Chapter 25 Homework Solutions

Chapter 25 Homework Solutions 1. The change in the potential energy of the proton is

?Y œ ; ?Z œ ˆ"Þ' ‚ "!"* C‰ˆ$& ‚ "!$ V‰ œ &Þ' ‚ "!"& J

By conservation of energy, the final kinetic energy of the proton must be &Þ' ‚ "!"& J, so ˆ&Þ' ‚ "!"& J‰ œ

" 7@# #

where 7 œ "Þ'(# ‚ "!#( kg. Solving for @ gives a speed of #Þ' ‚ "!' mÎs .

2. The potential at a distance of %Þ$ nm from a proton is Z œ

" ; " a"Þ' ‚ "!"* Cb œ œ !Þ$$& V %1%! < %1%! a%Þ$ ‚ "!* mb

Therefore, the initial potential energy of the electron is

Y œ ;Z œ ˆ"Þ' ‚ "!"* C‰a!Þ$$& Vb = &Þ$' ‚ "!#! J

When the electron gets very far from the proton, the potential energy will be zero, which represents an increase of &Þ$' ‚ "!#! J. Therefore, the kinetic energy must decrease by the same amount. The initial kinetic energy of the electron is O œ

" " # 7@# œ ˆ*Þ"" ‚ "!$" kg‰ˆ%Þ! ‚ "!& mÎs‰ œ (Þ#* ‚ "!#! J # #

Therefore the final kinetic energy will be a(Þ#* ‚ "!#! Jb  a&Þ$' ‚ "!#! Jb œ "Þ*$ ‚ "!#! J, so "Þ*$ ‚ "!#! J œ

" 7@# #

Solving for @ gives a final velocity of #Þ" ‚ "!& mÎs .

3. For points outside the sphere, the potential of a charged sphere obeys the same formula as the potential around a point charge: Z œ

" ; %1%!