the strength of the magnetic field at various distances from the wires. (b) There will be little or no magnetic field detected due to Ampère’s law. (c) The magnetic field detected will be much larger since there is a net current.

Making Connections 23. The input current varies in magnitude causing a magnetic field in the voice coil that is fixed to the cone. The varying field in the voice coil causes the coil (and cone) to move back and forth when interacting with the cylindrical tubular magnets. The cone causes compressions and rarefactions in the air, which in turn produces sound. 24. When a sound wave interacts with the C-shaped diaphragm it causes the magnetic material in the rod to vibrate. According to Lenz’s law the current produced in the coils around the magnet produce a magnetic field to resist this motion. The windings are setup so that the current in each coil is in the same direction at all times. These coils are connected to an external circuit (amplifier) to change the current into sound.

Extension 25. Students might design a velocity selector combined with a mass spectrometer as shown on page 435 in the text.

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Chapter 8 Magnetic Fields and Electromagnetism 503

UNIT 3 PERFORMANCE TASK FIELD THEORY AND TECHNOLOGY (Pages 430–431) The Unit 3 Performance Task is research-based and open-ended. It provides students with an opportunity to select an emerging or established technological device or process of interest whose operation involves the use of at least two of the following: gravitational, magnetic, and electric fields. In addition to selecting the device or process and displaying an understanding of the relevant concepts, the student evaluates the social and economic impacts of the technology, and presents the findings in a suitable form chosen by the student. Technologies chosen may include satellites, TV picture tubes, mass spectrometers, particle accelerators, nuclear generators, speakers and amplifiers, relays, electromagnets, maglev technologies, etc. Responses to the Analysis and Evaluation questions will depend on the specific technology chosen and the type of research conducted. The student may submit any one of the following products for assessment/evaluation: • written report • poster presentation • science fair booth • audio-visual presentation • video clip • computer simulation • Web site The product will contain answers to the Analysis questions given in the textbook. While much of the information can be obtained from the Internet, textbooks, library books, magazines, and newspaper articles may also be useful sources of information. Students can refer to Appendix A3, text pages 765–766, for suggestions about decision making and defining, researching, and analyzing issues.

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Unit 3 Performance Task 505

UNIT 3 SELF QUIZ (Pages 432–433)

True/False 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

T F Only the acceleration is zero. F The field line only indicates the direction of the force on the charge. T T T F The oil droplet falls at a constant speed due to air friction. T F The kinetic energy will decrease and the potential energy will increase. T F A magnetic field cannot change the speed, it can only change the velocity under this condition. F There will be no magnetic force. T F The wires will attract each other.

Multiple Choice q1 q2 15. (b) F2 = 2 2 d    2 k

F , middle one forces in opposite directions. 4 17. (c) Electrons are negative, therefore, the field is to the right. Then, F = qε 16. (b) Force from farthest neighbor is

ε=

F q

1.6 ×10 −16 N 1.6 ×10 −19 C G ε = 1.0 × 103 N/C [right]

ε=

18. (d) 1  19. (a) E = eε  d  2  v 20. (b) mv = m   + 2m (v2 ) 2 21. (a) 22. (d) 23. (e) We need the angle. 24. (c) m ( 2v )

2

25. (a) q 2vB =

2r

26. (c)

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Completion 27. (a) field (b) inverse square (c) magnetic 28. (a) perpendicular (b) coaxial (c) electric potential energy (d) energy, momentum 29. parabolic, circular 30. decrease in magnitude, increases, increases or decreases

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Unit 3 Self Quiz 507

UNIT 3 REVIEW (Pages 434–437)

Understanding Concepts 1.

The doctors and nurses should wear conducting shoes to allow charge to escape from their bodies. If they wore insulating shoes and built up charge on their bodies, it could cause a spark that could be dangerous to the patient or equipment used. 2. Students should draw diagrams for each part of the question to help them calculate the answer. q1 = +6.0 ¯ 10–5 C q2 = –2.0 ¯ 10–5 C d = 36 cm = 0.36 m (a) q3 = +5.0 ¯ 10–5 C d = 18 cm = 0.18 m G F =?

