# Electric field outside an infinite wire

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 7.57.

The electric field inside a resistor with a constant cross-section is given in Griffiths’s Example 7.1 as

$\displaystyle \mathbf{E}=\frac{I\rho}{A}\hat{\mathbf{z}} \ \ \ \ \ (1)$

where in this case ${\rho}$ is the resistivity (not the charge density!), ${I}$ is the current and ${A}$ is the cross-sectional area of the resistor. If we consider the simple case where the cross section is circular and make the resistor very long (so we can call it a wire with radius ${a}$), then

$\displaystyle \mathbf{E}=\frac{I\rho}{\pi a^{2}}\hat{\mathbf{z}} \ \ \ \ \ (2)$

Surprisingly, the problem of finding the field outside the wire is not completely solved. If we assume that the current returns through a superconducting coaxial cylinder of radius ${b}$ around the wire, and that the magnetic field is constant in time, so that ${\nabla\times\mathbf{E}=0}$ and thus ${\mathbf{E}=-\nabla V}$, then we can apply Laplace’s equation in cylindrical coordinates to find the potential. In order to solve it, we need boundary conditions. Since the coaxial cylinder is a perfect conductor, ${V=0}$ there. On the surface of the inner wire, we have

$\displaystyle V\left(a,z\right)=-\frac{I\rho z}{\pi a^{2}} \ \ \ \ \ (3)$

The problem is, what is ${V}$ in between the two cylinders? The solution we derived earlier was for the case where ${V}$ was independent of ${z}$:

$\displaystyle V(r,\phi)=A\ln r+B+\sum_{n=1}^{\infty}r^{n}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right)+\sum_{n=-\infty}^{-1}r^{n}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (4)$

However, in the special case where ${V\left(r,z\right)=zf\left(r\right)}$, this solution also works for finding ${f}$ (since using separation of variables on all three cylindrical variables separates the ${z}$ from the rest of ${V}$). Since there is no ${\phi}$ dependence, both sums disappear and we are left with

 $\displaystyle V\left(r=b\right)$ $\displaystyle =$ $\displaystyle 0=A\ln b+B\ \ \ \ \ (5)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle -A\ln b \ \ \ \ \ (6)$

On the wire:

 $\displaystyle V\left(a\right)$ $\displaystyle =$ $\displaystyle -\frac{I\rho z}{\pi a^{2}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\ln a+B\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\ln\frac{a}{b}\ \ \ \ \ (9)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle -\frac{I\rho z}{\pi a^{2}\ln\left(a/b\right)} \ \ \ \ \ (10)$

So

$\displaystyle V\left(r,z\right)=-\frac{I\rho z\ln\left(r/b\right)}{\pi a^{2}\ln\left(a/b\right)} \ \ \ \ \ (11)$

The field is then

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{I\rho}{\pi a^{2}\ln\left(a/b\right)}\left[\frac{z}{r}\hat{\mathbf{r}}+\ln\frac{r}{b}\hat{\mathbf{z}}\right] \ \ \ \ \ (13)$

The surface charge density on the wire is found from the difference in radial components of the field. Inside the wire there is no radial component, so

$\displaystyle \sigma=\epsilon_{0}\Delta E_{r}=\frac{I\rho\epsilon_{0}z}{\pi a^{3}\ln\left(a/b\right)} \ \ \ \ \ (14)$

As Griffiths remarks, the results for the field and surface charge are ‘peculiar’ (I would say ‘wrong’) since they depend on ${z}$, which shouldn’t be the case for an infinite wire with no variation along its length. It is surely impossible since the results depend on where we set the origin, which is completely arbitrary.

# Electric potential from a steady current

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 7.41.

Mark A. Heald (1984), American Journal of Physics, 52, 522.

Ohm’s law is derived from the assumption that in a conducting medium ${\mathbf{J}=\sigma\mathbf{E}}$. If the currents are steady, then the charge density is independent of time, so ${\nabla\cdot\mathbf{J}=\frac{d\rho}{dt}=0}$, which implies that (if ${\sigma}$ is constant throughout the material) ${\nabla\cdot\mathbf{E}=-\nabla^{2}V=0}$, and we can use the various methods we derived earlier to solve Laplace’s equation and determine the potential. One example of this is given here.

We have a cylindrical pipe of conductor of radius ${a}$, with a very thin slot cut down the side. A battery is attached to either side of the slot so that one edge of the slot is maintained at potential ${+V_{0}/2}$ and the other at ${-V_{0}/2}$. This results in a uniform, steady current flowing around the circumference of the pipe. Since the conductivity of the pipe is uniform, the potential at an angle ${\phi}$ on the surface of the pipe is

$\displaystyle V\left(a,\phi\right)=\frac{V_{0}}{2\pi}\phi \ \ \ \ \ (1)$

for ${-\pi<\phi<+\pi}$. Our job is to find the potential both inside and outside the cylinder.

We can approach this problem using the series solution to Laplace’s equation in cylindrical coordinates that we worked out earlier:

$\displaystyle V(r,\phi)=A\ln r+B+\sum_{n=1}^{\infty}r^{n}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right)+\sum_{n=1}^{\infty}\frac{1}{r^{n}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (2)$

with the boundary condition given above.

Inside the cylinder, to keep ${V}$ finite we must throw away the log term and the series in inverse powers of ${r}$, so inside we have

$\displaystyle V_{r

Similarly, outside we get

$\displaystyle V_{r>a}=B_{o}+\sum_{n=1}^{\infty}\frac{1}{r^{n}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (4)$

At the boundary, we must have a continuous potential satisfying the boundary condition, so

$\displaystyle B_{i}+\sum_{n=1}^{\infty}a^{n}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right)=\frac{V_{0}}{2\pi}\phi=B_{o}+\sum_{n=1}^{\infty}\frac{1}{a^{n}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (5)$

At this point, we would like to be able to equate the coefficients of the sin and cos terms, but the boundary condition depends on ${\phi}$ and not explicitly on sines or cosines. However, one of the nice things about Fourier series (which these series are, really) is that we can express any function in terms of them, so we can try to find a series such that

$\displaystyle \frac{V_{0}}{2\pi}\phi=\sum_{n=1}^{\infty}F_{n}\sin n\phi \ \ \ \ \ (6)$

