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.

  • $\begingroup$ 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
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    $\begingroup$ @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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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