# Dielectric shell surrounding conducting sphere

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.24.

As an exercise in applying Laplace’s equation to a problem with dielectrics, suppose we have a conducting sphere of radius ${a}$ surrounded by a spherical shell of dielectric of outer radius ${b}$ and susceptibility ${\epsilon}$, with the whole system in a uniform electric field ${\mathbf{E}_{0}}$.

The general solution to Laplace’s equation for the potential in spherical coordinates is

$\displaystyle V(r,\theta)=\sum_{n=0}^{\infty}\left(A_{n}r^{n}+\frac{B_{n}}{r^{n+1}}\right)P_{n}(\cos\theta) \ \ \ \ \ (1)$

where ${P_{l}}$ is the degree-${l}$ Legendre polynomial.

Inside the sphere, the potential is constant (since the field is zero inside a conductor), so we might as well take it to be zero.

In the region ${a, the solution is the general one given above. For ${r>b}$, in order to avoid an infinite field for large ${r}$, we have to drop the ${A_{l}}$ terms except for ${A_{1}}$ since we need a potential that gives a constant field. If we take the field to lie in the ${z}$ direction, then since ${z=r\cos\theta=rP_{1}\left(\cos\theta\right)}$ we have for this region

$\displaystyle V_{r>b}=-E_{0}rP_{1}+\sum_{n=0}^{\infty}\frac{C_{n}}{r^{n+1}}P_{n}\left(\cos\theta\right) \ \ \ \ \ (2)$

Note that we’ve used a different set of coefficients ${C_{n}}$ here, since the solution in this region is distinct from that for the region ${a. That is, the coefficients ${B_{n}\ne C_{n}}$.

We can obtain conditions on the coefficients by equating terms of the various Legendre polynomials in the boundary conditions. Continuity of the potential at ${r=a}$ gives the condition

$\displaystyle A_{n}a^{n}+\frac{B_{n}}{a^{n+1}}=0 \ \ \ \ \ (3)$

Continuity at ${r=b}$ gives

 $\displaystyle A_{1}b+\frac{B_{1}}{b^{2}}$ $\displaystyle =$ $\displaystyle -E_{0}b+\frac{C_{1}}{b^{2}}\ \ \ \ \ (4)$ $\displaystyle A_{n}b^{n}+\frac{B_{n}}{b^{n+1}}$ $\displaystyle =$ $\displaystyle \frac{C_{n}}{b^{n+1}}\qquad(n\ne1) \ \ \ \ \ (5)$

Finally, we can use the condition at the boundary of two dielectrics to get, since there is no free charge at the boundary:

 $\displaystyle \epsilon_{1}\frac{\partial V_{1}}{\partial n}$ $\displaystyle =$ $\displaystyle \epsilon_{2}\frac{\partial V_{2}}{\partial n}\ \ \ \ \ (6)$ $\displaystyle \epsilon A_{n}nb^{n-1}-\epsilon\frac{\left(n+1\right)B_{n}}{b^{n+2}}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{\left(n+1\right)C_{n}}{b^{n+2}}\qquad(n\ne1)\ \ \ \ \ (7)$ $\displaystyle \epsilon A_{1}-2\epsilon\frac{B_{1}}{b^{3}}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}E_{0}-2\epsilon_{0}\frac{C_{1}}{b^{3}} \ \ \ \ \ (8)$

Consider first the terms for ${n\ne1}$. We get

 $\displaystyle B_{n}$ $\displaystyle =$ $\displaystyle -a^{2n+1}A_{n}\ \ \ \ \ (9)$ $\displaystyle A_{n}\left(b^{n}-\frac{a^{2n+1}}{b^{n+1}}\right)$ $\displaystyle =$ $\displaystyle \frac{C_{n}}{b^{n+1}}\ \ \ \ \ (10)$ $\displaystyle C_{n}$ $\displaystyle =$ $\displaystyle A_{n}\left(b^{2n+1}-a^{2n+1}\right) \ \ \ \ \ (11)$

Since both ${B_{n}}$ and ${C_{n}}$ are proportional to ${A_{n}}$, the only way the third boundary condition (at the dielectric/vacuum boundary) above can be satisfied is if ${A_{n}=B_{n}=C_{n}=0}$ for ${n\ne1}$. We are therefore left with the ${n=1}$ terms. For these we get

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle -a^{3}A_{1}\ \ \ \ \ (12)$ $\displaystyle A_{1}\left(b-\frac{a^{3}}{b^{2}}\right)$ $\displaystyle =$ $\displaystyle -E_{0}b+\frac{C_{1}}{b^{2}}\ \ \ \ \ (13)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle A_{1}\left(b^{3}-a^{3}\right)+b^{3}E_{0} \ \ \ \ \ (14)$

