Electrical energy stored by charged concentric spherical shells

[phantomvommand]

I thought up of this problem myself, so I do not have solutions. I would appreciate if you could correct my approach to solving this problem.

Firstly, the charge induced on the inner surface of shell B is -q, and so the charge on the outer surface of shell B is Q+q.

The energy stored can be calculated as the potential energy stored by a shell capacitor of charge q, as given by (q^2/8pi e0)(1/a - 1/b). Then, we have to add the potential energy stored in the E-field outside the 2 shells, which is (1/4pi e0) (Q+q)/b.

Does my approach look right?

 

[etotheipi]

I think the method is okay, but I wonder if the potential energy outside the second sphere should instead be $$U = \frac{(Q+q)^2}{8 \pi \varepsilon_0 b}$$

 

[phantomvommand]

I think the method is okay, but I wonder if the potential energy outside the second sphere should instead be $$U = \frac{(Q+q)^2}{8 \pi \varepsilon_0 b}$$

You are right. I forgot that the shell theorem does not apply to energy and potential.

 

[etotheipi]

Potential energy no, but it does apply to potential! To get the above you just need to write the energy of a second spherical capacitor from radius $b$ to radius infinity (exactly the same way as you calculated the first contribution to the energy, between the shells). But if you're interested, you can also find the energy another way!

Take, for example, just the region outside the outer shell. From Gauss you see that the electric potential for $r \in [b, \infty)$ is just $\phi(r) = \frac{k(Q+q)}{r}$, i.e. the same as for a point charge $Q + q$ at the origin, due to the spherical symmetry. The associated electric field is $\mathbf{E} = \frac{k(Q+q)}{r^2} \mathbf{e}_r$. And the norm-squared of that vector is $|\mathbf{E}|^2 = \frac{k^2(Q+q)^2}{r^4}$

So denote by $\Omega \subseteq \mathbb{R}^3$ the set of points with radii $r \in [b, \infty)$, then given the volume element $dV = r^2 dr do$ in spherical coordinates you have$$
\begin{align*}

U_2 &= \frac{\varepsilon_0}{2} \int_{\Omega} |\mathbf{E}|^2 dV \\ \\

&= \frac{\varepsilon_0 k^2 (Q+q)^2}{2} \int_{0}^{4\pi} \int_{b}^{\infty} r^{-2} dr do = \frac{\varepsilon_0 k^2 (Q+q)^2}{2} \frac{4\pi}{b} = \frac{(Q+q)^2}{8\pi \varepsilon_0 b}

\end{align*}
$$Of course, you could also do the same integral in the region between the two shells, which would give you the electric energy $U_1$. And the sum $U_1 + U_2$ gives you what you're after.