We can use the usual procedure for finding ${F_{n}}$: multiply both sides by ${\sin m\phi}$ and integrate:

$\displaystyle \frac{V_{0}}{2\pi}\int_{-\pi}^{\pi}\phi\sin m\phi d\phi=\sum_{n=1}^{\infty}F_{n}\int_{-\pi}^{\pi}\sin m\phi\sin n\phi d\phi \ \ \ \ \ (7)$

Since the sine function is orthogonal on the interval of integration, we get

$\displaystyle \frac{V_{0}}{2\pi}\int_{-\pi}^{\pi}\phi\sin m\phi d\phi=F_{m}\int_{-\pi}^{\pi}\sin^{2}m\phi d\phi \ \ \ \ \ (8)$

Working out the integrals, we get

$\displaystyle \frac{\left(-1\right)^{m+1}V_{0}}{m}=\pi F_{m} \ \ \ \ \ (9)$

so

$\displaystyle \frac{V_{0}}{2\pi}\phi=\frac{V_{0}}{\pi}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}\sin n\phi \ \ \ \ \ (10)$

Now we can equate coefficients, so we get for ${r

 $\displaystyle B_{i}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle B_{n}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle A_{n}$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{n+1}}{na^{n}}\ \ \ \ \ (13)$ $\displaystyle V_{r $\displaystyle =$ $\displaystyle -\frac{V_{0}}{\pi}\sum_{n=1}^{\infty}\left(-\frac{r}{a}\right)^{n}\frac{\sin n\phi}{n} \ \ \ \ \ (14)$

Similar reasoning on the outside gives

 $\displaystyle B_{o}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (15)$ $\displaystyle D_{n}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (16)$ $\displaystyle C_{n}$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{n+1}a^{n}}{n}\ \ \ \ \ (17)$ $\displaystyle V_{r>a}$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{\pi}\sum_{n=1}^{\infty}\left(-\frac{a}{r}\right)^{n}\frac{\sin n\phi}{n} \ \ \ \ \ (18)$

To get the answer given in Griffiths, we need to sum the series. Maple is able to do this explicitly, and we get

 $\displaystyle V_{r>a}$ $\displaystyle =$ $\displaystyle \frac{V_{0}}{\pi}\arctan\left(\frac{\frac{a}{r}\sin\phi}{1+\frac{a}{r}\cos\phi}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V_{0}}{\pi}\arctan\left(\frac{a\sin\phi}{r+a\cos\phi}\right) \ \ \ \ \ (20)$

The potential inside the cylinder is the same except with ${a}$ and ${r}$ swapped, so we have:

$\displaystyle V_{r

We can check that these formulas do indeed satisfy the boundary condition using a couple of trig identities. Using double angle formulas we have

 $\displaystyle \sin\phi$ $\displaystyle =$ $\displaystyle 2\sin\frac{\phi}{2}\cos\frac{\phi}{2}\ \ \ \ \ (22)$ $\displaystyle \cos\phi$ $\displaystyle =$ $\displaystyle 2\cos^{2}\frac{\phi}{2}-1 \ \ \ \ \ (23)$

so at ${r=a}$, we have

 $\displaystyle V\left(a,\phi\right)$ $\displaystyle =$ $\displaystyle \frac{V_{0}}{\pi}\arctan\left(\frac{2a\sin\frac{\phi}{2}\cos\frac{\phi}{2}}{a\left(1+2\cos^{2}\frac{\phi}{2}-1\right)}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V_{0}}{\pi}\arctan\frac{\sin\frac{\phi}{2}}{\cos\frac{\phi}{2}}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V_{0}}{2\pi}\phi \ \ \ \ \ (26)$

The surface charge density on the cylinder can be found from the radial component of the electric field at the surface, which is

 $\displaystyle E_{r}$ $\displaystyle =$ $\displaystyle -\left.\frac{\partial V}{\partial r}\right|_{r=a}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm\frac{V_{0}\sin\phi}{2\pi a\left(1+\cos\phi\right)} \ \ \ \ \ (28)$

with the plus sign for the inner surface and the minus sign for the outer surface.

The surface charge density is ${\epsilon_{0}E_{r}}$, or

$\displaystyle \sigma=\pm\frac{\epsilon_{0}V_{0}\sin\phi}{2\pi a\left(1+\cos\phi\right)} \ \ \ \ \ (29)$

(The answer given in Griffiths’s question is clearly wrong; the charge density does not depend on ${r}$ (what Griffiths calls ${s}$), and this answer agrees with that given in Heald’s paper.)

# Magnetostatic boundary conditions; Laplace’s equation

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.15.

From the magnetostatic boundary conditions on the magnetic field ${\mathbf{B}}$ we can work out the boundary conditions on the auxiliary field ${\mathbf{H}}$. First, we need the divergence of ${\mathbf{H}}$:

$\displaystyle \nabla\cdot\mathbf{H}=\frac{1}{\mu_{0}}\nabla\cdot\mathbf{B}-\nabla\cdot\mathbf{M}=-\nabla\cdot\mathbf{M} \ \ \ \ \ (1)$

Using a similar argument to that for the electric field, we can see that ${\mathbf{H}}$ and ${\mathbf{M}}$ have equal and opposite discontinuites at a boundary, so

$\displaystyle H_{\perp}^{above}-H_{\perp}^{below}=-\left(M_{\perp}^{above}-M_{\perp}^{below}\right) \ \ \ \ \ (2)$

Using the same argument as for magnetostatic boundary conditions, the boundary condition on the component of ${\mathbf{H}}$ parallel to the boundary is

$\displaystyle \mathbf{H}_{\parallel}^{above}-\mathbf{H}_{\parallel}^{below}=\mathbf{K}_{f}\times\hat{\mathbf{n}} \ \ \ \ \ (3)$

where ${\mathbf{K}_{f}}$ is the free surface current density on the boundary.

The quantity ${\mathbf{H}_{\parallel}}$ is written as a vector since it lies in the tangent plane to the surface and has a direction given by the cross product ${\mathbf{K}_{f}\times\hat{\mathbf{n}}}$.

