# 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.

# Legendre polynomials: generation by Gram-Schmidt process

Required math: calculus, vectors

Required physics: none

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.25.

Starting with the functions ${1,x,x^{2},x^{3}}$ we can apply the Gram-Schmidt orthogonalization procedure to generate some polynomials that are orthonormal on the interval ${x\in[-1,1]}$. (In this post, we use ${f_{1},f_{2}}$ etc for the original vectors and primed terms ${f_{1}',f_{2}'}$ etc for the orthonormal vectors obtained after Gram-Schmidt. This is different from the earlier post where we used a subscript 1 for the original and subscript 2 for the orthonormal vectors. Apologies for the inconsistency.)

The first function is

$\displaystyle f_{1}=1 \ \ \ \ \ (1)$

To normalize a function over an interval we divide it by the square root of the integral of its square modulus over that interval so here we divide ${f_{1}}$ by ${\sqrt{\int_{-1}^{1}1^{2}\cdot dx}=\sqrt{2}}$:

$\displaystyle f_{1}'=\frac{1}{\sqrt{2}} \ \ \ \ \ (2)$

Over a symmetric interval, the functions 1 and ${x}$ are already orthogonal (1 is even and ${x}$ is odd), so we can normalize ${f_{2}=x}$ directly to get

$\displaystyle f_{2}'=\sqrt{\frac{3}{2}}x \ \ \ \ \ (3)$

To get ${f_{3}}$ we begin by noting that ${x^{2}}$ and ${x}$ are orthogonal (even versus odd again), so the Gram-Schmidt process reduces to

 $\displaystyle f_{3}$ $\displaystyle =$ $\displaystyle x^{2}-\langle f_{1}'|x^{2}\rangle|f_{1}'\rangle\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x^{2}-\frac{1}{3} \ \ \ \ \ (5)$

Normalizing this gives

$\displaystyle f_{3}'=\sqrt{\frac{5}{8}}(3x^{2}-1) \ \ \ \ \ (6)$

Finally, noting that ${x^{3}}$ is orthogonal to ${x^{2}}$ and constants, we have

 $\displaystyle f_{4}$ $\displaystyle =$ $\displaystyle x^{3}-\langle f_{2}'|x^{3}\rangle|f_{2}'\rangle\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x^{3}-\frac{3}{5}x \ \ \ \ \ (8)$

Normalizing gives

$\displaystyle f_{4}'=\sqrt{\frac{7}{8}}(5x^{3}-3x) \ \ \ \ \ (9)$

Apart from the normalization, these orthonormalized polynomials are the same as the Legendre polynomials.

# Laplace’s equation – charged line segment

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.38.

An example of using the general series solution to Laplace’s equation. We have a linear charge density ${\lambda}$ along the ${z}$ axis between ${z=-a}$ and ${z=+a}$. Find the potential for any position where ${r>a}$.

We can approach this problem using the same technique as we used for finding the potential due to a charged disk. We worked out the potential for points in the ${xy}$ plane earlier (see Example 2 in this earlier post). Using the coordinates suitable for the current problem, we have

 $\displaystyle V(r,\pi/2)$ $\displaystyle =$ $\displaystyle \frac{2\lambda}{4\pi\epsilon_{0}}\left[\ln\left(a+\sqrt{r^{2}+a^{2}}\right)-\ln r\right]\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\lambda}{2\pi\epsilon_{0}}\ln\left[\frac{a}{r}+\sqrt{1+\frac{a^{2}}{r^{2}}}\right] \ \ \ \ \ (2)$

We can expand the log term in powers of ${1/r}$ and get

$\displaystyle V(r,\pi/2)=\frac{\lambda}{2\pi\epsilon_{0}}\left[\frac{a}{r}-\frac{1}{6}\frac{a^{3}}{r^{3}}+\frac{3}{40}\frac{a^{5}}{r^{5}}+\ldots\right] \ \ \ \ \ (3)$

This series must match the general form of Laplace’s equation for ${r>a}$. This consists entirely of the terms in inverse powers of ${r}$, since we require the potential to be finite for large ${r}$. That is, we must have

$\displaystyle V(r,\theta)=\sum_{l=0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (4)$

By comparing the series, we see that terms in odd ${l}$ must vanish, so ${B_{l}=0}$ if ${l}$ is odd. The second series must match the earlier one for the special case of ${\theta=\pi/2}$, that is when ${\cos\theta=0}$. Looking up tables of the Legendre polynomials, we find for the first few:

 $\displaystyle P_{0}(0)$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (5)$ $\displaystyle P_{2}(0)$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\ \ \ \ \ (6)$ $\displaystyle P_{4}(0)$ $\displaystyle =$ $\displaystyle \frac{3}{8} \ \ \ \ \ (7)$

Therefore, we have

$\displaystyle V(r,\pi/2)=\frac{B_{0}}{r}-\frac{B_{2}}{2r^{3}}+\frac{3}{8}\frac{B_{4}}{r^{5}}+\ldots \ \ \ \ \ (8)$

Comparing the two series, we get

 $\displaystyle B_{0}$ $\displaystyle =$ $\displaystyle \frac{\lambda a}{2\pi\epsilon_{0}}\ \ \ \ \ (9)$ $\displaystyle B_{2}$ $\displaystyle =$ $\displaystyle \frac{\lambda}{2\pi\epsilon_{0}}\frac{a^{3}}{3}\ \ \ \ \ (10)$ $\displaystyle B_{4}$ $\displaystyle =$ $\displaystyle \frac{\lambda}{2\pi\epsilon_{0}}\frac{a^{5}}{5} \ \ \ \ \ (11)$

The general solution is then

 $\displaystyle V(r,\theta)$ $\displaystyle =$ $\displaystyle \frac{\lambda}{2\pi\epsilon_{0}}\left[\frac{a}{r}+\frac{a^{3}}{3r^{3}}P_{2}(\cos\theta)+\frac{a^{5}}{5r^{5}}P_{4}(\cos\theta)+\ldots\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\lambda}{2\pi\epsilon_{0}}\frac{a}{r}\left[1+\frac{a^{2}}{3r^{2}}P_{2}(\cos\theta)+\frac{a^{4}}{5r^{4}}P_{4}(\cos\theta)+\ldots\right] \ \ \ \ \ (13)$

