Electric Field inside a solid of uniform charge density

[Sistine]

Homework Statement

What is the electric field inside a a solid of uniform chage density
i.e.

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\int_V\rho(\mathbf{r}')\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|}dV'$$

What is the electric field at $$\mathbf{r}'=\mathbf{r}$$ if $$\mathbf{r}\in V$$

Homework Equations

The Attempt at a Solution

I tried taking the limit as $$\mathbf{r}\to \mathbf{r}'$$ of $$\mathbf{E}(\mathbf{r})$$. Can we then take the limit under the integral sign? I tried using Riemannian sums to prove this but still I'm not sure. I then got that $$\mathbf{E}(\mathbf{r}')=\infty$$ but I don't think that this is correct.

 

[Sistine]

I have thought again about the problem. I know that all the charge within the solid will work to repel itself so that the all the charge tends towards the surface. So that by Gauss' law the electric field within the surface is zero. But I still don't know what the electric field will be on the surface.

 

[turin]

... a solid of uniform chage density ...

I know that all the charge within the solid will work to repel itself so that the all the charge tends towards the surface. So that by Gauss' law the electric field within the surface is zero.

It doesn't matter that the charge repels itself. If it is prevented from moving, then it cannot go to the surface. You are given that the charge density is uniform, thus requiring you to deal with charge inside the surface. You are trying to apply the concept of an electrical conductor to an insulator.

However, you are on to something. Think more about Gauss' Law, and spherical symmetry, specifically when you have charge inside the surface.