In the special case where there is no free current anywhere, then

$\displaystyle \nabla\times\mathbf{H}=\mathbf{J}_{f}=0 \ \ \ \ \ (4)$

which means that the vector field ${\mathbf{H}}$ can be written as the gradient of a scalar field ${W}$:

$\displaystyle \mathbf{H}=-\nabla W \ \ \ \ \ (5)$

As an example of using this, we can derive the field due to a uniformly magnetized sphere. The magnetization is ${\mathbf{M}=M\hat{\mathbf{z}}}$ within the sphere and zero outside. Since ${\mathbf{M}}$ is constant everywhere except at the boundary, then

$\displaystyle \nabla\cdot\mathbf{H}=-\nabla^{2}W=0 \ \ \ \ \ (6)$

everywhere except at the boundary. This is Laplace’s equation and in spherical coordinates, the general solution is

$\displaystyle W(r,\theta)=\sum_{l=0}^{\infty}\left(A_{l}r^{l}+\frac{B_{l}}{r^{l+1}}\right)P_{l}(\cos\theta) \ \ \ \ \ (7)$

We can follow the usual procedure for finding the coefficients by imposing boundary conditions on ${W}$. Just as in the electrostatic case, the potential must be continuous at the boundary, and must be finite everywhere. This means that inside the sphere, we must have ${B_{l}=0}$ and outside the sphere ${A_{l}=0}$ so that

 $\displaystyle W_{in}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}A_{l}r^{l}P_{l}(\cos\theta)\ \ \ \ \ (8)$ $\displaystyle W_{out}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}(\cos\theta) \ \ \ \ \ (9)$

At the boundary we must have ${W_{in}=W_{out}}$ and equating coefficients of each Legendre polynomial we get

$\displaystyle B_{l}=A_{l}R^{2l+1} \ \ \ \ \ (10)$

We can now consider the derivative of ${W}$ in the ${r}$ direction, which gives the normal component of ${\mathbf{H}}$ as ${-\partial W/\partial r}$. Using 2 we have

 $\displaystyle \left.\frac{\partial W}{\partial r}\right|_{out}-\left.\frac{\partial W}{\partial r}\right|_{in}$ $\displaystyle =$ $\displaystyle M_{\perp}^{above}-M_{\perp}^{below}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -M\hat{\mathbf{r}}\cdot\hat{\mathbf{z}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -M\cos\theta \ \ \ \ \ (13)$

Taking the derivatives of the series above, we get for ${l=1}$ after using 10

 $\displaystyle -\frac{l+1}{R^{l+2}}B_{l}$ $\displaystyle =$ $\displaystyle lA_{l}R^{l-1}-M\ \ \ \ \ (14)$ $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle \frac{M}{3}\ \ \ \ \ (15)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{M}{3}R^{3} \ \ \ \ \ (16)$

For ${l\ne1}$ we have

 $\displaystyle -\frac{l+1}{R^{l+2}}B_{l}$ $\displaystyle =$ $\displaystyle lA_{l}R^{l-1}\ \ \ \ \ (17)$ $\displaystyle -\left(l+1\right)R^{l-1}A_{l}$ $\displaystyle =$ $\displaystyle lA_{l}R^{l-1}\ \ \ \ \ (18)$ $\displaystyle \left(2l+1\right)A_{l}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (19)$

Since this must be true for all ${l\ne1}$ we must have ${A_{l}=B_{l}=0}$ for these cases. Thus

 $\displaystyle W_{in}$ $\displaystyle =$ $\displaystyle \frac{M}{3}r\cos\theta\ \ \ \ \ (20)$ $\displaystyle W_{out}$ $\displaystyle =$ $\displaystyle \frac{M}{3}\frac{R^{3}}{r^{2}}\cos\theta \ \ \ \ \ (21)$

Taking the negative gradient, we get, since ${r\cos\theta=z}$

 $\displaystyle \mathbf{H}_{in}$ $\displaystyle =$ $\displaystyle -\frac{M}{3}\hat{\mathbf{z}}\ \ \ \ \ (22)$ $\displaystyle \mathbf{H}_{out}$ $\displaystyle =$ $\displaystyle \frac{MR^{3}}{3r^{3}}\left(2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\boldsymbol{\theta}}\right) \ \ \ \ \ (23)$

From this we can get the field ${\mathbf{B}=\mu_{0}\left(\mathbf{H}+\mathbf{M}\right)}$ (remember ${\mathbf{M}=0}$ outside the sphere):

 $\displaystyle \mathbf{B}_{in}$ $\displaystyle =$ $\displaystyle \frac{2\mu_{0}M}{3}\hat{\mathbf{z}}\ \ \ \ \ (24)$ $\displaystyle \mathbf{B}_{out}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}MR^{3}}{3r^{3}}\left(2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\boldsymbol{\theta}}\right) \ \ \ \ \ (25)$

# Dipole in dielectric sphere

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.34.

Another example of solving a boundary value problem in a system with linear dielectrics. Suppose we have an ideal dipole ${\mathbf{p}}$ at the centre of a sphere of dielectric (dielectric constant ${\epsilon_{r}}$) and radius ${R}$. What is the potential at any point (inside or outside the sphere)?

We can use a similar approach to that of the problem of a dielectric cylinder in an electric field. We first specify the boundary conditions. At the surface of the sphere, the potential must be continuous, so we have

$\displaystyle V_{in}(R)=V_{out}(R) \ \ \ \ \ (1)$

As we saw in the cylinder problem, the condition on the normal derivative of the potential is, since on the outside of the sphere, ${\epsilon=\epsilon_{0}}$:

$\displaystyle \epsilon\left.\frac{\partial V_{in}}{\partial r}\right|_{r=R}=\epsilon_{0}\left.\frac{\partial V_{out}}{\partial r}\right|_{r=R} \ \ \ \ \ (2)$

Since there is no external field, we must have ${V\rightarrow0}$ as ${r\rightarrow\infty}$. Finally, as ${r\rightarrow0}$, the potential must behave like that of an ideal dipole, so, assuming that ${\mathbf{p}}$ points in the ${+z}$ direction:

$\displaystyle \lim_{r\rightarrow0}V_{in}(r)=\frac{p\cos\theta}{4\pi\epsilon r^{2}} \ \ \ \ \ (3)$

Note that we’re using ${\epsilon}$ rather than ${\epsilon_{0}}$ in this formula, since the dipole is inside the dielectric and the potential is reduced by a factor of ${\epsilon_{r}}$, where ${\epsilon=\epsilon_{0}\epsilon_{r}}$.