There are two forces on this charge both directed toward the negative charge. The magnitude of this force is: kq q kq q F = 12 3 + 22 3 r13 r23

(9.0 ×10 N ⋅ m /C )(6.0 ×10 C )(5.0 ×10 C ) + (9.0 ×10 N ⋅ m /C )( 2.0 ×10 C )(5.0 ×10 C ) = 9

2

−5

2

−5

9

2

(0.18 m )

−5

2

−5

( 0.18 m )

2

2

G F = 1.1 ×103 N [toward the negative charge] The force on the third small charge is 1.1 ¯ 103 N [toward the negative charge]. (b) There are two forces on this charge, one directed toward the negative charge and the other away. Assign the positive direction toward the negative charge.

The magnitude of this force is: kq q kq q F = − 12 3 + 22 3 r13 r23

(9.0 ×10 N ⋅ m /C )(6.0 ×10 C )(5.0 ×10 C ) + (9.0 ×10 Nm /C )(2.0 ×10 C )(5.0 ×10 C ) 9

=−

2

−5

2

(0.54 m )

2

−5

9

2

−5

2

−5

(0.18 m )

2

G F = 3.7 × 10 2 N [toward the negative charge] The force on the third small charge is 3.7 ¯ 102 N [toward the negative charge]. (c) The two electric fields will be in opposite directions. We assign the positive direction away from the positive charge.

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There are two forces on this charge both directed toward the negative charge. The magnitude of this force is: kq kq ε = 213 − 223 r13 r23

(9.0 ×10 N ⋅ m /C )(6.0 ×10 C ) − (9.0 ×10 N ⋅ m /C )(2.0 ×10 C ) 9

=

2

−5

2

9

2

(0.18 m )

−5

2

(0.54 m )

2

2

G ε = 1.6 × 107 N/C [away from the positive charge] The magnitude of the electric field is 1.6 ¯ 107 N/C [away from the positive charge]. (d)

The distance from the point to the charges is r=

(0.18 m ) + (0.18 m ) 2

2

r = 0.255 m The magnitude of the electric field from the positive charge is: kq ε1 = 21 r1

(9.0 ×10 N ⋅ m /C )(6.0 ×10 C ) = 9

2

−5

2

(0.255 m )

2

ε1 = 8.30 × 106 N/C For the negative charge:

ε2 =

kq2 r12

(9.0 ×10 N ⋅ m /C )(2.0 ×10 C ) 9

=

2

−5

2

(0.255 m )

2

ε 2 = 2.77 × 106 N/C The total electric field is:

ε=

(8.30 ×10 N/C ) + (2.77 ×10 6

2

6

N/C ) − 2 (8.30 × 106 N/C )( 2.77 × 106 N/C ) cos 45° 2

ε = 6.6 × 106 N/C The total electric field is 6.6 ¯ 106 N/C. (e) By symmetry this point is on the line joining the charges outside of the two charges. It is closer to the negative charge since it is the smaller charge and to ensure the fields are in opposite directions. If the distance from the negative charge is x then the distance from the positive charge is 0.36 + x. The two electric fields must be equal in magnitude to cancel, therefore: k ( 2.0 ×10 −5 C ) k (6.0 × 10−5 C ) = 2 x2 (0.36 + x )

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Unit 3 Review 509

If we cancel and cross-multiply then:

(0.36 + x )

2

= 3x 2

2 x 2 − 0.72 x − 0.1296 = 0 Using the quadratic formula: x= =

3.

−b ± b 2 − 4ac 2a − ( −0.72 ) ±

( −0.72 ) − 4 ( 2 )( −0.1296 ) 2 (2) 2

x = 0.49 m Therefore, the point at which the magnitude of the electric field is zero is 0.49 m. k = 9.0 × 109 N ⋅ m 2 /C 2 q1 = 1.0 ×10 −6 C q2 = −1.0 ×10−6 C q3 = 2.0 ×10 −6 C r12 = 1.0 m F =? This is a special triangle and so the other distances are 2.0 m and kq q F = 12 2 r12

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )(1.0 ×10 C ) 9

=

3 m. The magnitude of the forces along the sides is

2

−6

2

−6

(1.0 m )

2

F = 0.90 N F=

kq1 q3 r132

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )(2.0 ×10 C ) = 9

2

−6

2

(

3m

−6

)

2

F = 0.60 N F=

kq2 q3 r232

(9.0 ×10 N ⋅ m /C )(1.0 ×10 C )(2.0 ×10 C ) 9

=

2

−6

2

−6

( 2.0 m )

