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Forum index » Science and Technology » Physics » Electromagnetics
electrostatic and conductor
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Ludovico
science forum beginner


Joined: 20 May 2006
Posts: 11

PostPosted: Sat Jul 15, 2006 1:54 pm    Post subject: electrostatic and conductor Reply with quote

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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