We can write this in terms of the total charge in the line segment, which is ${Q=2a\lambda}$:

$\displaystyle V(r,\theta)=\frac{Q}{4\pi\epsilon_{0}r}\left[1+\frac{a^{2}}{3r^{2}}P_{2}(\cos\theta)+\frac{a^{4}}{5r^{4}}P_{4}(\cos\theta)+\ldots\right] \ \ \ \ \ (14)$

# 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

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

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

$\displaystyle V_{out}(r,\theta)=\sum_{l=0}^{\infty}\frac{C_{l}}{r^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (2)$

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

$\displaystyle \sum_{l=0}^{\infty}\left[A_{l}a^{l}+\frac{B_{l}}{a^{l+1}}\right]P_{l}(\cos\theta)=V_{0} \ \ \ \ \ (3)$

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

 $\displaystyle A_{0}+\frac{B_{0}}{a}$ $\displaystyle =$ $\displaystyle V_{0}\ \ \ \ \ (4)$ $\displaystyle B_{l}$ $\displaystyle =$ $\displaystyle -a^{2l+1}A_{l}\ \ \ (l>0) \ \ \ \ \ (5)$

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

$\displaystyle \sum_{l=0}^{\infty}\left[A_{l}b^{l}+\frac{B_{l}}{b^{l+1}}\right]P_{l}(\cos\theta)=\sum_{l=0}^{\infty}\frac{C_{l}}{r^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (6)$

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

$\displaystyle A_{l}+\frac{B_{l}}{b^{l+1}}=\frac{C_{l}}{b^{l+1}} \ \ \ \ \ (7)$

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

$\displaystyle A_{0}+\frac{B_{0}}{b}=\frac{C_{0}}{b} \ \ \ \ \ (8)$

and for ${l>0}$:

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

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

 $\displaystyle \left.\frac{\partial V}{\partial r}\right|_{out}-\left.\frac{\partial V}{\partial r}\right|_{in}$ $\displaystyle =$ $\displaystyle -\frac{\sigma}{\epsilon_{0}}\ \ \ \ \ (11)$ $\displaystyle \sum_{l=0}^{\infty}\left[-(l+1)\frac{C_{l}}{b^{l+2}}-lA_{l}b^{l-1}+(l+1)\frac{B_{l}}{b^{l+2}}\right]P_{l}(\cos\theta)$ $\displaystyle =$ $\displaystyle -\frac{k}{\epsilon_{0}}\cos\theta \ \ \ \ \ (12)$

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

 $\displaystyle -\frac{C_{0}}{b^{2}}+\frac{B_{0}}{b^{2}}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle -2\frac{C_{1}}{b^{3}}-A_{1}+2\frac{B_{1}}{b^{3}}$ $\displaystyle =$ $\displaystyle -\frac{k}{\epsilon_{0}}\ \ \ \ \ (14)$ $\displaystyle -(l+1)\frac{C_{l}}{b^{l+2}}-lA_{l}b^{l-1}+(l+1)\frac{B_{l}}{b^{l+2}}$ $\displaystyle =$ $\displaystyle 0\qquad l>1 \ \ \ \ \ (15)$

The first of these equations gives us

$\displaystyle B_{0}=C_{0} \ \ \ \ \ (16)$

Substituting this into 8 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:

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

This gives us, from the above relations

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle -\frac{a^{3}k}{3\epsilon_{0}}\ \ \ \ \ (19)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle \frac{k}{3\epsilon_{0}}\left(b^{3}-a^{3}\right) \ \ \ \ \ (20)$

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:

$\displaystyle V(r,\theta)=\begin{cases} \frac{aV_{0}}{r}+\frac{k}{3\epsilon_{0}}\left(r-\frac{a^{3}}{r^{2}}\right)\cos\theta & a\le r\le b\\ \frac{aV_{0}}{r}+\frac{k}{3\epsilon_{0}}\left(\frac{b^{3}-a^{3}}{r^{2}}\right)\cos\theta & r>b \end{cases} \ \ \ \ \ (21)$

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):

 $\displaystyle -\frac{\sigma}{\epsilon_{0}}$ $\displaystyle =$ $\displaystyle \left.\frac{\partial V}{\partial r}\right|_{r=a}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{a}+\frac{k}{\epsilon_{0}}\cos\theta\ \ \ \ \ (23)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{V_{0}\epsilon_{0}}{a}-k\cos\theta \ \ \ \ \ (24)$

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:

 $\displaystyle Q$ $\displaystyle =$ $\displaystyle 4\pi a^{2}\frac{V_{0}\epsilon_{0}}{a}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi\epsilon_{0}aV_{0} \ \ \ \ \ (26)$

For large distances, from the above formula

$\displaystyle V\rightarrow\frac{aV_{0}}{r} \ \ \ \ \ (27)$

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

 $\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{Q}{4\pi\epsilon_{0}r}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{aV_{0}}{r} \ \ \ \ \ (29)$

# Dipoles

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 3.4.2 & Problem 3.27-3.29

We’ve seen that the potential of a charge distribution can be written as a multipole expansion in terms of Legendre polynomials:

$\displaystyle V(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\int\rho(\mathbf{r}')P_{n}(\cos\theta')r'^{n}d^{3}\mathbf{r}' \ \ \ \ \ (1)$

where ${\mathbf{r}}$ is the vector from the origin to the observation point, ${\mathbf{r}'}$ is the vector from the origin to the point with charge density ${\rho(\mathbf{r}')}$ and ${\theta'}$ is the angle between ${\mathbf{r}}$ and ${\mathbf{r}'}$.

The ${n=0}$ term in this expansion is the monopole term, and the potential for that reduces to the equivalent potential for a point charge.

$\displaystyle V_{0}(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}r}\int\rho(\mathbf{r}')d^{3}\mathbf{r}'=\frac{Q}{4\pi\epsilon_{0}r} \ \ \ \ \ (2)$

The ${n=1}$ term is the dipole term, and dominates if the total charge is zero (that is, if there are equal amounts of positive and negative charge).