2

F = 0.45 N For the +1.0 C charge: F=

(0.45 N ) + (0.90 N ) 2

2

− 2 (0.45 N )(0.90 N ) cos 60°

F = 0.779 N or 0.78 N

510 Unit 3 Electric, Gravitational, and Magnetic Fields

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Using the sine law: sin θ sin 60° = 0.90 0.779  0.90 (sin 60° )  θ = sin −1   0.779   θ = 90° Therefore, the force on the charge is 0.78 N [30º to the triangle]. For the – 1.0 C charge: F=

(0.60 N ) + (0.90 N ) 2

2

F = 1.1 N  0.90  θ = tan −1    0.60  θ = 56° Therefore, the force on the charge is 1.1 N [56º to the triangle]. For the +2.0 C charge: F=

(0.45 N ) + (0.60 N ) 2

2

− 2 (0.45 N )(0.60 N ) cos150°

F = 1.01 N, or 1.0 N Using the sine law:

4.

sin θ sin150° = 0.60 1.01  0.60 (sin150° )  θ = sin −1   1.01   θ = 17° Therefore the force on the charge is 1.0 N [163º to the triangle]. r = 2.64 ¯ 10–11 m G F =? The force on either electron is due to the electric force from the nucleus and the other electron and they are in opposite directions. Define the positive direction toward the nucleus. kq q kq q F = − 12 2 + 22 3 r12 r23 =

kq2 q3 kq1q2 − 2 r232 r12

(9.0 ×10 N ⋅ m /C )(1.6 ×10 C )(3.2 ×10 ( 2.64 ×10 m ) 9

F=

2

−19

2

−11

2

−19

C)

(9.0 ×10 N ⋅ m /C )(1.6 ×10 C )(1.6 ×10 (5.28 ×10 m ) 9



2

−19

2

−11

−19

C)

2

G F = 5.8 × 10−7 N [toward the nucleus] The electric force on either electron is 5.8 × 10–7 N [toward the nucleus]. 5. (a) No, for a conductor in electrostatic equilibrium the electric field lines are perpendicular to the surface of the conductor. If they are not then there is a component of the field parallel to the surface that causes the charges to move. (b) Yes, the charges are held in place (not free to move across the surface) and so the field lines need not be perpendicular to the surface.

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Unit 3 Review 511

6.

7.

m = 7.0 g = 7.0 ¯ 10–3 kg q = 1.5 µC = 1.5 ¯ 10–6 C θ = 8° ε=? The vertical and horizontal components of the tension must cancel the gravitational and electric forces respectively. T cos8° = mg T=

mg cos8°

Therefore, T sin 8° = qε T sin 8° q mg tan 8° = q

ε=

(7.0 ×10 =

−3

kg ) (9.8 N/kg )( tan 8° )

1.5 × 10−6 C ε = 6.7 × 103 N/C The electric field is 6.7 ¯ 103 N/C. 8. (a) Charge q2 is positive and charge q1 is negative.

(b) By examining the number of field lines out of q2 and into q1, the ratio is 9.

r = 4.0 ¯ 10–14 m v=?

512 Unit 3 Electric, Gravitational, and Magnetic Fields

16 = 4. 4

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At the point of closest approach the two electrons must have the same velocity and momentum must be conserved at all times, therefore: m(2v ) − mv = (2m)v ′ v′ =

v 2

Also, the total energy must be conserved, therefore: 2

2 1 1 1 2  v  ke m ( 2v ) + mv 2 = ( 2m )   + 2 2 2 r 2

5 2 1 2 ke 2 mv = mv + 2 4 r 9 2 ke 2 mv = 4 r v= = =

4ke 2 9mr 2e k 3 mr

2 (1.6 × 10−19 C ) 3

9.0 ×109 N ⋅ m 2 /C 2

(9.11×10

−31

kg )( 4.0 × 10 −14 m )

v = 5.3 × 107 m/s The initial speed of each electron is 5.3 ¯ 107 m/s. 10. (a) Yes, otherwise there would be a force on the particle and it would accelerate. (b) Not necessarily, if the charge is moving parallel to the magnetic field lines then the magnetic force on the charge would be zero and it would move at a constant velocity. 11. V = 2.4 ¯ 103 V B = 0.60 T θ = 90° (a) F = ? First we need the speed of the electron before it enters the field. Using conservation of energy we have: 1 qV = mv 2 2 2qV v= m =