Plugging these into the third boundary condition gives

 $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle \frac{-3\epsilon_{0}b^{3}E_{0}}{\epsilon\left(b^{3}+2a^{3}\right)+2\epsilon_{0}\left(b^{3}-a^{3}\right)}\ \ \ \ \ (15)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{3\epsilon_{0}a^{3}b^{3}E_{0}}{\epsilon\left(b^{3}+2a^{3}\right)+2\epsilon_{0}\left(b^{3}-a^{3}\right)} \ \ \ \ \ (16)$

In terms of the dielectric constant ${\epsilon_{r}=\epsilon/\epsilon_{0}}$ we get

 $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle \frac{-3b^{3}E_{0}}{\epsilon_{r}\left(b^{3}+2a^{3}\right)+2\left(b^{3}-a^{3}\right)}\ \ \ \ \ (17)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{3a^{3}b^{3}E_{0}}{\epsilon_{r}\left(b^{3}+2a^{3}\right)+2\left(b^{3}-a^{3}\right)} \ \ \ \ \ (18)$

The potential inside the dielectric shell is therefore

 $\displaystyle V_{a $\displaystyle =$ $\displaystyle A_{1}r\cos\theta+\frac{B_{1}}{r^{2}}\cos\theta\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3b^{3}E_{0}\cos\theta}{\epsilon_{r}\left(b^{3}+2a^{3}\right)+2\left(b^{3}-a^{3}\right)}\left[-r+\frac{a^{3}}{r^{2}}\right] \ \ \ \ \ (20)$

The field can be found from the gradient

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial V}{\partial r}\hat{\mathbf{r}}-\frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3b^{3}E_{0}}{\epsilon_{r}\left(b^{3}+2a^{3}\right)+2\left(b^{3}-a^{3}\right)}\left[\left(1+2\frac{a^{3}}{r^{3}}\right)\cos\theta\hat{\mathbf{r}}+\left(-1+\frac{a^{3}}{r^{3}}\right)\sin\theta\hat{\theta}\right] \ \ \ \ \ (23)$

This reduces to the situation of a conducting sphere in a uniform field if we set ${\epsilon_{r}=1}$ (effectively replacing the dielectric by a vacuum). In that case, the potential reduces to

$\displaystyle V_{\epsilon_{r}=1}=-E_{0}r\cos\theta+\frac{a^{3}}{r^{2}}E_{0}\cos\theta \ \ \ \ \ (24)$

The first term is just the applied field, and the second term arises from the induced charge on the conductor.

## 8 thoughts on “Dielectric shell surrounding conducting sphere”

1. georgianajackal

Had been working on this problem since last night and finally I just put down the books at 1:30 am, started working again at 4:00 am in the morning. Finally, done it now but I had to take help from here. it was 6:00 am then.
so with help I could do it in a quarter to thee hours. Goodness knows how much longer would have it taken without all this help.
Thanks Growescience.

2. Pingback: Dielectric shell surrounding conducting sphere | physics

3. Forrest Shriver

I believe I caught an error. In equation 7, in the second term on the LHS, should the term in the denominator not be $\frac{1}{b^{n+2}}$, instead of $\frac{1}{b^{n+1}}$ ? Sorry if I’m mistaken.

4. Georgia

How are you able to take the potential inside the sphere to be zero, whilst also defining the potential in the limit that r-> infinity as E_0 rcos(theta)?

I would have thought that would result in the potential at large r being E_0 rcos(theta) – some constant that is required to make the potential inside the sphere zero?

There was an error in my equation 2, in that the solution for ${r>b}$ should have a different set of coefficients ${C_{n}}$ (instead of ${B_{n}}$), which I’ve now fixed. I think the source of your confusion is that we have a different solution function in each of the three zones, and we make them all consistent with each other by matching them up at the boundaries. Thus we are free to set ${V=0}$ for ${r and also ${V_{r>b}\rightarrow-E_{0}rP_{1}}$ for large ${r}$, provided that we impose the boundary conditions. The solution ${V_{r>b}}$ is a different function from ${V_{r, so we shouldn’t include any information about ${V_{r in the series for ${V_{r>b}}$.
If you mean equation 4 (please use equation numbers when asking about an equation!), that’s the ${n=1}$ term at ${r=b}$, which comes from equating the corresponding terms in equations 1 and 2.