 $\displaystyle V_{1}(\mathbf{r})$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\int\rho(\mathbf{r}')P_{1}(\cos\theta')r'd^{3}\mathbf{r}'\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\int r'\rho(\mathbf{r}')\cos\theta'd^{3}\mathbf{r}' \ \ \ \ \ (4)$

Since ${\theta'}$ is the angle between ${\mathbf{r}}$ and ${\mathbf{r}'}$, we have ${\hat{\mathbf{r}}\cdot\mathbf{r}'=r'\cos\theta'}$, where ${\hat{\mathbf{r}}}$ is the unit vector in the ${\mathbf{r}}$ direction. Since this unit vector is fixed for a given ${\mathbf{r}}$, we can take it outside the integral and get

$\displaystyle V_{1}(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\cdot\int\mathbf{r}'\rho(\mathbf{r}')d^{3}\mathbf{r}' \ \ \ \ \ (5)$

The integral is now a property of the charge distribution alone, without any reference to ${\mathbf{r}}$ (although it does depend on the choice of origin). The integral is a vector and depends on ${\mathbf{r}'}$, which varies as you move around within the charge distribution. In practice, if the integrand doesn’t possess any symmetry that makes it easy to integrate, the only practical way of doing such an integral is to convert to rectangular coordinates, since the unit vectors in the rectangular system are constants and can be taken outside the integral.

The integral is called the dipole moment of the charge distribution, and is usually given the symbol ${\mathbf{p}}$:

 $\displaystyle \mathbf{p}$ $\displaystyle \equiv$ $\displaystyle \int\mathbf{r}'\rho(\mathbf{r}')d^{3}\mathbf{r}'\ \ \ \ \ (6)$ $\displaystyle V_{1}(\mathbf{r})$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\cdot\mathbf{p} \ \ \ \ \ (7)$

For ${n}$ point charges, the dipole moment can be written as

$\displaystyle \mathbf{p}=\sum_{i=1}^{n}q_{i}\mathbf{r}_{i}' \ \ \ \ \ (8)$

Example 1. We have four point charges as follows. Charge ${3q}$ placed at ${(x,y,z)=(0,0,a)}$, ${q}$ at ${(0,0,-a)}$, ${-2q}$ at ${(0,-a,0)}$ and ${-2q}$ at ${(0,a,0)}$. We can use 8 to work out the dipole moment.

 $\displaystyle \mathbf{p}$ $\displaystyle =$ $\displaystyle qa(3\hat{\mathbf{z}}-\hat{\mathbf{z}}-2\hat{\mathbf{y}}+2\hat{\mathbf{y}})\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2qa\hat{\mathbf{z}} \ \ \ \ \ (10)$

From here, we can get the dipole potential

$\displaystyle V_{1}(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}r^{2}}\mathbf{p}\cdot\hat{\mathbf{r}} \ \ \ \ \ (11)$

In spherical coordinates, since ${\mathbf{p}}$ points along the z axis, we have ${\mathbf{p}\cdot\hat{\mathbf{r}}=2qa\hat{\mathbf{z}}\cdot\hat{\mathbf{r}}=2qa\cos\theta}$, and

$\displaystyle V_{1}(\mathbf{r})=\frac{2qa\cos\theta}{4\pi\epsilon_{0}r^{2}} \ \ \ \ \ (12)$

Example 2. We have a spherical shell of radius ${R}$ centred at the origin with a surface charge density of ${\sigma=k\cos\theta}$ for a constant ${k}$ (in spherical coordinates), and we want to find the dipole moment.

We can use the symmetry of the situation to note that if we converted the integral to rectangular coordinates, then since the charge density depends only on ${\theta}$, a charge element at location ${(x,y,z)}$ will have an equal element at ${(-x,-y,z)}$, so the components of the dipole moment in the ${x}$ and ${y}$ directions will be zero. The component of ${\mathbf{r}'}$ in the ${z}$ direction is ${R\cos\theta}$, so the moment is

 $\displaystyle \mathbf{p}$ $\displaystyle =$ $\displaystyle k\hat{\mathbf{z}}\int_{0}^{2\pi}\int_{0}^{\pi}\left(\cos\theta'\right)(R\cos\theta')(R^{2}\sin\theta')d\theta'd\phi'\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{3}\pi kR^{3}\hat{\mathbf{z}} \ \ \ \ \ (14)$

The dipole component in the potential is therefore

 $\displaystyle V_{1}(\mathbf{r})$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\mathbf{p}\cdot\hat{\mathbf{r}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{kR^{3}}{3\epsilon_{0}r^{2}}\cos\theta \ \ \ \ \ (16)$

Referring back to 1 and using the fact that ${\sigma=k\cos\theta=kP_{1}(\cos\theta)}$, we can use the fact that the Legendre polynomials are orthogonal to conclude that this must also be the only non-zero term in the multipole expansion.

Example 3. The classic dipole consists of two equal and opposite charges a distance ${d}$ apart. To make it easier to analyze in spherical coordinates, we place the dipole with both charges on the ${z}$ axis, and measure the vector ${\mathbf{r}}$ from the midpoint of the distance between them. Thus we have a charge ${+q}$ at ${z=d/2}$ and ${-q}$ at ${z=-d/2}$. We can then define ${\mathbf{r}_{+}}$ and ${\mathbf{r}_{-}}$ to be the vectors from the two charges to the observation point. With the geometry as given, the angle ${\theta}$ is the angle between the ${z}$ axis and ${\mathbf{r}}$, so it’s the usual spherical angle ${\theta}$. It is also the angle between ${\mathbf{r}}$ and the vector to ${+q}$, so that ${\pi-\theta}$ is the angle between ${\mathbf{r}}$ and the vector to ${-q}$.