With these conditions, we must solve Laplace’s equation in spherical coordinates, so we can quote the general form of the solution:

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

Inside the sphere, we have

$\displaystyle V_{in}=\frac{p\cos\theta}{4\pi\epsilon r^{2}}+\sum_{l=0}^{\infty}A_{l}r^{l}P_{l}\left(\cos\theta\right) \ \ \ \ \ (5)$

Outside, we have

$\displaystyle V_{out}=\sum_{l=0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}\left(\cos\theta\right) \ \ \ \ \ (6)$

Applying the continuity condition 1 and equating coefficients of the ${P_{l}}$, we get

 $\displaystyle \frac{p}{4\pi\epsilon R^{2}}+A_{1}R$ $\displaystyle =$ $\displaystyle \frac{B_{1}}{R^{2}}\ \ \ \ \ (7)$ $\displaystyle A_{l}R^{l}$ $\displaystyle =$ $\displaystyle \frac{B_{l}}{R^{l+1}}\quad(l\ne1) \ \ \ \ \ (8)$

We can therefore express the ${B_{l}}$ in terms of ${A_{l}}$:

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{p}{4\pi\epsilon}+A_{1}R^{3}\ \ \ \ \ (9)$ $\displaystyle B_{l}$ $\displaystyle =$ $\displaystyle R^{2l+1}A_{l}\quad\left(l\ne1\right) \ \ \ \ \ (10)$

Using the condition on the derivatives 2 and equating coefficients as before, we get

 $\displaystyle -\frac{p}{2\pi R^{3}}+\epsilon A_{1}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{2B_{1}}{R^{3}}\ \ \ \ \ (11)$ $\displaystyle \epsilon A_{l}lR^{l-1}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{\left(l+1\right)B_{l}}{R^{l+2}}\quad\left(l\ne1\right) \ \ \ \ \ (12)$

Substituting for ${B_{l}}$ into the second equation gives

$\displaystyle \epsilon A_{l}lR^{2l+1}=-\epsilon_{0}\left(l+1\right)A_{l}R^{2l+1} \ \ \ \ \ (13)$

The only way this can be satisfied is if ${A_{l}=0}$ for ${l\ne1}$, so we get ${A_{l}=B_{l}=0}$ for ${l\ne1}$.

For the ${l=1}$ case, we can solve the two equations in ${A_{1}}$ and ${B_{1}}$ to get (using ${\epsilon=\epsilon_{0}\epsilon_{r}}$):

 $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle \frac{p}{2\pi\epsilon R^{3}}\left[\frac{\epsilon-\epsilon_{0}}{\epsilon+2\epsilon_{0}}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2p}{4\pi R^{3}\epsilon}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right)\ \ \ \ \ (15)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{p}{4\pi\epsilon}\left[1+2\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p}{4\pi\epsilon_{0}}\left(\frac{3}{\epsilon_{r}+2}\right) \ \ \ \ \ (17)$

The potential is

 $\displaystyle V_{in}(r)$ $\displaystyle =$ $\displaystyle \frac{p\cos\theta}{4\pi\epsilon r^{2}}+\frac{2pr\cos\theta}{4\pi R^{3}\epsilon}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p\cos\theta}{4\pi\epsilon r^{2}}\left[1+2\frac{r^{3}}{R^{3}}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right)\right]\ \ \ \ \ (19)$ $\displaystyle V_{out}(r)$ $\displaystyle =$ $\displaystyle \frac{p\cos\theta}{4\pi\epsilon_{0}r^{2}}\left(\frac{3}{\epsilon_{r}+2}\right) \ \ \ \ \ (20)$

# 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.

# Dielectric cylinder in uniform electric field

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.22.

In our original investigation into electrostatic boundary conditions in a vacuum, we found that the electric field is discontinuous across a layer of surface charge. The discontinuity was

$\displaystyle E_{\perp}^{above}-E_{\perp}^{below}=\frac{\sigma}{\epsilon_{0}} \ \ \ \ \ (1)$

We can do a similar analysis for a dielectric by using the displacement instead of the bare electric field. We know that

$\displaystyle \int_{A}\mathbf{D}\cdot d\mathbf{a}=Q_{f} \ \ \ \ \ (2)$

where ${Q_{f}}$ is the free charge enclosed by the surface of integration. Therefore, if we consider a dielectric with a free surface charge of ${\sigma_{f}}$, we can use a small cylindrical Gaussian surface with ends on either side of the surface to say that

$\displaystyle D_{\perp}^{above}-D_{\perp}^{below}=\sigma_{f} \ \ \ \ \ (3)$

Thus if there is no free surface charge on the dielectric, the perpendicular component of ${\mathbf{D}}$ is continuous across the surface. The dielectric may be polarized, since polarization induces only a bound surface charge, and that has no effect on the continuity of ${D_{\perp}}$.

In a linear dielectric, ${\mathbf{D}=\epsilon\mathbf{E}}$ and as usual ${\mathbf{E}=-\nabla V}$, so we can convert this into a condition on the potential on each side of the surface layer:

$\displaystyle \epsilon_{1}\frac{\partial V_{1}}{\partial n}-\epsilon_{2}\frac{\partial V_{2}}{\partial n}=\sigma_{f} \ \ \ \ \ (4)$

As an example, suppose we have a long cylinder of dielectric that is in a uniform electric field ${\mathbf{E}_{0}}$, where the field is perpendicular to the axis of the cylinder. We can find the potential everywhere, and thus find the field inside the cylinder.

