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 1


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

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


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