2 (1.6 × 10 −19 C )( 2.4 × 103 V ) 9.1 × 10 −31 kg

v = 2.9 ×107 m/s To calculate the force: F = qvB sin θ

= (1.6 × 10−19 C )( 2.9 × 107 m/s ) (0.60 T )(sin 90° )

F = 2.8 ×10 −12 N The magnetic force on the electron is 2.8 ¯ 10–12 N. (b) The electron will move with uniform circular motion while it is in the magnetic field. (c) No, it will move out of the field eventually since it follows a circular path once it enters the field. This circular path will lead it back out of the field at some point. 12. l = 25 cm = 0.25 m θ = 90° B = 0.18 T F = 0.14 N I=? Copyright © 2003 Nelson

Unit 3 Review 513

F = BlI sin θ I= =

F Bl sin θ

(0.14 N ) (0.18 T )(0.25 m ) sin 90°

I = 3.1 A The current in the wire is 3.1 A. 13. r = 3.0 m I = 1.2 A B=? (a) According to the right-hand rule the field from each wire is pointing down and since the fields have the same magnitude at the midpoint then I   B1 = 2  µ0   2π r 

 ( 4π × 10−7 T ⋅ m/A ) (1.2 A )   = 2   2π )(1.5 m ) (   B1 = 3.2 ×10 −7 T The magnetic field midway between the two wires is 3.2 ¯ 10–9 T [down]. (b) The field will be zero since the magnetic fields of equal magnitude are in opposite directions according to the righthand rule. 14. L = 12 cm = 0.12 m N=5 B = 1.6 ¯ 10–2 T I=? BL I= µ0 N

(1.6 ×10 T ) (0.12 m ) (4π ×10 T ⋅ m/A ) (500 ) −2

=

−7

I = 3.1 A The current in the coil is 3.1 A. 15. l = 4.00 m d = 8.00 cm = 8.00 ¯ 10–2 m F = 2.80 ¯ 10–5 N I=?  I  Since F = IlB sin θ and B = µ0  :  2π d  µ IIl F= 0 1 2 2π d µ I (2I ) l = 0 2π d µ0 I 2 l = πd I= =

π Fd µ0 l π ( 2.80 × 10 −5 N )(8.00 × 10−2 m )

( 4π ×10

−7

T ⋅ m/A ) ( 4.00 m )

I = 1.18 A The current in one conductor is 1.18 A. The other current is 2I, which is 2.36 A.

514 Unit 3 Electric, Gravitational, and Magnetic Fields

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16. ε = 1.2 ¯ 104 V/m B = 0.20 T m = 2.2 ¯ 10–26 kg (a) v = ? Since the particle passes through undeflected, the magnetic force up must be equal in magnitude to the electric force down. FM = FE evB = eε ε v= B 1.2 ×10 4 V/m = 0.20 T v = 6.0 ×10 4 m/s The speed of the particle is 6.0 ¯ 104 m/s. (b) r = ? mv 2 evB = r mv r= eB (2.2 ×10−26 kg )(6.0 ×104 m/s ) = (1.6 ×10−19 C ) (0.20 T ) r = 4.1× 10 −2 m The radius of the path of the particle is 4.1 ¯ 10–2 m. 17. Let the subscript I represent the inner conductor, and the subscript O represent the outer conductor. II = 1.0 A [into the page] IO = 1.5 A = 1.5 A – 1.0 A = 0.50 A [out of the page] G BA = ? G BB = ? According to Ampère’s law the magnitude of the magnetic field at A is:  I  BA = µ0  I   2π r  =

(4π ×10

−7

T ⋅ m/A ) (1.0 A )

( 2π ) (1.0 × 10−3 m )

G BA = 2.0 × 10−4 T [down]

According to Ampère’s law the magnitude of the magnetic field at B is:  I  BB = µ0  O   2π r 

( 4π ×10 =

−7

T ⋅ m/A ) (0.50 A )

( 2π ) ( 4.0 ×10−3 m )

G BB = 2.5 ×10−5 T [up] The magnetic field at A is 2.0 ¯ 10–4 T [down], and at B is 2.5 ¯ 10–5 T [up].

Applying Inquiry Skills 18. Students need a method to measure the electric charge such as the deflection of a pith ball electroscope from the vertical. Factors to be tested could include temperature of the air, humidity, dust particles in the air, etc.