Using the law of cosines for the sides of a triangle, we get

 $\displaystyle r_{+}$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}+\frac{d^{2}}{4}-2r\frac{d}{2}\cos\theta}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r\sqrt{1-\frac{d}{r}\cos\theta+\left(\frac{d}{2r}\right)^{2}}\ \ \ \ \ (18)$ $\displaystyle r_{-}$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}+\frac{d^{2}}{4}-2r\frac{d}{2}\cos\left(\pi-\theta\right)}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r\sqrt{1+\frac{d}{r}\cos\theta+\left(\frac{d}{2r}\right)^{2}} \ \ \ \ \ (20)$

The potential at point ${\mathbf{r}}$ is therefore

$\displaystyle V(\mathbf{r})=\frac{q}{4\pi\epsilon_{0}}\left(\frac{1}{r_{+}}-\frac{1}{r_{-}}\right) \ \ \ \ \ (21)$

By substituting for ${r_{+}}$ and ${r_{-}}$, we can then use a Taylor expansion on the variable ${d/r}$ (which will be small if the charges are close together and we move far away from them) to get the multipole expansion. We get, for the first few terms:

 $\displaystyle V_{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (22)$ $\displaystyle V_{1}$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{\cos\theta}{r^{2}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{P_{1}(\theta)}{r^{2}}\ \ \ \ \ (24)$ $\displaystyle V_{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (25)$ $\displaystyle V_{3}$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{5\cos^{3}\theta-3\cos\theta}{8r^{4}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{P_{3}(\cos\theta)}{4r^{4}}\ \ \ \ \ (27)$ $\displaystyle V_{4}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (28)$ $\displaystyle V_{5}$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{63\cos^{5}(\theta)-70\cos^{3}\theta+15\cos\theta}{128r^{6}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{P_{5}(\cos\theta)}{16r^{6}} \ \ \ \ \ (30)$

Since ${r/r_{+}}$ is the generating function for the Legendre polynomials, and since ${\cos(\pi-\theta)=-\cos\theta}$, then ${r/r_{-}}$ is the generating function for Legendre polynomials with each term in the sum multiplied by ${(-1)^{n}}$. Therefore, since ${V}$ as given in 21 takes the difference between these two generating functions, all the terms involving even polynomials will cancel out, leaving only the odd terms, as we’ve seen in the first few explicit terms above.

For large enough ${r}$, the dipole term dominates, and we can say that for large distances, the potential of a dipole of two point charges goes as ${1/r^{2}}$.

# Multipole expansion in electrostatics

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 3.4.1 & Problem 3.26

For a given charge distribution, we can write down a multipole expansion, which gives the potential as a series in powers of ${1/r}$, where ${r}$ is the distance from the origin to the observation point.

We know that the potential in general is

$\displaystyle V(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}}\int\rho(\mathbf{r}')\frac{d^{3}\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|} \ \ \ \ \ (1)$

In the integral, ${\mathbf{r}'}$ is the position of charge element ${\rho(\mathbf{r}')d^{3}\mathbf{r}'}$. From the law of cosines

$\displaystyle |\mathbf{r}-\mathbf{r}'|=\sqrt{r^{2}+r'^{2}-2rr'\cos\theta'} \ \ \ \ \ (2)$

where ${\theta'}$ is the angle between ${\mathbf{r}}$ and ${\mathbf{r}'}$. We can rewrite this as

 $\displaystyle \frac{1}{|\mathbf{r}-\mathbf{r}'|}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{r^{2}+r'^{2}-2rr'\cos\theta'}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r}\frac{1}{\sqrt{1+\frac{r'^{2}}{r^{2}}-2\frac{r'}{r}\cos\theta'}} \ \ \ \ \ (4)$

From the theory of Legendre polynomials, it is known that the last factor in this expression is a generating function for the polynomials. That is, if we write the square root as an power series, we get

$\displaystyle \frac{1}{\sqrt{1+\frac{r'^{2}}{r^{2}}-2\frac{r'}{r}\cos\theta'}}=\sum_{n=0}^{\infty}P_{n}(\cos\theta')\left(\frac{r'}{r}\right)^{n} \ \ \ \ \ (5)$

The coefficient of ${\left(\frac{r'}{r}\right)^{n}}$ in the series is the Legendre polynomial ${P_{n}(\cos\theta')}$. This can be verified for the first few terms by calculating the Taylor series expansion of the square root term about ${r'/r=0}$. This is tedious to do by hand, but using Maple, we get, defining ${s\equiv r'/r}$:

$\displaystyle \frac{1}{\sqrt{1+s^{2}-2s\cos\theta'}}=1+s\cos\theta'+s^{2}\left(\frac{3}{2}\cos^{2}\theta'-\frac{1}{2}\right)+s^{3}\left(\frac{5}{2}\cos^{3}\theta'-\frac{3}{2}\cos\theta'\right)+\ldots \ \ \ \ \ (6)$

It is important to note that the angle ${\theta'}$ is equivalent to the angle ${\theta}$ in spherical coordinates only if the observation point ${\mathbf{r}}$ lies on the ${z}$ axis, since that is the only configuration where the angle between the observation vector and a charge element corresponds to the spherical coordinate angle ${\theta}$. (A more general multipole expansion uses spherical harmonics rather than just Legendre polynomials, but that’s a topic for a more advanced post.)

With this restriction, we can substitute the series expansion back into 1 to get

 $\displaystyle V(\mathbf{r})$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r}\int\rho(\mathbf{r}')\sum_{n=0}^{\infty}P_{n}(\cos\theta')\left(\frac{r'}{r}\right)^{n}d^{3}\mathbf{r}'\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\int\rho(\mathbf{r}')P_{n}(\cos\theta')r'^{n}d^{3}\mathbf{r}' \ \ \ \ \ (8)$

The first few terms in this series have special names. The ${n=0}$ term is

$\displaystyle \frac{1}{4\pi\epsilon_{0}r}\int\rho(\mathbf{r}')d^{3}\mathbf{r}'=\frac{Q}{4\pi\epsilon_{0}r} \ \ \ \ \ (9)$

where ${Q}$ is the total charge. This is called the monopole term, and shows that to a first approximation, the potential of any charge distribution is just the potential of a point charge with the same total charge.

The next term in the series is

$\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\int r'P_{1}(\cos\theta')\rho(\mathbf{r}')d^{3}\mathbf{r}'=\frac{1}{4\pi\epsilon_{0}r^{2}}\int r'\cos\theta'\rho(\mathbf{r}')d^{3}\mathbf{r}' \ \ \ \ \ (10)$

This is called the dipole term.

