0
$\begingroup$

Gauss's Law shows that the electric field everywhere inside a spherical shell of uniform charge density is $0$. Suppose we have a surface which divides space into two disjoint regions (an interior and exterior). If the electric field is $0$ everywhere interior to the surface, does it follow that the surface is a spherical shell of uniform charge density?

Edit: I am stupid. As Alfred Centauri pointed out, a zero electric field everywhere means there is no restriction on the surface. So let me impose the condition that the surface does not have charge zero.

$\endgroup$
  • $\begingroup$ The differential form of Gauss' law is $\nabla \cdot {\vec E} = \rho / \epsilon_0$. If $\vec E = 0$ everywhere inside the spherical shell, then how do you have a charge there, i.e. $\rho \ne 0$? $\endgroup$ – Sofia Mar 2 '15 at 2:14
  • $\begingroup$ Is this the question as it is asked on your homework assignment? $\endgroup$ – garyp Mar 2 '15 at 2:15
  • $\begingroup$ @JoshuaBenabou is the charge distributed only on the surface of the spherical shell? $\endgroup$ – Sofia Mar 2 '15 at 2:18
  • $\begingroup$ @Sofia: I don't know vector calculus too well and in any case I am familiar only with the integral form of Gauss's Law. I don't understand your question. Is it not true that the electric field is zero everywhere inside a spherical shell of uniform charge density. According to my textbook this is this case. $\endgroup$ – Joshua Benabou Mar 2 '15 at 2:22
  • 1
    $\begingroup$ @JoshuaBenabou aha! It's O.K. But if the surface that you take as separating between interior and exterior, can be any surface, right? As long as interior to this surface $\vec E = 0$, it means that, if there are charges, they are outside this surface, right? $\endgroup$ – Sofia Mar 2 '15 at 2:27
1
$\begingroup$

No, it does not imply that the surface is spherical and charged uniformly.

Imagine a charged conducting shell of arbitrary shape. (An ellipsoid is a simple example.) Gauss' Law tells us that the charges in the conductor fly to the outside surface of the conductor, and the distribution of charges is such that the E-field inside is zero. But for a non-spherical conductor, the charge distribution is explicitly not spherical, and the charge distribution on it is not uniform.

$\endgroup$
  • $\begingroup$ I don't understand any of this. $\endgroup$ – Joshua Benabou Mar 2 '15 at 3:17
  • $\begingroup$ @garyp who spoke of a charged conductor here? What you want with an ellipsoidal conductor? $\endgroup$ – Sofia Mar 2 '15 at 3:20
  • $\begingroup$ @Sofia I'm simply providing an example of a non-spherical surface that has zero E-field within. Unless I don't understand the question, this is a direct response to the question in the last sentence of the first paragraph. $\endgroup$ – garyp Mar 2 '15 at 14:25
0
$\begingroup$

The other way of phrasing your question: if I have a surface of uniform charge density, and the field inside is zero everywhere, does it follow that the surface is a sphere?

The answer is "yes". Imagine we have a non spherical surface. We know that it is possible to have zero field inside any conductor regardless of shape. But we also know that the charge distribution on a non-spherical conductor is non-uniform. And finally we know that if we have a solution that meets the boundary conditions, that is THE solution.

It follows that a non spherical surface with uniform charge distribution dos not have zero field inside - only a spherical surface does.

$\endgroup$
  • $\begingroup$ I see. And how do you know that a nonsperical conductor has non uniform charge density. Apologies if its a elementary. $\endgroup$ – Joshua Benabou Mar 3 '15 at 5:51
  • $\begingroup$ @RobJeffries yes that was a typo. $\endgroup$ – Floris Mar 3 '15 at 11:42
  • $\begingroup$ @JoshuaBenabou if the charge on a non-spherical conductor is uniform, then the charges closer to the center of charge will have a different potential than the ones further out, and will experience a repulsive force. Charge will redistribute until the surface is an equipotential. Intuitively the charges like to "get as far from each other as possible". $\endgroup$ – Floris Mar 3 '15 at 11:45

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.