Finding the energy stored in an spherical shell, but integral diverges

Question

I am attempting to find the energy stored in assembling an spherical shell (denoted by $S$) uniformed distributed of total charge $q$, and radius $R$. To do so, I want to use the formula: $$W = \frac{1}{2} \epsilon_0 \int_{\text{S}}\sigma V da$$

The problem comes when I try to calculate $V$. Using Gauss's law to find $E$ and then $V$, it's easy to see that $V = \frac{q}{4\pi \epsilon_0 R}$ on the surface of the sphere. However when I try to calculate $V$ using an integral method, the integral diverges. The formula for $V$ is $$V = \frac{1}{4\pi \epsilon_0} \int_S \frac{\sigma}{|\vec{r}-\vec{r}\hspace{1pt}'|} da'$$

where $\vec{r}$ is the location of $\rho$ and $\vec{r}\hspace{1pt}'$ is the position of the the other charges which affects $\rho$. One can see that this integral diverges, which makes sense because we are assuming an continuous charge distribution, which means $|\vec{r}-\vec{r}\hspace{1pt}'| \to 0$.

I don't understand why this is? Using this integral methods, we should still arrive at the same answer as the Gauss's law method. Could someone please explain why this doesn't work/or a way to get around this issue?

Answer 1

The integral does not diverge. Although |r −r′|→0, $da'=\sin(\theta)d\theta d\phi$ approaches $0$ as well. Using spherical coordinates with the point at which the potential is being evaluated placed at the north pole we get: $$\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\sin(\theta)d\theta d\phi}{\sqrt{2-2\cos(\theta)}}=4\pi$$

Answer 2

We know that the electric field inside a uniformly charged spherical shell is zero, because all the charges lies on the outside surface. i.e.,

$$\vec E_{in}=0$$ and $$ \vec E_{out}=\frac{Q}{4\pi \epsilon_o r^2} \hat{r}$$

$\textit{Electrostatic Energy stored in the spherical shell is :}$

$$U=\frac{1}{2}\epsilon_o \iiint_V E^2 \, d\tau$$

$$U=\frac{1}{2}\epsilon_o \underbrace{\iiint_V E_{in}^2 \, d\tau}_{0}+\frac{1}{2}\epsilon_o \iiint_V E_{out}^2 \, d\tau$$

$$U=\frac{1}{2}\epsilon_o \iiint_V E_{out}^2 \, d\tau$$

$$=\frac{1}{2}\epsilon_o \iiint_V \left(\frac{Q}{4\pi \epsilon_o r^2 }\right)^2 \, 4\pi r^2 \, dr$$

$$=\frac{1}{2}\epsilon_o\left(\frac{Q}{4\pi \epsilon_o }\right)^2 \times 4\pi \int_R^{\infty} \frac{1}{r^2}\, dr$$

$$U=\frac{Q^2}{8\pi \epsilon_o R}$$