To solve this problem, we set up a cylindrical coordinate system with its axis centred on the axis of the cylinder. We can define the ${\phi=0}$ direction as the direction of the electric field. In addition to the boundary condition 4, we have the addition boundary condition that the potential must be continuous, and we also have the asymptotic condition that for large ${r}$, the electric field must tend to a constant. Since there is no free charge, ${\sigma_{f}=0}$ and the three conditions become

 $\displaystyle \epsilon_{1}\left.\frac{\partial V_{1}}{\partial n}\right|_{r=R}$ $\displaystyle =$ $\displaystyle \epsilon_{2}\left.\frac{\partial V_{2}}{\partial n}\right|_{r=R}\ \ \ \ \ (5)$ $\displaystyle V_{1}(R)$ $\displaystyle =$ $\displaystyle V_{2}(R)\ \ \ \ \ (6)$ $\displaystyle \lim_{r\rightarrow\infty}V_{2}(r)$ $\displaystyle =$ $\displaystyle -E_{0}r\cos\phi \ \ \ \ \ (7)$

Here, ${V_{1}}$ is the potential inside the cylinder and ${V_{2}}$ is the potential outside. ${R}$ is the radius of the cylinder. The last condition says that the potential for large ${r}$ must be ${-E_{0}\zeta}$ where ${\zeta\equiv r\cos\phi}$ is a coordinate measured along the ${\phi=0}$ direction. It would be confusing to call this coordinate ${z}$, since in cylindrical coordinates, ${z}$ denotes the direction along the axis of the cylinder, which we’ve already said is the axis of the dielectric. With this definition, the field is ${\mathbf{E}_{0}=-\left(dV/d\zeta\right)\hat{\zeta}=E_{0}\hat{\zeta}}$.

Since there is no free charge, the system satisfies Laplace’s equation, so we can make use of our solution in cylindrical coordinates. The general solution is

$\displaystyle V(r,\phi)=A\ln r+K+\sum_{n=1}^{\infty}r^{n}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right)+\sum_{n=1}^{\infty}\frac{1}{r^{n}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (8)$

Inside the cylinder we can throw away the log term and the ${1/r^{n}}$ terms to keep the potential finite at ${r=0}$. Thus inside we have

$\displaystyle V_{1}=K_{1}+\sum_{n=1}^{\infty}r^{n}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right) \ \ \ \ \ (9)$

Outside, we again throw away the log term. We can also throw away all the ${r^{n}}$ terms except for ${n=1}$, since we need the asymptotic behaviour referred to above. Thus we get

$\displaystyle V_{2}=K_{2}-E_{0}r\cos\phi+\sum_{n=1}^{\infty}\frac{1}{r^{n}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (10)$

We can now apply the two other boundary conditions above. First:

 $\displaystyle V_{1}\left(R\right)$ $\displaystyle =$ $\displaystyle V_{2}\left(R\right)\ \ \ \ \ (11)$ $\displaystyle K_{1}+\sum_{n=1}^{\infty}R^{n}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right)$ $\displaystyle =$ $\displaystyle K_{2}-E_{0}R\cos\phi+\sum_{n=1}^{\infty}\frac{1}{R^{n}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (12)$

From the uniqueness of series, we can equate coefficients of the various trig functions, so we get

 $\displaystyle K_{1}$ $\displaystyle =$ $\displaystyle K_{2}\ \ \ \ \ (13)$ $\displaystyle A_{n}R^{n}$ $\displaystyle =$ $\displaystyle \frac{C_{n}}{R^{n}}\ \ \ \ \ (14)$ $\displaystyle B_{n}R^{n}$ $\displaystyle =$ $\displaystyle \frac{D_{n}}{R^{n}}\qquad(n\ne1)\ \ \ \ \ (15)$ $\displaystyle RB_{1}$ $\displaystyle =$ $\displaystyle -E_{0}R+\frac{D_{1}}{R} \ \ \ \ \ (16)$

Since the value of the constant ${K_{1}}$ makes no difference to the field (it disappears when we take the derivative), we might as well take ${K_{1}=K_{2}=0}$.

Now from the derivative condition, we have, using ${\epsilon_{1}=\epsilon_{0}\epsilon_{r}}$, where ${\epsilon_{r}}$ is the dielectric constant, and ${\epsilon_{2}=\epsilon_{0}}$, since outside the dielectric we have a vacuum:

 $\displaystyle \epsilon_{1}\left.\frac{\partial V_{1}}{\partial n}\right|_{r=R}$ $\displaystyle =$ $\displaystyle \epsilon_{2}\left.\frac{\partial V_{2}}{\partial n}\right|_{r=R}\ \ \ \ \ (17)$ $\displaystyle \epsilon_{r}\left[\sum_{n=1}^{\infty}nR^{n-1}\left(A_{n}\sin n\phi+B_{n}\cos n\phi\right)\right]$ $\displaystyle =$ $\displaystyle -E_{0}\cos\phi-\sum_{n=1}^{\infty}\frac{n}{R^{n+1}}\left(C_{n}\sin n\phi+D_{n}\cos n\phi\right) \ \ \ \ \ (18)$

Again, equating coefficients, we get

 $\displaystyle \epsilon_{r}nA_{n}R^{n-1}$ $\displaystyle =$ $\displaystyle -\frac{n}{R^{n+1}}C_{n}\ \ \ \ \ (19)$ $\displaystyle \epsilon_{r}nB_{n}R^{n-1}$ $\displaystyle =$ $\displaystyle -\frac{n}{R^{n+1}}D_{n}\qquad(n\ne1)\ \ \ \ \ (20)$ $\displaystyle \epsilon_{r}B_{1}$ $\displaystyle =$ $\displaystyle -E_{0}-\frac{D_{1}}{R^{2}} \ \ \ \ \ (21)$

Combining these equations with those from the first boundary condition above, we get, in each of the first two equations, the condition that either ${\epsilon_{r}=-1}$ (impossible, since dielectric constants are always ${\ge1}$) or ${A_{n}=C_{n}=0}$ and, for ${n\ne1}$, ${B_{n}=D_{n}=0}$. We are therefore left with the last equation from each boundary condition, and we can solve these two equations to get

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle -\frac{2E_{0}}{\epsilon_{r}+1}\ \ \ \ \ (22)$ $\displaystyle D_{1}$ $\displaystyle =$ $\displaystyle \frac{\epsilon_{r}-1}{\epsilon_{r}+1}R^{2}E_{0} \ \ \ \ \ (23)$

We therefore get for the potentials

 $\displaystyle V_{1}$ $\displaystyle =$ $\displaystyle -\frac{2E_{0}}{\epsilon_{r}+1}r\cos\phi\ \ \ \ \ (24)$ $\displaystyle V_{2}$ $\displaystyle =$ $\displaystyle \frac{\epsilon_{r}-1}{\epsilon_{r}+1}\frac{R^{2}E_{0}}{r}\cos\phi-E_{0}r\cos\phi \ \ \ \ \ (25)$

From this we can get the field inside the cylinder

 $\displaystyle V_{1}$ $\displaystyle =$ $\displaystyle -\frac{2E_{0}}{\epsilon_{r}+1}\zeta\ \ \ \ \ (26)$ $\displaystyle \mathbf{E}_{in}$ $\displaystyle =$ $\displaystyle -\frac{dV_{1}}{d\zeta}\hat{\zeta}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\epsilon_{r}+1}\mathbf{E}_{0} \ \ \ \ \ (28)$

Rather surprisingly, the field inside the cylinder is uniform.

