0
$\begingroup$

Suppose I have a point charge sitting in the $z>0$ half-space in front of a (infinitely large) conducting plate lying in the $x,y$ plane. I know how to calculate the field in the upper half-space using mirror images.

But how does the field look like on the other side of the plate? Is it zero?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

But how does the field look like on the other side of the plate? Is it zero?

No there are charges induced on both the surfaces equal in magnitude but opposite sign so as to cancel the effect of electric field of the charge q(your point charge) going through the sheet.

Now because Negative charge is attracted towards positive charge the side of sheet facing the charge q will accumulate a charge -q with the centre of charge at the point you calculate using mirror symmetry and the positive induced charge on the other side isolated and uniformly distributed.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ One could add that this is a manifestation of the "screening" phenomenon. $\endgroup$
    – Roger V.
    Commented Nov 17, 2020 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.