## Laplace’s equation: conducting sphere and shell of charge

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.37.

Another example of using the series solution to Laplace’s equation to find the potential using boundary conditions.

We have a conducting sphere of radius ${a}$ maintained at a constant potential ${V_{0}}$. Concentric with this sphere is a larger spherical shell of radius ${b}$ with a surface charge density of ${\sigma(\theta)=k\cos\theta}$. Find the potential in the two regions (i) ${a\le r\le b}$ and (ii) ${r>b}$.

The general solution in terms of Legendre polynomials is

In the inner region, we must use the most general form of the solution, since ${r}$ tends neither to zero nor infinity. In the outer region, to keep the potential finite, we dispose of the terms in ${r^{l}}$, so we have

Note that we must use a different set of coefficients since the potential will have a different functional form outside the shell.

We can now apply the various boundary conditions. First, at ${r=a}$ we must have ${V=V_{0}}$ so

$\displaystyle \sum_{l=0}^{\infty}\left[A_{l}a^{l}+\frac{B_{l}}{a^{l+1}}\right]P_{l}(\cos\theta)=V_{0}$

From the orthogonality of the ${P_{l}}$, only the ${l=0}$ term on the left is non-zero, so we get

 $\displaystyle A_{0}+\frac{B_{0}}{a}$ $\displaystyle =$ $\displaystyle V_{0}$ $\displaystyle B_{l}$ $\displaystyle =$ $\displaystyle -a^{2l+1}A_{l}\ \ \ (l>0)$

Next, we look at ${r=b}$. Here, the potential must be continuous, so

$\displaystyle \sum_{l=0}^{\infty}\left[A_{l}b^{l}+\frac{B_{l}}{b^{l+1}}\right]P_{l}(\cos\theta)=\sum_{l=0}^{\infty}\frac{C_{l}}{r^{l+1}}P_{l}(\cos\theta)$

Equating coefficients of the ${P_{l}}$ we get

$\displaystyle A_{l}+\frac{B_{l}}{b^{l+1}}=\frac{C_{l}}{b^{l+1}}$

Using the above relation between ${A_{l}}$ and ${B_{l}}$ we get, first for ${l=0}$:

$\displaystyle A_{0}+\frac{B_{0}}{b}=\frac{C_{0}}{b}\ \ \ \ \ (3)$
and for ${l>0}$:

 $\displaystyle A_{l}\left(b^{l}-\frac{a^{2l+1}}{b^{l+1}}\right)$ $\displaystyle =$ $\displaystyle \frac{C_{l}}{b^{l+1}}$ $\displaystyle C_{l}$ $\displaystyle =$ $\displaystyle A_{l}\left(b^{2l+1}-a^{2l+1}\right)$

Finally, we use the surface charge density on the shell. From our earlier example of this type, we get

 $\displaystyle \left.\frac{\partial V}{\partial r}\right|_{out}-\left.\frac{\partial V}{\partial r}\right|_{in}$ $\displaystyle =$ $\displaystyle -\frac{\sigma}{\epsilon_{0}}$ $\displaystyle \sum_{l=0}^{\infty}\left[-(l+1)\frac{C_{l}}{b^{l+2}}-lA_{l}b^{l-1}+(l+1)\frac{B_{l}}{b^{l+2}}\right]P_{l}(\cos\theta)$ $\displaystyle =$ $\displaystyle -\frac{k}{\epsilon_{0}}\cos\theta$

Since ${P_{1}(\cos\theta)=\cos\theta}$, we can equate coefficients on both sides to get

 $\displaystyle -\frac{C_{0}}{b^{2}}+\frac{B_{0}}{b^{2}}$ $\displaystyle =$ $\displaystyle 0$ $\displaystyle -2\frac{C_{1}}{b^{3}}-A_{1}+2\frac{B_{1}}{b^{3}}$ $\displaystyle =$ $\displaystyle -\frac{k}{\epsilon_{0}}$ $\displaystyle -(l+1)\frac{C_{l}}{b^{l+2}}-lA_{l}b^{l-1}+(l+1)\frac{B_{l}}{b^{l+2}}$ $\displaystyle =$ $\displaystyle 0\qquad l>1$

The first of these equations gives us

$\displaystyle B_{0}=C_{0}$

Substituting this into 3 gives us ${A_{0}=0}$; ${B_{0}=aV_{0}=C_{0}}$.

The second equation can be reduced by substituting for ${B_{1}}$ and ${C_{1}}$ in terms of ${A_{1}}$ from the above relations:

 $\displaystyle \left(2\frac{b^{3}-a^{3}}{b^{3}}+1+2\frac{a^{3}}{b^{3}}\right)A_{1}$ $\displaystyle =$ $\displaystyle \frac{k}{\epsilon_{0}}$ $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle \frac{k}{3\epsilon_{0}}$

This gives us, from the above relations

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle -\frac{a^{3}k}{3\epsilon_{0}}$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle \frac{k}{3\epsilon_{0}}\left(b^{3}-a^{3}\right)$

The third equation can be solved by taking ${A_{l}=B_{l}=C_{l}=0}$ for all ${l>1}$. Since the solution to Laplace’s equation is unique, this must be the solution. Putting all this together, we get:

$\displaystyle V(r,\theta)=\begin{cases} \frac{aV_{0}}{r}+\frac{k}{3\epsilon_{0}}\left(r-\frac{a^{3}}{r^{2}}\right)\cos\theta & a\le r\le b\\ \frac{aV_{0}}{r}+\frac{k}{3\epsilon_{0}}\left(\frac{b^{3}-a^{3}}{r^{2}}\right)\cos\theta & r>b\end{cases}$

The induced surface charge on the conductor can be found from the normal derivative at the surface ${r=a}$ (the potential inside the conductor is constant, so its derivative is zero):

 $\displaystyle -\frac{\sigma}{\epsilon_{0}}$ $\displaystyle =$ $\displaystyle \left.\frac{\partial V}{\partial r}\right|_{r=a}$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{a}+\frac{k}{\epsilon_{0}}\cos\theta$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{V_{0}\epsilon_{0}}{a}-k\cos\theta$

If the sphere is grounded so that ${V_{0}=0}$, then the induced charge is just the negative of the charge on the shell. The total charge in the system is thus the surface area of the conducting sphere times the constant term:

 $\displaystyle Q$ $\displaystyle =$ $\displaystyle 4\pi a^{2}\frac{V_{0}\epsilon_{0}}{a}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi\epsilon_{0}aV_{0}$

For large distances, from the above formula

$\displaystyle V\rightarrow\frac{aV_{0}}{r}$

This is consistent with the total charge, since the potential for a point charge is

 $\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{Q}{4\pi\epsilon_{0}r}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{aV_{0}}{r}$