# Field of a polarized cylinder

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.13.

As another example of the use of the bound charges representation of a polarized object, we can look at a cylinder which contains a uniform polarization ${\mathbf{P}}$ perpendicular to its axis. In this case, ${\nabla\cdot\mathbf{P}=0}$ everywhere inside the cylinder. If we take the direction of ${\mathbf{P}}$ to be ${\phi=0}$ then ${\mathbf{P}\cdot\hat{\mathbf{n}}=P\cos\phi}$. We thus must find the potential of an infinite cylinder with a surface charge of

$\displaystyle \sigma_{b}=P\cos\phi \ \ \ \ \ (1)$

We’ve worked out the general solution to Laplace’s equation in cylindrical coordinates before, so we can use the results from there. The solution inside the cylinder is

$\displaystyle V_{in}=B_{in}+\sum_{n=1}^{\infty}\left[A_{n}r^{n}\sin n\phi+B_{n}r^{n}\cos n\phi\right] \ \ \ \ \ (2)$

while outside it is

$\displaystyle V_{out}=B_{out}+\sum_{n=1}^{\infty}\left[\frac{D_{n}}{r^{n}}\cos n\phi-\frac{C_{n}}{r^{n}}\sin n\phi\right] \ \ \ \ \ (3)$

From the boundary condition requiring the potential to be continuous at the surface of the cylinder we get the relations

 $\displaystyle B_{out}$ $\displaystyle =$ $\displaystyle B_{in}\ \ \ \ \ (4)$ $\displaystyle C_{n}$ $\displaystyle =$ $\displaystyle -A_{n}R^{2n}\ \ \ \ \ (5)$ $\displaystyle D_{n}$ $\displaystyle =$ $\displaystyle B_{n}R^{2n} \ \ \ \ \ (6)$

where ${R}$ is the radius of the cylinder.

From the condition on the derivative of the potential at the boundary, we get

$\displaystyle \sum_{n=1}^{\infty}\left[2nR^{n-1}A_{n}\right]\sin n\phi+\sum_{n=1}^{\infty}\left[2nR^{n-1}B_{n}\right]\cos n\phi=\frac{\sigma_{b}}{\epsilon_{0}} \ \ \ \ \ (7)$

From 1 we see that all coefficients of the sine terms are zero, as are all coefficients of cosine terms except for ${n=1}$. We therefore get

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{P}{2\epsilon_{0}}\ \ \ \ \ (8)$ $\displaystyle D_{1}$ $\displaystyle =$ $\displaystyle R^{2}B_{1}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{PR^{2}}{2\epsilon_{0}} \ \ \ \ \ (10)$

The potential in the two regions is thus

 $\displaystyle V_{in}$ $\displaystyle =$ $\displaystyle \frac{P}{2\epsilon_{0}}r\cos\phi\ \ \ \ \ (11)$ $\displaystyle V_{out}$ $\displaystyle =$ $\displaystyle \frac{PR^{2}}{2r\epsilon_{0}}\cos\phi \ \ \ \ \ (12)$

From this we can get the field by taking the negative gradient. Since ${r\cos\phi=x}$, the field inside the cylinder is

 $\displaystyle \mathbf{E}_{in}$ $\displaystyle =$ $\displaystyle -\nabla V_{in}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{P}{2\epsilon_{0}}\hat{\mathbf{x}} \ \ \ \ \ (14)$

The interior field is thus uniform, just as the field inside a uniformly polarized sphere is uniform.

Outside the cylinder we need to take the gradient in cylindrical coordiantes. We get

 $\displaystyle \mathbf{E}_{out}$ $\displaystyle =$ $\displaystyle -\frac{\partial V_{out}}{\partial r}\hat{\mathbf{r}}-\frac{1}{r}\frac{\partial V_{out}}{\partial\phi}\hat{\boldsymbol{\phi}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{R^{2}}{2r^{2}\epsilon_{0}}\left[P\cos\phi\hat{\mathbf{r}}+P\sin\phi\hat{\boldsymbol{\phi}}\right] \ \ \ \ \ (16)$

We can write this in terms of the polarization vector. If ${\mathbf{P}}$ points in the direction ${\phi=0}$ then we have

$\displaystyle \mathbf{P}=P\cos\phi\hat{\mathbf{r}}-P\sin\phi\hat{\boldsymbol{\phi}} \ \ \ \ \ (17)$

where the minus sign on the second term is because ${\hat{\boldsymbol{\phi}}}$ points in the direction of increasing ${\phi}$, which is clockwise from ${\phi=0}$. We have then

$\displaystyle P\cos\phi\hat{\mathbf{r}}+P\sin\phi\hat{\boldsymbol{\phi}}=2\left(\mathbf{P}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{P} \ \ \ \ \ (18)$

so the field is

$\displaystyle \mathbf{E}_{out}=\frac{R^{2}}{2r^{2}\epsilon_{0}}\left[2\left(\mathbf{P}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{P}\right] \ \ \ \ \ (19)$

# Laplace’s equation – Fourier series example 4

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 3.47.

