# What's the electric field in a sphere with charge $Q$ and an inverse linear/square law density of charge?

In Classical Electrodynamics by Jackson problem 1.4 asks to find the electric field as a function of the radius of a sphere charged with a charge Q and with a spherically symmetric density that goes as $$r^n$$ with $$n > -3$$.

I found the expression of $$\rho$$ by normalizing it to the total charge Q. I can calculate the electric field via Gauss law when $$n \geq 0$$ but in the other cases the integral of $$\rho$$ diverges. I was thinking of trying to exclude the singularity by excluding a sphere of radius $$\epsilon$$ and then making epsilon go to zero but not sure.

## 1 Answer

As you write, the expression for the charge density can be found by considering that the total charge within the sphere is $$Q$$. Indeed, if we define $$\rho(r) = A r^n$$ we have

$$Q = 4 \pi A \int_0^{R_s} r^n r^2 dr = 4 \pi A \int_0^{R_s} r^{n+2} dr$$

where $$R_s$$ is the radius of the sphere.

The Gauss law in this case takes the form

$$4 \pi R^2 E(R) = \frac{4 \pi}{\epsilon_0} \int_0^R \rho(r) r^2 dr$$

from which we get

$$E(R) = \frac{1}{\epsilon_0 R^2} \int_0^R \rho(r) r^2 dr = \frac{A}{\epsilon_0 R^2} \int_0^R r^{n+2} dr$$

The integral on the right-hand side diverges only for $$n \leq -3$$, which is not a issue given the text of the problem to solve.

For instance, if $$\rho(r) = A r^{-2}$$ we obtain

$$E(R) = \frac{A}{\epsilon_0 R}$$

whereas if $$\rho(r) = A r^{-1}$$ the field is

$$E(R) = \frac{A}{2 \epsilon_0}$$