# How the electric field inside a charged ball is zero

1. Oct 11, 2013

### FOIWATER

I am having issues with multiple gauss' law problems.
Assume we have a charged spherical surface. Inside the surface, E=0V/m. But why?
I understand that at the center it should be zero due to symmetry, but why would it be zero at some point say, inside the surface but not at the center?
Don't the flux lines radiate in all directions?

2. Oct 11, 2013

### jfizzix

If all of the charge is uniformly distributed at the surface, then the field everywhere inside will be zero. With Gauss's law, we know that the field inside the sphere would be zero because any smaller gaussian sphere would enclose no charge.
The reason for this is that the electric flux over the surface of the sphere would be zero, and this is only because there is no charge inside the sphere. this gaussian surface could take any shape, so we know that the electric field has to be zero everywhere inside the sphere.

However, if you have a charged sphere whose charge is distributed evenly throughout the whole volume of the sphere, then the electric field would be zero only at the center of the sphere, and would grow linearly in the radial direction until the surface of the sphere, at which point the field would fall off as the inverse square.

One last thing to note. If a sphere is made out of a conducting material, and charge is added to it, the charge will distribute itself evenly on the surface of the sphere, and not be anywhere inside the sphere. If there were charges inside the sphere, then there would be an electric field inside the sphere, which would keep moving the charges until the field was zero.

3. Oct 12, 2013

### FOIWATER

Yes, but how can we explain that it is zero not at the center without using gauss' law, where's the intuition?

It seems like the charge on the surface would cause flux inward as well

4. Oct 12, 2013

### vela

Staff Emeritus
Calculate the electric field at a point due to charge on an infinitesimal area of the sphere subtended by solid angle $d\Omega$ at the two opposite ends of the sphere. You'll see the contributions cancel out.

Last edited: Oct 12, 2013
5. Oct 13, 2013

### vanhees71

It's easier to show this with the local laws. For electrostatics there's a scalar potential,
$$\vec{E}=-\vec{\nabla} \Phi.$$
The equation for the potential is the Poisson equation,
$$\Delta \Phi=-\rho,$$
where $\rho$ is the charge density (in Heaviside-Lorentz CGS units).

Now if there are no charges inside and if the situation is spherically symmetric the above equation simplifies to an ordinary differential equation for $\phi=\phi(r)$, where $r=|\vec{x}|$, and there is a unique solution which is free of singularities at the origin. What does this solution mean for the electric field inside the sphere?

6. Oct 13, 2013

### Ibix

A qualitative argument:

Start at the centre. The force is zero - you can see that by slicing the sphere in any plane passing through your location at the centre and observing that the two parts are identical.

Now move a small distance in the z direction. Again you can split the sphere on a plane. If that plane is parallel to the z axis, again the situation is symmetrical so there is no force in the x or y directions.

But what about the z direction? Split the sphere in two with a plane parallel to the x-y plane through your location. You have two unequal parts. The part above you is smaller, but on average closer, than the part below you. As you move further from the centre, the part above you gets smaller, but its centre of mass gets closer. Obviously these two effects will act opposite ways on the magnitude of the force. That they cancel completely isn't obvious, but this argument makes the result plausible (at least to me).

7. Oct 14, 2013

### jfizzix

The intuition for a conducting sphere with uniform charge distributed on its surface comes from the fact that the electric field goes as the inverse square of the distance from the charges generating the field, and that the sphere is a 3-dimensional closed surface.

What this means is that inside the sphere the contribution from the charges on all sides of the sphere exactly cancel each other out.

Another way of visualizing this: If you were inside a large hollow sphere whose inner surface was uniformly radiating light, because the intensity of light also decreases as the inverse square of the distance from the source, you would see exactly the same brightness in all directions no matter where in the sphere you were.
In particular, the area in your field of vision goes as the square of your distance to that surface while the intensity of each point decreases as the inverse square, perfectly cancelling out with the amount of area in your field of vision.