For ${n=2}$, we get the quadrupole term

$\displaystyle \frac{1}{4\pi\epsilon_{0}r^{3}}\int r'^{2}P_{2}(\cos\theta')\rho(\mathbf{r}')d^{3}\mathbf{r}'=\frac{1}{4\pi\epsilon_{0}r^{3}}\int r'^{2}\left(\frac{3}{2}\cos^{2}\theta'-\frac{1}{2}\right)\rho(\mathbf{r}')d^{3}\mathbf{r}' \ \ \ \ \ (11)$

Finally, for ${n=3}$ we get the octopole term

$\displaystyle \frac{1}{4\pi\epsilon_{0}r^{4}}\int r'^{3}P_{3}(\cos\theta')\rho(\mathbf{r}')d^{3}\mathbf{r}'=\frac{1}{4\pi\epsilon_{0}r^{4}}\int r'^{3}\left(\frac{5}{2}\cos^{3}\theta'-\frac{3}{2}\cos\theta'\right)\rho(\mathbf{r}')d^{3}\mathbf{r}' \ \ \ \ \ (12)$

As an example, consider a solid sphere with a charge density

$\displaystyle \rho(\mathbf{r}')=k\frac{R}{r'^{2}}(R-2r')\sin\theta' \ \ \ \ \ (13)$

We can use the integrals above to find the first non-zero term in the series, and thus get an approximation for the potential. Note that we can do this only for points on the z axis.

By direct calculation, we have for the monopole term:

 $\displaystyle \frac{1}{4\pi\epsilon_{0}r}\int\rho(\mathbf{r}')d^{3}\mathbf{r}'$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r}\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}k\frac{R}{r'^{2}}(R-2r')\sin\theta'r'^{2}\sin\theta'd\phi'd\theta'dr'\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

since the integral over ${r'}$ gives zero. Thus the monopole term vanishes, as it always does if the total charge is zero.

For the dipole term, we get

 $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\int r'\cos\theta'\rho(\mathbf{r}')d^{3}\mathbf{r}'$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{2}}\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}k\frac{R}{r'^{2}}(R-2r')r'^{3}\sin\theta'\cos\theta'\sin\theta'd\phi'd\theta'dr'\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (17)$

This time, the integral over ${\theta'}$ gives zero, since the term ${\cos\theta'\sin^{2}\theta'}$ is odd relative to the interval ${[0,\pi]}$.

 $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{3}}\int r'^{2}\left(\frac{3}{2}\cos^{2}\theta'-\frac{1}{2}\right)\rho(\mathbf{r}')d^{3}\mathbf{r}'$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{3}}\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}k\frac{R}{r'^{2}}(R-2r')\sin\theta'\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle r'^{4}\left(\frac{3}{2}\cos^{2}\theta'-\frac{1}{2}\right)\sin\theta'd\phi'd\theta'dr'\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}r^{3}}\left(\frac{\pi^{2}kR^{5}}{48}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\pi kR^{5}}{192\epsilon_{0}r^{3}} \ \ \ \ \ (20)$

The octopole term comes out to zero, since the terms in ${\theta'}$ are again odd relative to the interval ${[0,\pi]}$.

# Laplace’s equation – charged sphere

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.22.

We’ve seen one example of the relation between surface charge and potential using the series solution to Laplace’s equation. We consider a spherical shell of radius ${R}$ and wish to find the potential from the surface charge on this shell. The general solution to Laplace’s equation is

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

Inside the sphere, to prevent an infinity at the origin, ${B_{l}=0}$, while outside, to prevent infinities in the other direction, ${A_{l}=0}$. In the earlier post, we found that:

 $\displaystyle \frac{\sigma(\theta)}{\epsilon_{0}}$ $\displaystyle =$ $\displaystyle \frac{1}{2R}\sum_{l=0}^{\infty}\left(2l+1\right)^{2}C_{l}P_{l}(\cos\theta)\ \ \ \ \ (2)$ $\displaystyle C_{l}$ $\displaystyle \equiv$ $\displaystyle \frac{2}{2l+1}A_{l}R^{l} \ \ \ \ \ (3)$

Simplifying, we get

$\displaystyle \frac{\sigma(\theta)}{\epsilon_{0}}=\sum_{l=0}^{\infty}(2l+1)A_{l}R^{l-1}P_{l}(\cos\theta) \ \ \ \ \ (4)$

From here, we can use the orthogonality of the Legendre polynomials to get an expression for ${A_{l}}$. Multiply both sides by ${P_{m}(\cos\theta)\sin\theta}$ and integrate from 0 to ${\pi}$:

 $\displaystyle \frac{1}{\epsilon_{0}}\int_{0}^{\pi}\sigma(\theta)P_{m}(\cos\theta)\sin\theta d\theta$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}(2l+1)A_{l}R^{l-1}\int_{0}^{\pi}P_{m}(\cos\theta)P_{l}(\cos\theta)\sin\theta d\theta\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}(2l+1)A_{l}R^{l-1}\frac{2}{2m+1}\delta_{lm}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2R^{m-1}A_{m} \ \ \ \ \ (7)$

So, reverting back to ${l}$ as the index:

$\displaystyle A_{l}=\frac{1}{2R^{l-1}\epsilon_{0}}\int_{0}^{\pi}\sigma(\theta)P_{l}(\cos\theta)\sin\theta d\theta \ \ \ \ \ (8)$

As an example, consider a spherical shell with a uniform charge density of ${+\sigma_{0}}$ in the region ${0\le\theta<\pi/2}$ and ${-\sigma_{0}}$ from ${\pi/2<\theta\le\pi}$. We can use 8 to work out the ${A_{l}}$ explicitly. First, note that ${\sin\theta}$ is an even function over the interval ${[0,\pi]}$ (that is, ${\sin(\pi/2+x)=\sin(\pi/2-x)}$). Over the same interval, ${P_{l}(\cos\theta)}$ is even if ${l}$ is even and odd if ${l}$ is odd. The specified charge distribution is odd. (Another way is to convert the integral by the substitution ${x=\cos\theta}$ and then the functions are even or odd in the traditional way, with respect to the origin ${x=0}$ as the mid-point.)