Here is another example of solving Laplace’s equation in rectangular coordinates. We have a rectangular pipe extending in the ${z}$ direction with boundaries at ${x=\pm b}$, ${y=0}$ and ${y=a}$. The potential on each face is held constant and satisfies

$\displaystyle V=\begin{cases} 0 & y=0\\ V_{0} & y=a\\ 0 & x=-b\\ 0 & x=b \end{cases} \ \ \ \ \ (1)$

The problem is to find ${V(x,y)}$ everywhere inside the pipe. We can use the separation of variables technique to arrive at a solution of form:

$\displaystyle V(x,y)=\left(Ae^{kx}+Be^{-kx}\right)\left(C\sin ky+D\cos ky\right) \ \ \ \ \ (2)$

Because the problem is symmetric in ${x}$ we must have ${V(-x,y)=V(x,y)}$, which means that ${A=B}$ and we can write

$\displaystyle V(x,y)=\left(C\sin ky+D\cos ky\right)\cosh kx \ \ \ \ \ (3)$

where we’ve absorbed the constants ${A}$ and ${B}$ into ${C}$ and ${D}$.

The boundary condition at ${y=0}$ means that ${D=0}$. At first glance we can’t do much with the boundary conditions at the other three faces. Let’s leave this solution for the moment and try a different approach.

In the original derivation of the separation of variables solution, we had the equations

 $\displaystyle \frac{1}{X}\frac{d^{2}X}{dx^{2}}$ $\displaystyle =$ $\displaystyle C_{1}\ \ \ \ \ (4)$ $\displaystyle \frac{1}{Y}\frac{d^{2}Y}{dy^{2}}$ $\displaystyle =$ $\displaystyle C_{2}\ \ \ \ \ (5)$ $\displaystyle \frac{1}{Z}\frac{d^{2}Z}{dz^{2}}$ $\displaystyle =$ $\displaystyle C_{3} \ \ \ \ \ (6)$

We made the implicit assumption that the constants were non-zero, requiring only that their sum is zero so as to satisfy the full Laplace’s equation:

$\displaystyle C_{1}+C_{2}+C_{3}=0 \ \ \ \ \ (7)$

In this problem, though, we can use a trick to get the solution. Since the problem doesn’t depend on ${z}$, we can consider the two equations

 $\displaystyle \frac{d^{2}X}{dx^{2}}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \frac{d^{2}Y}{dy^{2}}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (9)$

The general solution of this pair of equations is linear in each of ${x}$ and ${y}$ separately in order for the second derivatives to be zero. Thus we get

 $\displaystyle X(x)$ $\displaystyle =$ $\displaystyle c_{1}x+c_{2}\ \ \ \ \ (10)$ $\displaystyle Y(y)$ $\displaystyle =$ $\displaystyle c_{3}y+c_{4} \ \ \ \ \ (11)$

and the overall solution is thus

$\displaystyle V=\left(c_{1}x+c_{2}\right)\left(c_{3}y+c_{4}\right) \ \ \ \ \ (12)$

Applying the boundary conditions on ${y}$ we get

 $\displaystyle V(x,0)$ $\displaystyle =$ $\displaystyle c_{4}\left(c_{1}x+c_{2}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

This must be true for all ${x}$ so either ${c_{4}=0}$ or ${c_{1}=c_{2}=0}$. If we choose the latter case, then ${V=0}$ everywhere, which isn’t much use, so we’ll take ${c_{4}=0}$.

At the other boundary, we have

 $\displaystyle V(x,a)$ $\displaystyle =$ $\displaystyle c_{3}a\left(c_{1}x+c_{2}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle V_{0} \ \ \ \ \ (16)$

Again, this must be true for all ${x}$, so ${c_{1}=0}$ and ${c_{2}c_{3}=V_{0}/a}$. Thus a solution that satisfies the ${y}$ boundary conditions is

$\displaystyle V(x,y)=\frac{V_{0}}{a}y \ \ \ \ \ (17)$

Clearly this doesn’t satisfy the ${x}$ boundary conditions, but we can now return to our earlier solutions to the separation of variables problem to see how to fix this. If we find a solution that satisfies the new boundary conditions

$\displaystyle V=\begin{cases} 0 & y=0\\ 0 & y=a\\ -\frac{V_{0}}{a}y & x=-b\\ -\frac{V_{0}}{a}y & x=b \end{cases} \ \ \ \ \ (18)$

then we can add this solution to the solution we just found. This new solution will satisfy all four boundary conditions, since the sums of the potentials at each of the four boundaries come out to what was originally specified. We can now use the usual technique of building an infinite series of solutions and using the orthogonality of the sine functions to work out the coefficients. That is, returning to our first solution, we have, after satisfying the boundary condition at ${y=0}$:

$\displaystyle V(x,y)=C\sin ky\cosh kx \ \ \ \ \ (19)$

If we now require this solution to be zero at ${y=a}$, then we must have

$\displaystyle k=\frac{n\pi}{a} \ \ \ \ \ (20)$

for ${n=1,2,3,\ldots}$. We can then construct the sum of these terms and combine it with the other solution we found to give

$\displaystyle V(x,y)=\frac{V_{0}}{a}y+\sum_{n=1}^{\infty}C_{n}\sin\left(\frac{n\pi y}{a}\right)\cosh\left(\frac{n\pi x}{a}\right) \ \ \ \ \ (21)$

To satisfy the ${x}$ boundary conditions we must have

$\displaystyle \sum_{n=1}^{\infty}C_{n}\sin\left(\frac{n\pi y}{a}\right)\cosh\left(\frac{n\pi b}{a}\right)=-\frac{V_{0}}{a}y \ \ \ \ \ (22)$

for ${0. We now multiply both sides of this equation by ${\sin\frac{m\pi y}{a}}$ and integrate from 0 to ${a}$, using the orthogonality of the sine, that is

$\displaystyle \int_{0}^{a}\sin\frac{m\pi y}{a}\sin\frac{n\pi y}{a}=\begin{cases} 0 & n\ne m\\ \frac{a}{2} & n=m \end{cases} \ \ \ \ \ (23)$

Therefore

 $\displaystyle C_{n}\frac{a}{2}\cosh\frac{n\pi b}{a}$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{a}\int_{0}^{a}y\sin\frac{n\pi y}{a}dy\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{a}\left(-\frac{(-1)^{n}a^{2}}{\pi n}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{(-1)^{n}aV_{0}}{\pi n}\ \ \ \ \ (26)$ $\displaystyle C_{n}$ $\displaystyle =$ $\displaystyle \frac{2(-1)^{n}V_{0}}{n\pi\cosh(n\pi b/a)} \ \ \ \ \ (27)$

The final form of the potential is

$\displaystyle V(x,y)=\frac{V_{0}}{a}y+\frac{2V_{0}}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}\sin\left(n\pi y/a\right)\cosh\left(n\pi x/a\right)}{n\cosh(n\pi b/a)} \ \ \ \ \ (28)$

# Laplace’s equation – cylindrical shell with opposing charges

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.39.

