# How does Gauss's Law imply that the electric field is zero inside a hollow sphere?

Let's say that I have a hollow sphere of radius $R$. I wish to find the electric field inside it at some point.

Gauss's Law tells us that:

$$\iint{ \vec{E}(\vec{r}). \vec{dA} } = \frac{\sum q}{\epsilon}$$

Now my teacher and others taught me that in order to find the electric field one can draw a gaussian surface and apply this law and would get that the electric field is equal to zero because the charge enclosed is $0$.

My question is: Doesn't guass's law only finds the electric field "due to the charge enclosed" and since we draw the gaussian surface "inside the sphere" where there is no charge, wouldn't it be wrong to simply say that the electric field due to the "whole hollow sphere" is zero even though "we aren't drawing the gaussian surface around the charge"? I hope it makes sense.

• The 'hollow sphere', is it a CONDUCTING sphere? There is a lot you know about fields inside a conductor, that you don't know about a 'sphere'. – Whit3rd Aug 23 '16 at 5:39

## 1 Answer

Doesn't guass's law only finds the electric field "due to the charge enclosed"

No. The $\mathbf{E}$ in Gauss's Law is the electric field due to all charges, both inside and outside the Gaussian surface.

The reason that charges outside do not contribute to the total surface integral is the field they produce "contributes twice", once when the field "enters" and once when it "leaves" the surface. Gauss's Law tells us that these contributions must cancel out.

What does this look like inside a spherical shell? Well first we argue from symmetry considerations that

1. The field should depend only on $r$, the distance from the centre of the shell
2. The field should be directed radially

Now we can invoke Gauss's Law on a spherical surface of radius $r<R$ and get $$4\pi r^2 E = 0 \quad\Rightarrow\quad E = 0$$

Note that the symmetry argument here is important. If I break spherical symmetry by, say, adding a point charge at some point, then the field inside the shell will be the field produced by the charge I added by the principle of superposition. Simply saying that there is not charge inside the Gaussian surface without this extra requirement is not enough to say the field is $0$.

• In general you cannot use the symmetry argument because the electric field is zero in the volume enclosed by a closed surface conducting shell of any shape if there are no charges within that volume.. – Farcher Aug 22 '16 at 18:36
• @Farcher The situation you are describing for a conducting surface is very different from the case being discussed of a surface with a fixed charge. For a conducting surface. For a conducting surface you are considering the result of some external field and how the charges in the conductor move in response. Here we are considering a shell of charge which in not free to move and where there need be no external field. For a fixed distribution of charge, the symmetry requirement is necessary. – By Symmetry Aug 22 '16 at 18:50