Therefore, only odd ${l}$ terms will have a non-zero integral. For these terms, we need to work out

 $\displaystyle A_{l}$ $\displaystyle =$ $\displaystyle \frac{\sigma_{0}}{2R^{l-1}\epsilon_{0}}I_{l}\ \ \ \ \ (9)$ $\displaystyle I_{l}$ $\displaystyle \equiv$ $\displaystyle \int_{0}^{\pi/2}P_{l}(\cos\theta)\sin\theta d\theta-\int_{\pi/2}^{\pi}P_{l}(\cos\theta)\sin\theta d\theta \ \ \ \ \ (10)$

We can look up tables of Legendre polynomials and work out these integrals manually (they all involve powers of the cosine multiplied by a single sine, so the integrals are fairly straightforward, but tedious). Or we can use software to do the integrals, and we get

 $\displaystyle I_{0}$ $\displaystyle =$ $\displaystyle I_{2}=I_{4}=I_{6}=\ldots=0\ \ \ \ \ (11)$ $\displaystyle I_{1}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (12)$ $\displaystyle I_{3}$ $\displaystyle =$ $\displaystyle -\frac{1}{4}\ \ \ \ \ (13)$ $\displaystyle I_{5}$ $\displaystyle =$ $\displaystyle \frac{1}{8} \ \ \ \ \ (14)$

The potential, up to ${l=6}$, is therefore, for ${r:

$\displaystyle V(r,\theta)=\frac{\sigma_{0}}{2\epsilon_{0}}\left[rP_{1}(\cos\theta)-\frac{r^{3}}{4R^{2}}P_{3}(\cos\theta)+\frac{r^{5}}{8R^{4}}P_{5}(\cos\theta)+\ldots\right] \ \ \ \ \ (15)$

For ${r>R}$, we found in the earlier post that ${B_{l}=A_{l}R^{2l+1}}$ from the requirement that the potential be continuous at the boundary. Therefore

$\displaystyle B_{l}=\frac{\sigma_{0}}{2\epsilon_{0}}R^{l+2}I_{l} \ \ \ \ \ (16)$

so

$\displaystyle V(r,\theta)=\frac{\sigma_{0}}{2\epsilon_{0}}\left[\frac{R^{3}}{r^{2}}P_{1}(\cos\theta)-\frac{R^{5}}{4r^{4}}P_{3}(\cos\theta)+\frac{R^{7}}{8r^{6}}P_{5}(\cos\theta)+\ldots\right] \ \ \ \ \ (17)$

# Laplace’s equation – charged disk

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.21

In Example 3 in an earlier post, we found that the potential of a uniformly charged disk of radius ${R}$ can be worked out for points on the axis of the disk and is

$\displaystyle V(z)=\frac{\sigma}{2\epsilon_{0}}\left(\sqrt{z^{2}+R^{2}}-\left|z\right|\right) \ \ \ \ \ (1)$

In spherical coordinates, a point with coordinate ${z}$ on the axis has coordinates ${(r,0)}$ if ${z>0}$ or ${(r,\pi)}$ if ${z<0}$. We can therefore write

$\displaystyle V(r,0)=V(r,\pi)=\frac{\sigma}{2\epsilon_{0}}\left(\sqrt{r^{2}+R^{2}}-r\right) \ \ \ \ \ (2)$

This formula is valid only for points on the ${z}$ axis, but we can use it, together with the general solution of Laplace’s equation in terms of Legendre polynomials, to get an approximation for the potential off the ${z}$ axis.

We consider first the case of ${r>R}$. In this case the most general solution of Laplace’s equation in spherical coordinates is:

$\displaystyle V(r,\theta)=\sum_{l=0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (3)$

For ${\theta=0}$ we get, since ${P_{l}(0)=1}$ for all ${l}$:

$\displaystyle V(r,0)=\sum_{l=0}^{\infty}\frac{B_{l}}{r^{l+1}} \ \ \ \ \ (4)$

To find the coefficients ${B_{l}}$ we need to expand 2. We can expand it in powers of ${(R/r)}$:

 $\displaystyle V(r,0)$ $\displaystyle =$ $\displaystyle \frac{\sigma r}{2\epsilon_{0}}\left(\sqrt{1+\frac{R}{r^{2}}^{2}}-1\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma}{2\epsilon_{0}}\left(\frac{R^{2}}{2r}-\frac{R^{4}}{8r^{3}}+\ldots\right) \ \ \ \ \ (6)$

Comparing with 4, we get

 $\displaystyle B_{0}$ $\displaystyle =$ $\displaystyle \frac{\sigma R^{2}}{4\epsilon_{0}}\ \ \ \ \ (7)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle B_{2}$ $\displaystyle =$ $\displaystyle -\frac{\sigma R^{4}}{16\epsilon_{0}} \ \ \ \ \ (9)$

so the general formula for the potential off the axis is

$\displaystyle V(r,\theta)=\frac{\sigma}{2\epsilon_{0}}\left[\frac{R^{2}}{2r}P_{0}(\cos\theta)-\frac{R^{4}}{8r^{3}}P_{2}(\cos\theta)+\ldots\right] \ \ \ \ \ (10)$

Since the odd polynomial terms all vanish, and the even Legendre polynomials are even functions, this expansion is also valid for the region ${z<0}$, where ${\pi/2<\theta\le\pi}$.

For the case ${r, we have

$\displaystyle V(r,\theta)=\sum_{l=0}^{\infty}A_{l}r^{l}P_{l}(\cos\theta) \ \ \ \ \ (11)$

When ${\theta=0,}$ we get

$\displaystyle V(r,0)=\sum_{l=0}^{\infty}A_{l}r^{l} \ \ \ \ \ (12)$

We can now expand 2 in powers of ${r/R}$, and we get

 $\displaystyle V(r,0)$ $\displaystyle =$ $\displaystyle \frac{\sigma}{2\epsilon_{0}}\left(R\sqrt{1+\frac{r^{2}}{R^{2}}}-r\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma}{2\epsilon_{0}}\left(R-r+\frac{r^{2}}{2R}-\frac{r^{4}}{8R^{3}}+\ldots\right) \ \ \ \ \ (14)$

