Net charge of non-uniformly charged insulated sphere

Question

So assume there is an insulated sphere with a non-uniform charge density and radius $R$. It has a constant electric field of $E$.

Here is my current line of thinking:

We can pick a Gaussian surface at radius $r < R$.

That would give $ E(4 \pi r^2) = \frac{q(r)}{\epsilon_o} $, where $q(r)$ is a function which defines the charge enclosed by the Gaussian surface.

Assuming this is right (It seems too easy) can't we pick a point r so close to R, such that we can say $r=R$ in our calculations? that would give:

$q(R) = \frac{ER^2}{k_e} $

Is this right?

Answer 1

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}(r)}{\epsilon_0} $$

with $Q_{enc}(r)$ the charge enclosed by the surface of radius $r$.

If you pick $r = R$ you get $Q_{enc}(R) = 4\pi R^2E\epsilon_0$.

I'm not entirely comfortable that this is consistent with a physical distribution of charge. Because $E$ is constant everywhere, it must be that $Q_{enc}(r) \propto r^2$. But then charge density is infinite at the centre.