As another example of applying the solution to Laplace’s equation in cylindrical coordinates, we consider the following problem. We are given a cylindrical non-conducting shell or radius ${R}$ carrying a charge density of ${\sigma_{0}}$ on its upper half (${-\pi<\phi<0}$) and ${-\sigma_{0}}$ on its lower half (${0<\phi<\pi)}$. Find the potential everywhere.

We can begin in the same manner as for the other problem involving a cylindrical shell that we solved earlier. The solution is the same up until the point where we introduce the surface charge.

Thus, outside the shell, we have

$\displaystyle V_{out}=B_{out}+\sum_{n=1}^{\infty}\left[\frac{D_{n}}{r^{n}}\cos n\phi-\frac{C_{n}}{r^{n}}\sin n\phi\right] \ \ \ \ \ (1)$

Inside the shell, we have

$\displaystyle V_{in}=B_{in}+\sum_{n=1}^{\infty}\left[A_{n}r^{n}\sin n\phi+B_{n}r^{n}\cos n\phi\right] \ \ \ \ \ (2)$

Since the potential is continuous over a surface charge, we must have ${V_{out}(R)=V_{in}(R)}$, so we get

$\displaystyle B_{out}+\sum_{n=1}^{\infty}\left[\frac{D_{n}}{R^{n}}\cos n\phi-\frac{C_{n}}{R^{n}}\sin n\phi\right]=B_{in}+\sum_{n=1}^{\infty}\left[A_{n}R^{n}\sin n\phi+B_{n}R^{n}\cos n\phi\right] \ \ \ \ \ (3)$

Equating coefficients of the sine and cosine, we get

 $\displaystyle B_{out}$ $\displaystyle =$ $\displaystyle B_{in}\ \ \ \ \ (4)$ $\displaystyle C_{n}$ $\displaystyle =$ $\displaystyle -A_{n}R^{2n}\ \ \ \ \ (5)$ $\displaystyle D_{n}$ $\displaystyle =$ $\displaystyle B_{n}R^{2n} \ \ \ \ \ (6)$

The outward derivative of the potential is discontinuous across a surface charge, and we have

$\displaystyle \left.\frac{\partial V}{\partial r}\right|_{out}-\left.\frac{\partial V}{\partial r}\right|_{in}=-\frac{\sigma}{\epsilon_{0}} \ \ \ \ \ (7)$

Plugging in the formulas for ${V_{out}}$ and ${V_{in}}$, we get

$\displaystyle \sum_{n=1}^{\infty}\left[\frac{-nR^{2n}A_{n}}{R^{n+1}}-nR^{n-1}A_{n}\right]\sin n\phi+\sum_{n=1}^{\infty}\left[\frac{-nR^{2n}B_{n}}{R^{n+1}}-nR^{n-1}B_{n}\right]\cos n\phi=\begin{cases} \frac{\sigma_{0}}{\epsilon_{0}} & -\pi<\phi<0\\ -\frac{\sigma_{0}}{\epsilon_{0}} & 0<\phi<\pi \end{cases} \ \ \ \ \ (8)$

Since the surface charge is an odd function of ${\phi}$, we can eliminate the cosine terms, since the cosine is an even function. Therefore, we have ${B_{n}=D_{n}=0}$. We are free to choose the potential at infinity to be any constant, so we might as well take it to be zero, in which case we have ${B_{in}=B_{out}=0}$. We are therefore left with, after simplifying the term in brackets:

$\displaystyle -2\sum_{n=1}^{\infty}nR^{n-1}A_{n}\sin n\phi=\begin{cases} \frac{\sigma_{0}}{\epsilon_{0}} & -\pi<\phi<0\\ -\frac{\sigma_{0}}{\epsilon_{0}} & 0<\phi<\pi \end{cases} \ \ \ \ \ (9)$

To find the ${A_{n}}$, we can use the fact that the set of ${\sin n\phi}$ functions is orthogonal over the interval ${[-\pi,\pi]}$. That is

$\displaystyle \int_{-\pi}^{\pi}\sin m\phi\sin n\phi d\phi=\begin{cases} 0 & n\ne m\\ \pi & n=m \end{cases} \ \ \ \ \ (10)$

Therefore we can multiply both sides by ${\sin m\phi}$ and integrate to get

 $\displaystyle -2\pi mR^{m-1}A_{m}$ $\displaystyle =$ $\displaystyle \frac{\sigma_{0}}{\epsilon_{0}}\left[\int_{-\pi}^{0}\sin m\phi d\phi-\int_{0}^{\pi}\sin m\phi d\phi\right] \ \ \ \ \ (11)$

The integrals in brackets on the right come out to

$\displaystyle \left[\int_{-\pi}^{0}\sin m\phi d\phi-\int_{0}^{\pi}\sin m\phi d\phi\right]=\begin{cases} 0 & m\;\mathrm{even}\\ -\frac{4}{m} & m\;\mathrm{odd} \end{cases} \ \ \ \ \ (12)$

Therefore (changing the index from ${m}$ to ${n}$ for convenience):

$\displaystyle A_{n}=\begin{cases} 0 & n\;\mathrm{even}\\ \frac{2\sigma_{0}}{\pi\epsilon_{0}n^{2}R^{n-1}} & n\;\mathrm{odd} \end{cases} \ \ \ \ \ (13)$

We thus get

$\displaystyle C_{n}=\begin{cases} 0 & n\;\mathrm{even}\\ -\frac{2\sigma_{0}R^{n+1}}{\pi\epsilon_{0}n^{2}} & n\;\mathrm{odd} \end{cases} \ \ \ \ \ (14)$

The final formula for the potential is

$\displaystyle V(r,\phi)=\begin{cases} \frac{2\sigma_{0}}{\pi\epsilon_{0}}\sum_{n\; odd}^{\infty}\frac{r^{n}}{n^{2}R^{n-1}}\sin n\phi & rR \end{cases} \ \ \ \ \ (15)$

# 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 ${Q=2a\lambda}$:

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