Comparing this to 12, we get

 $\displaystyle A_{0}$ $\displaystyle =$ $\displaystyle \frac{\sigma R}{2\epsilon_{0}}\ \ \ \ \ (15)$ $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle -\frac{\sigma}{2\epsilon_{0}}\ \ \ \ \ (16)$ $\displaystyle A_{2}$ $\displaystyle =$ $\displaystyle \frac{\sigma}{4\epsilon_{0}R}\ \ \ \ \ (17)$ $\displaystyle A_{3}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (18)$

The off-axis expansion thus begins:

$\displaystyle V(r,\theta)=\frac{\sigma}{2\epsilon_{0}}\left[RP_{0}(\cos\theta)-rP_{1}(\cos\theta)+\frac{r^{2}}{2R}P_{2}(\cos\theta)+\ldots\right] \ \ \ \ \ (19)$

In this case, since there is a ${P_{1}}$ term, the same expansion isn’t valid for ${\theta=\pi}$. However, because the odd polynomials are odd functions, for ${\theta=\pi}$ we have

$\displaystyle V(r,\pi)=\sum_{l=0}^{\infty}(-1)^{l}A_{l}r^{l} \ \ \ \ \ (20)$

The only change we need to make in the expansion is thus to change the sign of ${A_{1}}$ so we get

$\displaystyle V(r,\theta)=\frac{\sigma}{2\epsilon_{0}}\left[RP_{0}(\cos\theta)+rP_{1}(\cos\theta)+\frac{r^{2}}{2R}P_{2}(\cos\theta)+\ldots\right] \ \ \ \ \ (21)$

All higher odd polynomials have ${A_{l}=0}$ so this is the only change that is required in the entire expansion.

# Laplace’s equation in spherical coordinates: surface charge

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 3.19, 3.18.

We found the general solution of Laplace’s equation in spherical coordinates to be:

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

If the potential is specified as ${V_{0}(\theta)}$ on the surface of a sphere, and there are no charges inside or outside the sphere, we can use the relations between the potential and the surface charge density to find that charge density on the sphere. Since the electric field is discontinuous across a charge layer, we have

$\displaystyle E_{\perp}^{above}-E_{\perp}^{below}=\frac{\sigma}{\epsilon_{0}} \ \ \ \ \ (2)$

and the field is related to the potential by

$\displaystyle \mathbf{E}=-\nabla V \ \ \ \ \ (3)$

we can use the radial symmetry of the problem to deduce that, at the surface of the sphere where ${r=R}$,

$\displaystyle \left.\frac{\partial V}{\partial r}\right|_{out}-\left.\frac{\partial V}{\partial r}\right|_{in}=-\frac{\sigma}{\epsilon_{0}} \ \ \ \ \ (4)$

For this problem, we know that inside the sphere, ${B_{l}=0}$ and outside the sphere, ${A_{l}=0}$. So we have

 $\displaystyle \left.\frac{\partial V}{\partial r}\right|_{in}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}lA_{l}r^{l-1}P_{l}(\cos\theta)\ \ \ \ \ (5)$ $\displaystyle \left.\frac{\partial V}{\partial r}\right|_{out}$ $\displaystyle =$ $\displaystyle -\sum_{l=0}^{\infty}(l+1)\frac{B_{l}}{r^{l+2}}P_{l}(\cos\theta) \ \ \ \ \ (6)$

At the boundary, we must have

$\displaystyle \sum_{l=0}^{\infty}(l+1)\frac{B_{l}}{R^{l+2}}P_{l}(\cos\theta)+\sum_{l=0}^{\infty}lA_{l}R^{l-1}P_{l}(\cos\theta)=\frac{\sigma(\theta)}{\epsilon_{0}} \ \ \ \ \ (7)$

However, since the potential is also continuous across a charge layer, we must have

$\displaystyle V_{0}(\theta)=\sum_{l=0}^{\infty}A_{l}R^{l}P_{l}(\cos\theta)=\sum_{l=0}^{\infty}\frac{B_{l}}{R^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (8)$

Since the Legendre polynomials are orthogonal, these two series must be equal term by term, so we get

$\displaystyle A_{l}R^{l}=\frac{B_{l}}{R^{l+1}} \ \ \ \ \ (9)$

Multiplying 8 both sides by ${P_{m}(\cos\theta)\sin\theta}$ and integrating from 0 to ${\pi}$, we get, using the orthogonality of the polynomials

 $\displaystyle \int_{0}^{\pi}V_{0}(\theta)P_{m}(\cos\theta)\sin\theta d\theta$ $\displaystyle =$ $\displaystyle \frac{2}{2m+1}A_{m}R^{m}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle C_{m} \ \ \ \ \ (11)$

Putting this back into 7 we get

 $\displaystyle \frac{\sigma(\theta)}{\epsilon_{0}}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\left[\frac{2l+1}{2R}(l+1)+\frac{2l+1}{2R}l\right]C_{l}P_{l}(\cos\theta)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =\frac{1}{2R}$ $\displaystyle \sum_{l=0}^{\infty}\left(2l+1\right)^{2}C_{l}P_{l}(\cos\theta) \ \ \ \ \ (13)$

We can apply this result to find the charge distribution for the case where ${V_{0}=k\cos3\theta}$. By using trig identities, we can expand the cosine and express ${V_{0}}$ in terms of the ${P_{l}}$:

$\displaystyle V_{0}(\theta)=\frac{k}{5}\left(-3P_{1}(\cos\theta)+8P_{3}(\cos\theta)\right) \ \ \ \ \ (14)$

From here, we see that ${\sigma}$ has contributions only from ${l=1}$ and ${l=3}$, and we get from the above formula for the ${C_{l}}$ coefficients:

 $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle -\frac{3k}{15}\ \ \ \ \ (15)$ $\displaystyle C_{3}$ $\displaystyle =$ $\displaystyle \frac{8k}{35}\ \ \ \ \ (16)$ $\displaystyle \sigma(\theta)$ $\displaystyle =$ $\displaystyle \frac{k\epsilon_{0}}{R}\left[-\frac{9}{5}P_{1}(\cos\theta)+\frac{56}{5}P_{3}(\cos\theta)\right] \ \ \ \ \ (17)$

# Laplace’s equation in spherical coordinates: examples 1

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 3.17, 3.18.

