Ludovico science forum beginner
Joined: 20 May 2006
Posts: 11
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Posted: Sat Jul 15, 2006 1:57 pm Post subject:
problem with charges
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Hi there, I've the following problem.
A charge q_1 is at distance "a" from an infinite planar conductor slab
(plane z=0, position of the charge: x=y=0, z=a).
Compute the work that is necessary to carry a second charge q_2
slowly from the infinite to the point x=y=0, z=b, b>a.
My solution is
First of all I have observed that the work that is necessary
to carry a second charge q_2 from the infinite to the point x=y=0, z=b,
is " U_f - U_i " where U_f stands for the final electrostatic energy of
the system q_2 plus q_1 plus slab and U_i stands for the initial
electrostatic energy, i.e., the energy of q_1 plus the slab.
Now U_i = -q_1^2 / (4a)
and
U_f = -q_1^2 / (4a) + 1/2 q_2 ( q_1 / (b-a) + q' / (b+a) )
where
q' = -q_1 is an image charge, position of q_1 is (0,0,-a)
then the result for is
Work = U_f-U_i = -q_1^2 / (4a) + 1/2 q_1q_2 / (b-a) -1/2 q_1q_2 / (b+a)
- (-1) q_1^2 / (4a) =
1/2 q_1q_2 / (b-a) -1/2 q_1q_2 / (b+a).
What do you think about? Is my solution correct?
Regards, Ludo |
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