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## Homework Statement

Compute, in the following two ways, the electrostatic energy [itex]W[/itex] of the uniformly charged solid sphere of radius [itex]a[/itex] (charge density [itex]\rho[/itex]) that is surrounded concentrically by a uniformly charged thin

spherical shell of radius [itex]b[/itex] (surface charge density [itex]\sigma[/itex]), where the charge densities satisfy [itex]\frac {4\pi a^3}{3}\rho + 4\pi^2 \sigma = 0[/itex] That is, the sum of all the charge is zero.

(a) Compute [itex]W=\int \frac {\epsilon_0 E^2}{2}d\tau[/itex].

(b) Compute [itex]W=\int \frac {\rho V}{2}d\tau[/itex].

## Homework Equations

[itex]V = -\int E \cdot dl[/itex]

## The Attempt at a Solution

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

In part (b) I found [itex]V[/itex] from [itex]E[/itex]. Integrating over the volume between the shell and sphere I got [itex]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})[/itex]. Then integrating over the sphere I got [itex]W_2 = \frac {2\pi a^5 \rho^2}{45 \epsilon_0}[/itex].

It seems that I went wrong finding the volume integral using [itex]\int \frac {\rho V}{2}d\tau[/itex]. My setup was

[itex]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) [/itex].

Can anyone spot my mistake?