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Unit 3 Review 515

19. (a) l = 35 cm = 0.35 m m = 12 g = 1.2 ¯ 10–2 kg θ = 14° From the diagram, T cos 7º = mg. Therefore, FE = T sin 7°  mg  =  sin 7°  cos 7°  = mg tan 7°

= (1.2 ×10 −2 kg ) (9.8 N/kg )( tan 7° ) FE = 1.4 × 10−2 N To determine the distance between the wires: d = 2 ( 7.0 × 10−2 m ) (sin 7° ) d = 1.7 × 10 −2 m To calculate the current in each wire: 2π dF I2 = µo l I=

( 2π ) (1.7 ×10−2 m )(1.4 × 10−2 N ) (4π ×10−7 T ⋅ m/A ) (0.35 m )

I = 58 A The current in each wire is 58 A, and must be in opposite directions. (b) Test each of the following variables, one at a time: current in the wire, length of the conductor, and distance between the wires. Calculations similar to those above can be used to determine the force. 20. Test the effect of the magnitude of the charge, the speed of the charge, and the angle between the velocity and the magnetic field. The charged ball could be suspended from an insulating thread and dropped from various heights through a magnetic field. 21. Since the liquids are conductors we can cause charge to flow through it. If this current is perpendicular to an external magnetic field it will experience a force. This magnetic force will push the liquid through the pipe if both the current and field are perpendicular to the length of the pipe. The pipes can pass near the nuclear material and absorb heat.

Making Connections 22. (a) Yes, the current in the lines is alternating and according to the law of electromagnetic induction the varying magnetic field of the wire will induce a current in the loop. The energy will come from the power lines, according to Lenz’s law, and the energy in the power lines will decrease. (b) Yes, since the total sale of energy will be less than the energy produced. The power company will be able to estimate fairly accurately the expected power loss in the wires. When they find out it is more than expected they will know there is a problem. (c) Yes, the frequency of the current induced in the loop will be the same as the current in the lines and the magnitude of the current in the loop will be proportional to the magnitude of the current in the line. The current in the loop can be transformed into sound and the conversation detected. 23. The screen becomes charged and attracts dust particles in the air by induction. These particles then attach to the screen very quickly because the charge on the screen is large. 24. (a) The negatively charged oxygen atom will be attracted to positive charge and the positively charged hydrogen atoms will be attracted to negative charges. (b) It doesn’t matter if the charge is positive or negative, because both will attract water molecules in the air. When water molecules come into contact with the surface some excess charge is transferred to the water molecule and the charge on the object will therefore decrease. (c) The ion is either positively or negatively charged and will attract water molecules. These water molecules will attract other water molecules. When enough become attached together, they form a water droplet.

516 Unit 3 Electric, Gravitational, and Magnetic Fields

Copyright © 2003 Nelson

25. Students will find that capacitors can be used to store electric energy that can be released very quickly in a burst of current. This energy can be used to form a brief burst of light or a short electrical signal. 26. When the magnetic tape moves past the gap in the ferromagnetic core the field lines are directed through the metal and the coil wrapped around it. The changing magnetic field in the coil induces a current in the coil with a magnitude and frequency related to the magnetic field of the tape and the speed at which it is moved. The speakers then convert the varying current into sound. 27. A degausser uses a strong magnetic field to change the orientation of the magnetic dipoles in each domain. Typically it consists of a coil with a large current in it. 28. Birds often fly over large distances in North America to migrate south and north. One theory suggests birds use the orientation of the magnetic field lines of the earth to guide them to different areas.

Extension 29. Typically the total energy of the charges will remain constant and the potential energy between the charges will increase when the distance decreases if the charges are similar and decrease if the charges are opposite in sign. The kinetic energy graphs for each particle for a range of values of y are shown. As y increases the percentage of kinetic energy transferred to the second charge decreases.

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Unit 3 Review 517

30. Electromagnetic radiation and the gravitational effects between distant objects are the main factors used to investigate the properties of distant objects. Students can study a wide range of topics. 31. The third charge must be placed at (a) or (b) since it should be closer to the smaller charge and the forces must be in opposite directions to cancel. We can determine which one by equating the two forces on the third charge q3. kqq3 k 4qq3 = 2 r2 ( r + 3)

( r + 3)

2

= 4r 2

r 2 + 6r + 9 = 4r 2 3r 2 − 6r − 9 = 0

3 ( r 2 − 2r − 3) = 0

( r + 1)( r − 3) = 0

r = 3 cm Since r must be positive, the answer is (a).

518 Unit 3 Electric, Gravitational, and Magnetic Fields

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