Chapter 25 Solutions. 25.1. ∆V = –14.0 V and. Q = –NA e = –(6.02 ∞ 1023)(1.60
∞ 10–19 C) = –9.63 ∞ 104 C. ∆V = W. Q. , so W = Q(∆V) = (–9.63 ∞ 104 ...
Chapter 25 Solutions ∆V = –14.0 V and
25.1
Q = –NA e = –(6.02 ∞ 1023)(1.60 ∞ 10–19 C) = –9.63 ∞ 104 C W ∆V = Q , so W = Q(∆V) = (–9.63 ∞ 104 C)(–14.0 J/C) = 1.35 ΜJ *25.8
(a)
&∆V& = Ed = (5.90 ∞ 103 V/m)(0.0100 m) = 59.0 V
(b)
1 2
2
mv f = &q(∆V)&;
1 2
(9.11 ∞ 10–31) vf2 = (1.60 ∞ 10–19)(59.0)
vf = 4.55 ∞ 106 m/s
*25.10
B
C
B
A
A
C
VB − VA = − ∫ E ⋅ ds = − ∫ E ⋅ ds − ∫ E ⋅ ds 0.500
VB − VA = (−Ecos180°)
∫
0.400
dy − (Ecos90.0°)
−0.300
VB – VA = (325)(0.800) = +260 V
∫
dx
−0.200
25.17
(a)
Since the charges are equal and placed symmetrically, F = 0
(b)
Since F = qE = 0, E = 0
(c)
2 –6 q 9 N ∙ m 2.00 ∞ 10 C V = 2ke r = 2 8.99 ∞ 10 C 2 0.800 m
V = 4.50 ∞ 104 V = 45.0 kV
*25.22
Ue = q4V1 + q4V2 + q4V3 = q4
1 4π∈ ∈0
Ue
(
= 10.0 × 10 −6 C
q1 q2 q3 r + r + r 2 3 1
) (8.99 × 10 9 N ⋅m 2/ C2 ) 0.6001 m + 0.1501 m + 2
(0.600 m)2 + (0.150 m)2 1
Ue= 8.95 J V = a + bx = 10.0 V + (−7.00 V m )x
25.36 (a)
At x = 0, V = 10.0 V At x = 3.00 m, V = − 11.0 V At x = 6.00 m , V =
(b)
E=−
− 32.0 V
dV = −b = − (−7.00 V m )= 7.00 N C in + x direction dx
25.50
No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric field in the wire to push more charge to the outer sphere. Charge Q is on the outer sphere. Therefore,