Example 1. A simple example of Laplace’s equation in spherical coordinates is that of a spherical shell of radius ${R}$ with a constant potential ${V_{0}}$ over its surface. We want to find the potential inside and outside the sphere.

The general solution in spherical coordinates was found in the last post:

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

Inside the sphere, all ${B_{l}}$ are zero to prevent an infinity at the origin, so we get

$\displaystyle V(r,\theta)=\sum_{l=0}^{\infty}A_{l}r^{l}P_{l}(\cos\theta) \ \ \ \ \ (2)$

At the spherical boundary, ${r=R}$ so we get

$\displaystyle V_{0}=\sum_{l=0}^{\infty}A_{l}R^{l}P_{l}(\cos\theta) \ \ \ \ \ (3)$

The only Legendre polynomial that doesn’t depend on ${\theta}$ is ${P_{0}=1}$, so it is only the ${l=0}$ term that is non-zero, and we get ${A_{0}=V_{0}}$, so inside the sphere, ${V=V_{0}}$ everywhere.

Outside the sphere, all ${A_{l}=0}$ to prevent the potential increasing to infinity for large ${r}$. Again, the potential must satisfy the boundary condition at ${r=R}$, and we get

$\displaystyle V_{0}=\sum_{l=0}^{\infty}\frac{B_{l}}{R^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (4)$

As above, we discard all terms except for ${l=0}$, which gives ${B_{0}=RV_{0}}$, and

$\displaystyle V=\frac{RV_{0}}{r} \ \ \ \ \ (5)$

Example 2. We saw that the coefficients ${A_{l}}$ and ${B_{l}}$ can be found by working out integrals, but in some special cases, it is easier to match up terms in the series on both sides of the equation. This happens if we can express ${V}$ as a series of cosines (admittedly, this doesn’t happen very often, but they are popular student exercises).

For example, suppose we have a spherical shell of radius ${R}$ on which the potential is ${V(\theta)=k\cos3\theta}$. Using some trig identities, we can convert the cosine term.

 $\displaystyle \cos3\theta$ $\displaystyle =$ $\displaystyle \cos(2\theta+\theta)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos2\theta\cos\theta-\sin2\theta\sin\theta\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle (2\cos^{2}\theta-1)\cos\theta-2\sin^{2}\theta\cos\theta\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\cos^{3}\theta-\cos\theta-2(1-\cos^{2}\theta)\cos\theta\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\cos^{3}\theta-3\cos\theta \ \ \ \ \ (10)$

We can now apply this to the general solution 1. Inside the sphere, the ${B_{l}}$ terms are all zero to prevent an infinity at the origin, so we get at the boundary:

 $\displaystyle V(R,\theta)$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}A_{l}R^{l}P_{l}(\cos\theta)\ \ \ \ \ (11)$ $\displaystyle k\left(4\cos^{3}\theta-3\cos\theta\right)$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}A_{l}R^{l}P_{l}(\cos\theta) \ \ \ \ \ (12)$

Since the only terms appearing in the potential are of degree 1 and 3, only ${P_{1}}$ and ${P_{3}}$ appear in the series on the right. From tables of Legendre polynomials, we have

 $\displaystyle P_{1}(\cos\theta)$ $\displaystyle =$ $\displaystyle \cos\theta\ \ \ \ \ (13)$ $\displaystyle P_{3}(\cos\theta)$ $\displaystyle =$ $\displaystyle \frac{5}{2}\cos^{3}\theta-\frac{3}{2}\cos\theta \ \ \ \ \ (14)$

Matching up terms for the 3rd degree term, we get

 $\displaystyle 4k$ $\displaystyle =$ $\displaystyle \frac{5}{2}A_{3}R^{3}\ \ \ \ \ (15)$ $\displaystyle A_{3}$ $\displaystyle =$ $\displaystyle \frac{8k}{5R^{3}} \ \ \ \ \ (16)$

With this value of ${A_{3}}$, the ${l=3}$ term in the series contributes a term ${-\frac{12}{5}k\cos\theta}$, so combining this with the ${l=1}$ term and equating this to the degree 1 term on the LHS, we get

 $\displaystyle -3k$ $\displaystyle =$ $\displaystyle A_{1}R-\frac{12}{5}k\ \ \ \ \ (17)$ $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle -\frac{3k}{5R} \ \ \ \ \ (18)$

The potential inside the sphere is thus:

$\displaystyle V_{in}(r,\theta)=\frac{k}{5}\left(-3\frac{r}{R}P_{1}(\cos\theta)+8\frac{r^{3}}{R^{3}}P_{3}(\cos\theta)\right) \ \ \ \ \ (19)$

Outside the sphere, we can use the technique, except this time it is the ${A_{l}}$ terms that are all zero to avoid an infinite potential for large ${r}$. We get, at the boundary:

 $\displaystyle V(R,\theta)$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\frac{B_{l}}{R^{l+1}}P_{l}(\cos\theta)\ \ \ \ \ (20)$ $\displaystyle k\left(4\cos^{3}\theta-3\cos\theta\right)$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\frac{B_{l}}{R^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (21)$

For the 3rd degree term:

 $\displaystyle 4k$ $\displaystyle =$ $\displaystyle \frac{5B_{3}}{2R^{4}}\ \ \ \ \ (22)$ $\displaystyle B_{3}$ $\displaystyle =$ $\displaystyle \frac{8kR^{4}}{5} \ \ \ \ \ (23)$

The ${l=3}$ term contributes a term of ${-\frac{12}{5}k\cos\theta}$ as before, so combining this with the ${l=1}$ term, we get

 $\displaystyle -3k$ $\displaystyle =$ $\displaystyle \frac{B_{1}}{R^{2}}-\frac{12}{5}k\ \ \ \ \ (24)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle -\frac{3kR^{2}}{5} \ \ \ \ \ (25)$

The outside potential is

$\displaystyle V_{out}(r,\theta)=\frac{k}{5}\left(-\frac{3R^{2}}{r^{2}}P_{1}(\cos\theta)+\frac{8R^{4}}{r^{4}}P_{3}(\cos\theta)\right) \ \ \ \ \ (26)$