2
$\begingroup$

In a conducting sphere, the net charge is evenly distributed at the surface because of repulsion while the inner electric field remains zero as charges cancel inside. Here is an example, where the net charge is the sum of the positive charges evenly distributed at the surface, which are what create the outer electric field:

enter image description here

However, even if the inner positive charge is moved, the outer surface charge distribution remains unchanged: enter image description here

Why exactly does this happen? I assume it's because the perturbation in the inner electric field (which for an instant creates a net inner electric field) reaches the inner electrons before reaching the surface positive charges, and by the time the perturbation in the field would have reached the positive charges on the surface, there's already been a redistribution of charge inside. Is this correct?

Doesn't the surface distribution change at all, even if for an instant?

I'm looking for a more intuitive/visual explanation, like the idea above.

$\endgroup$
1
$\begingroup$

As you know by Gauss Law, electric field inside a conductor is zero.Thus in what ever orientation the negative charge is induced in the inner surface of the conductor it creates no electric field through the conductor thus it cannot affect the orientation of the positive charge on the outer surface. This is because a charge can only be moved or displaced by electric field so no inner charge can manipulate the charge distribution on the outer surface. See the image attached below,

enter image description here

$\endgroup$
19
  • 1
    $\begingroup$ No, it's a property of conductor(can be proved by Gauss law) that it doesn't allow any electric field to pass through it, it more or less acts like a shield. $\endgroup$ Sep 25 at 8:28
  • 1
    $\begingroup$ There is nothing to do with the positive charge inside $\endgroup$ Sep 25 at 8:28
  • 1
    $\begingroup$ Yes, you can explain it that way too..I think! $\endgroup$ Sep 25 at 8:32
  • 1
    $\begingroup$ Whatever the size of cavity be the result will always be same.....the negative charge density may vary on the inner surface... $\endgroup$ Sep 25 at 8:37
  • 1
    $\begingroup$ Try out yourself, though there is no option to vary cavity size simphy.com/weblets/conductor-field-lines $\endgroup$ Sep 25 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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