Electrostatic Energy of Sphere in Shell

[Dionysus~]

Homework Statement

Compute, in the following two ways, the electrostatic energy $W$ of the uniformly charged solid sphere of radius $a$ (charge density $\rho$) that is surrounded concentrically by a uniformly charged thin
spherical shell of radius $b$ (surface charge density $\sigma$), where the charge densities satisfy $\frac {4\pi a^3}{3}\rho + 4\pi^2 \sigma = 0$ That is, the sum of all the charge is zero.
(a) Compute $W=\int \frac {\epsilon_0 E^2}{2}d\tau$.
(b) Compute $W=\int \frac {\rho V}{2}d\tau$.

Homework Equations

$V = -\int E \cdot dl$

The Attempt at a Solution

For part (a) the electric field between the shell and the sphere is $E = \frac {a^3 \rho}{3\epsilon_0 r^2}$. Plug it into the equation and integrate using spherical coordinates. I got $W_1 = \frac {2\pi a^5 \rho^2}{9\epsilon_0}(1 - \frac {a}{b})$. Then the electric field inside the sphere is $E=\frac {r\rho}{3\epsilon_0}$. Integrating over the sphere I got $W_2 = \frac {2\pi a^5 \rho^2}{45 \epsilon_0}$.

In part (b) I found $V$ from $E$. Integrating over the volume between the shell and sphere I got $W_1 = \frac {2\pi a^3 \rho^2}{18\epsilon_0} b^2 - \frac {2\pi a^5 \rho^2}{6\epsilon_0} + \frac {2\pi a^5 \rho^2}{9\epsilon_0}(\frac {a}{b})$. Then integrating over the sphere I got $W_2 = \frac {2\pi a^5 \rho^2}{45 \epsilon_0}$.

It seems that I went wrong finding the volume integral using $\int \frac {\rho V}{2}d\tau$. My setup was
$W_1 = \frac {1}{2} \frac {a^3 \rho^2}{3 \epsilon_0}\int_{0}^{2\pi} \int_0 ^\pi \int_a ^b \left (\frac {1}{r} - \frac {1}{b} \right ) (r^2 \sin \theta dr d\theta d\phi)$.
Can anyone spot my mistake?

 

[anathema]

Hello!

It looks like you may have made a mistake in your setup for the volume integral. The correct setup should be:

W_1 = \frac{1}{2} \int_0^{2\pi} \int_0^\pi \int_a^b \frac{\rho}{2} \left(\frac{1}{r^2} - \frac{1}{b^2}\right) r^2 \sin \theta dr d\theta d\phi

This is because the volume integral for electrostatic energy is given by W = \frac{1}{2} \int \rho \phi d\tau, where \phi is the electric potential. In your setup, you have multiplied by \rho again, which is unnecessary. Also, the limits of integration for r should be from a to b, as the volume between the sphere and shell extends from the inner radius a to the outer radius b.

I hope this helps clear up your confusion! Let me know if you have any other questions.