# Gauss' law and ions?

My text book says that with we have a singly ionized sodium atom net charge +e and if we choose a spherical surface centered on the ion and large enough to contain it all we do not need to know the distribution of charge inside the ion. And using Gauss' law can find the the electric field outside the sphere radius R as $E(R)=-\frac{e}{4\pi \varepsilon_0R^2}$.

Here is my problem surly we need to know that the charge is distributed evenly around the center?? If we this was not the case then all the charge could, say, be all at one side of the sphere and thus the electric field at that side will be greater then the electric field at the other side of the sphere will it not??

## 1 Answer

Yes indeed. To get the result you stated from Gauss' Law you must asume that the charge is distributed in such a way that you have a spherically symetric field. How you get to that field doesnt matter though. So the charge might all be concentrated at the center or all lying on the surface of a sphere: it doesnt matter as long as the field is spherically symmetric.