# Why is the electric potential on the surface of a sphere not infinite?

By using Gauss' Law, it can be shown that a uniformly charged hollow sphere can be treated as a point charge lying at its centre with a charge equal to that of the sphere. Owing to this fact, the potential at the sphere's surface becomes equal to ${V}\mathrm{{=}}\frac{Q}{4{\mathit{\pi}\mathit{\epsilon}}_{0}R}$

However, from what I've learnt in my class, the potential at any point due to a number of bodies can also be calculated as the sum of individual potential of each body at that point. So, I tried using this fact to evaluate the potential at the sphere's surface and encountered a huge problem. The answer that I calculated had an undefined term in it, which is clearly absurd. I have included my calculations in the images attached below.

So could someone please explain to me what am I doing wrong here?

Preliminaries

I will make use of a couple of results

1. If $|t|<1$ then

$$\frac{1}{\sqrt{1 - 2t x + x^2}} = \sum_n P_n(x)t^n \tag{1}$$

where $P_n(x)$ are the Legendre polynomials. With $P_0 = 1$

2. These polynomials are orthogonal

$$\int_{-1}^1dx\;P_n(x)P_m(x) = \frac{2}{2n + 1}\delta_{mn} \tag{2}$$

3. If $\alpha > 1$ then $1/\alpha < 1$ and

\begin{eqnarray} \int_{-1}^1dx\;\frac{1}{\sqrt{1 -2\alpha x + \alpha^2}} &=& \frac{1}{\alpha}\int_{-1}^1dx\;\frac{1}{\sqrt{1 -2(1/\alpha) x + (1/\alpha)^2}}\\ &\stackrel{(1)}{=}& \frac{1}{\alpha}\int_{-1}^1 dx\;\sum_n [P_n(x)\times \underbrace{1}_{P_0(x)}](1/\alpha)^n \\ &\stackrel{(2)}{=}& \frac{1}{\alpha} \sum_n (1/\alpha)^n\frac{2}{2n+1}\delta_{n0} = \frac{2}{\alpha} \tag{3} \end{eqnarray}

4. If you call $\mathbf{r} = \alpha R \hat{z}$ with $\alpha > 1$, and $\mathbf{r}'$ a point on a sphere of radius $R$, then

\begin{eqnarray} \mathbf{r}' - \mathbf{r} &=& R\sin\theta\cos\phi\hat{x} + R\sin\theta\sin\phi\hat{y} + R\cos\theta\hat{z} - \alpha R \hat{z}\\ |\mathbf{r}' - \mathbf{r}|^2 &=& R^2\cos^\theta\cos^\phi + R^2\cos^\theta\sin^\phi + R^2\cos^2\theta + \alpha^2 - 2 \alpha\cos\theta \\ &=& 1 + \alpha^2 - \alpha\cos\theta = 1 + \alpha^2 - \alpha x\tag{4} \end{eqnarray}

where $x = \cos\theta$

5. The volume charge density of a spherical shell with radius $R$ and charge $Q$ is

$$\rho(\mathbf{r}) = \delta(r - R)\frac{Q}{4\pi R^2} \tag{5}$$

as you can easily verify by noticing that $\int d^3\mathbf{r}\; \rho(\mathbf{r}) = Q$

Solving the problem

Now we have all the elements to find the potential. I will calculate the potential on a point along the $z$ axis outside the sphere, let us call that point $\mathbf{r} = \alpha R\hat{z}$, with $\alpha > 1$. The potential at this location is

\begin{eqnarray} V(\mathbf{r}) &=& \frac{1}{4\pi\epsilon_0}\int d^3\mathbf{r}'\; \frac{\rho(\mathbf{r}')}{|\mathbf{r}' - \mathbf{r}|} \\ &\stackrel{(5)}{=}& \frac{1}{4\pi\epsilon_0} \frac{Q}{4\pi R^2}\int d\Omega dr\; \frac{r^2\delta(r - R)}{|\mathbf{r}' - \mathbf{r}|} \\ &\stackrel{(4)}{=}& \frac{1}{4\pi\epsilon_0}\frac{Q}{4\pi R}\int_0^{2\pi}d\phi\int_{-1}^1 dx \frac{1}{1 + \alpha^2 -2\alpha x} \\ &\stackrel{(3)}{=}& \frac{1}{4\pi\epsilon_0} \frac{Q}{4\pi R}\times 2\pi \times \frac{2}{\alpha} = \frac{1}{4\pi\epsilon_0} \frac{Q}{\alpha R}\\ &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{|\mathbf{r}|} \end{eqnarray}

• Thank you for taking the time out for writing this detailed answer. Unfortunately for me, I am a high school student so I have a very limited knowledge of calculus right now. Instead of asking you to explain all the steps you wrote down, I'll just request you to please explain the error in my calculations. – Anindya Mahajan Nov 28 '16 at 19:07
• @AnindyaMahajan It boils down to the fact that the potential between two points is proportional to the inverse of their separation. So you need to calculate the distance between $P$ and the "small" charge $Rd\theta$. The other problem is that if you want to locate this charge on the sphere you will need two coordinates $\theta$ and $\phi$, so far you are only using one $\theta$ – caverac Nov 28 '16 at 19:12

It looks like you are not using the right distance in your formula for dV. The distance should be measured from the point P to dQ along a straight path. From the law of cosines you should be able to see that the distance you need in the denominator is not R(1-Cos[theta]) but rather R*Sqrt[2-2Cos[theta]].

This is a much more difficult integral to solve (see the answer by caverac). My earlier answer missed the point of what you are trying to do, so I deleted it.

• Ah, silly me! , you are indeed correct. I needed to take the distance of the point from the differential ring, and not its centre. Using Pythagoras theorem, the distance came out just like you said. Anyway, the integral now formed is not much different than it was before. It can still be solved using the method I used before, that is by making use of u-substitution (${1}\mathrm{{-}}\cos{\mathrm{(}}{x}{\mathrm{)}}\mathrm{{=}}{u}$ ) – Anindya Mahajan Nov 28 '16 at 20:57
• Ah, silly me! Yes, you are indeed correct. I needed to take the distance of the point from the differential ring, and not its centre. Found the distance using Pythagoras theorem and it was equal to what you said. Anyway, the integral now formed is not much different than it was before. It can still be solved using the method I used before, that is by making use of u-substitution (${1}\mathrm{{-}}\cos{\mathrm{(}}{x}{\mathrm{)}}\mathrm{{=}}{u}$ ) – Anindya Mahajan Nov 28 '16 at 20:59