1
$\begingroup$

There is a point charge inside a neutral conducting shell. My understanding is that when we move the charge from point K to L along the path, we experience no force since the electric field inside due to the sphere is zero. I have a couple of questions regarding this system.

figure

1) Is my understanding that the electric field due to the shell is zero inside correct? If the answer is a yes, then..

2) If we moved the charge really fast along the path, would there still be no field due to shell only?

3) When we are moving the charge, there must be a redistribution of charges on the conducting shell. Does that mean the electrostatic self energy of the shell change? If it changes, how come it is so, when we are doing no work on the charge inside? (Again constant speed)

4) While the redistribution of charges is in progress, do the electrons gain extra kinetic energy cause they are changing their positions around the sphere? And where does this energy come from in terms of energy conservation?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

1) No - see this answer.

2, 3, and 4) N/A.

Improve this answer
$\endgroup$
5
  • $\begingroup$ If its charge is not much less than the charge of the shell, and we move it really slowly, does it experience a coulomb force while moving? $\endgroup$
    – physicsguy19
    Commented May 5, 2018 at 20:31
  • $\begingroup$ what if the shell is neutral for example and the charge is a big one? $\endgroup$
    – physicsguy19
    Commented May 5, 2018 at 20:34
  • $\begingroup$ @physicsguy19 Sorry, I realized my answer was completely wrong and changed it. $\endgroup$
    – tparker
    Commented May 5, 2018 at 20:49
  • $\begingroup$ if the charge is at the centre of the shell, no force on the charge. if its not, everything changes.. is that correct? $\endgroup$
    – physicsguy19
    Commented May 5, 2018 at 21:31
  • $\begingroup$ @physicsguy19 Correct. If the charge isn't at the center of the shell, then it feels an radial force and gets attracted toward the nearest wall. $\endgroup$
    – tparker
    Commented May 5, 2018 at 21:34
-1
$\begingroup$

Your statement 1) does not hold if there is a bound charge inside the sphere. As to 4) the actor that moves the charge around is providing energy that is dissipated as Ohmic heat.

Improve this answer
$\endgroup$
0

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.