# Charge enclosed by a Gaussian surface inside an uniformly charged thin sphere [duplicate]

Why is the electric field due to a charge enclosed by a Gaussian surface inside an uniformly charged thin spherical shell, zero?

• There is no charge inside the sphere, right? Or do you mean electric field? If you mean electric field, then it depends on what you accept as true. Do you accept Gauss's law (in which case you have your answer) or do you accept Coulomb's Law (which can be derived from Gauss's Law)? It's harder to prove using Coulomb's Law, but not impossible. – Jared Jul 4 '15 at 5:54
• Do you mean electric field? – Cicero Jul 4 '15 at 6:01
• Yes. Electric field. Edited. – user84796 Jul 4 '15 at 6:44
• @user84796 But you didn't answer my question. In fact your question still doesn't really make sense. I think you mean to ask why is the electric field zero at all points inside of the thin spherical shell. It's important that you have spherical symmetry in this case because if the shape was not spherical, this would not be the case. – Jared Jul 4 '15 at 7:15
• See Newton's shell theorem. Possible duplicates: physics.stackexchange.com/q/150238/2451 and links therein. – Qmechanic Jul 4 '15 at 8:20

OK, I'm going to add an answer based on the possible duplicate and the other answer given. I'm going to assume we accept Gauss's Law here (which makes things much easier rather than trying to prove using Coulomb's law or Newton's theory of gravity from the duplicate answer). First, you need to appreciate the symmetry of the situation.

First off, we know that since the enclosed charge is zero that the electric flux through the enclosed Gaussian surface is zero. This does not help us determine the electric field--at all! It isn't until we reason that, through symmetry, that the electric field must be the same at all points along this Gaussian surface that we can use Gauss's Law to determine the electric field along that surface! I use the above figure to illustrate the symmetry. Notice that I rotated the figure by $45^\circ$. We see through this rotation that the originally right point now coincides with the originally left point! The figure has not changed therefore the value of the electric field must be equal at those two points!

It is through this symmetry that we reason that the electric field must be equal everywhere along that inside circle. Therefore the flux equation now becomes:

$$\Phi_E = \oint \vec{E}\circ d\vec{A} = EA = 0 \rightarrow E = 0$$

This is not the case when we do not have spherical symmetry. So here is another example:

So we do the exact same rotation so that now the originally left point coincides with the originally right point. However, now the symmetry does not produce the same original figure and thus we can no longer argue that the electric field for the two positions should be the same. Therefore we can no longer argue that:

$$\Phi_e = \oint \vec{E}\circ d\vec{A} = EA$$

(although the net electric flux is certainly still $0$)

We cannot argue this because the electric field is not constant over our internal surface and thus it's not true that the electric field (in this case) is identically zero everywhere inside the thin cube of uniform charge.

And no, it's not because I chose a sphere vs. a square as the Gaussian surface in the second case. Although we can argue that the magnitude of the net flux on each side of an internal cube would be equal (but the electric field is not constant along those surfaces).

Why is the electric field due to a charge enclosed by a Gaussian surface inside an uniformly charged thin spherical shell, zero?

I'm not quite clear about your question, but I will volunteer this answer: because there's no net force.

Have a look at the Wikipedia Coulomb force article, and note this: "An electric field is a vector field that associates to each point in space the Coulomb force experienced by a test charge". Imagine you set a positron down exactly in the middle of your sphere. The positron doesn't move to the left or the right, because the electrostatic forces are in equilibrium. Then if you set the positron down an inch to the left, the forces are still in equilibrium because of the geometry of the sphere. Draw circles to appreciate this. There's less than half of the circle pulling left, and more than half pulling right. Set it down another inch to the left, and the forces are still in equilibrium. And so on.

• If you set a test charge at the center of a thin cube (with uniform charge), it would experience no net force (i.e. no electric field). However, if you put a test charge "off-center" there would be a net force (a net electric field). I don't see how your arguments distinguish the case of a cube shape vs. a spherical shape. – Jared Jul 4 '15 at 8:25
• It's like the Shell theorem Jared. For your cube, imagine the force on your particle between two parallel plates, like so: | • |, then imagine it between horizontal and vertical plates only, then imagine setting it down to the left in each situation. – John Duffield Jul 4 '15 at 9:16