# Electric field within a cavity inside a dielectric

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.16

We can use superposition to work out the electric field within a small cavity that is hollowed out inside a block of dielectric. We’ll take the electric field within the dielectric to be ${\mathbf{E}_{0}}$ (which need not be constant), and the polarization to be ${\mathbf{P}}$ (again, not necessarily constant). The displacement is then ${\mathbf{D}_{0}=\epsilon_{0}\mathbf{E}_{0}+\mathbf{P}}$. In the small region of interest, we’ll take ${\mathbf{P}}$ pointing upwards, so that the field it induces points downwards.

Now if we hollow out a small sphere (assumed to be small enough that the field and polarization can be taken as constant within it), we can work out the field within the hollow sphere. The trick is that the polarization within empty space must be zero, so we can simulate the situation by superimposing a sphere with equal and opposite polarization ${-\mathbf{P}}$ on top of the dielectric. We’ve worked out the field within a uniformly polarized sphere earlier, and with polarization ${-\mathbf{P}}$ this field is

$\displaystyle \mathbf{E}_{s}=\frac{1}{3\epsilon_{0}}\mathbf{P} \ \ \ \ \ (1)$

Thus the net field within a spherical cavity is

$\displaystyle \mathbf{E}=\mathbf{E}_{0}+\mathbf{E}_{s}=\mathbf{E}_{0}+\frac{1}{3\epsilon_{0}}\mathbf{P} \ \ \ \ \ (2)$

Since the net polarization within the cavity is zero, the displacement is

$\displaystyle \mathbf{D}=\epsilon_{0}\mathbf{E}_{0}+\frac{1}{3}\mathbf{P}=\mathbf{D}_{0}-\frac{2}{3}\mathbf{P} \ \ \ \ \ (3)$

Now suppose the cavity is shaped like a long thin needle parallel to ${\mathbf{P}}$. Again, we superimpose a needle with the opposite polarization. We can work out the bound charges, and since the polarization within the needle is constant, ${\rho_{b}=\nabla\cdot\left(-\mathbf{P}\right)=0}$ and since the polarization is parallel to the axis, ${\sigma_{b}=-\mathbf{P}\cdot\hat{\mathbf{n}}=0}$ on the sides of the needle. On the ends of the needle, ${\sigma_{b}}$ is non-zero, but if the needle is long enough with small end points, this will contribute a very small field so we can approximate the situation by saying that the electric field is not significantly modified within the cavity:

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \mathbf{E}_{0}\ \ \ \ \ (4)$ $\displaystyle \mathbf{D}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\mathbf{E}_{0}=\mathbf{D}_{0}-\mathbf{P} \ \ \ \ \ (5)$

Finally, we consider a thin, circular wafer shaped cavity perpendicular to the polarization. In this case, the bound volume charge is again zero, but the surface charge is

$\displaystyle \sigma_{b}=-\mathbf{P}\cdot\hat{\mathbf{n}}=\pm P \ \ \ \ \ (6)$

with the plus sign on the bottom of the wafer and the minus sign on top. The field due to the bottom (positive) surface is (by Gauss’s law) ${P/2\epsilon_{0}}$ upwards, and the field due to the top (negative) surface is also ${P/2\epsilon_{0}}$ upwards, so the total field within the wafer is

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \mathbf{E}_{0}+\frac{1}{\epsilon_{0}}\mathbf{P}\ \ \ \ \ (7)$ $\displaystyle \mathbf{D}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\mathbf{E}=\mathbf{D}_{0} \ \ \ \ \ (8)$

Thus the needle leaves ${\mathbf{E}}$ unchanged and the wafer leaves ${\mathbf{D}}$ unchanged.

# Conducting sphere half-embedded in dielectric plane

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.36.

An interesting special case of a dielectric problem is one in which a conducting sphere is partially embedded in an infinite half-space of dielectric. That is, we have a metal sphere of radius ${R}$ with its centre at the origin, and the space ${z<0}$ is completely filled with a linear dielectric with susceptibility ${\chi_{e}}$, so that the lower hemisphere of the conductor is embedded in the dielectric with the top hemisphere in vacuum.

If we hold the sphere at a potential of ${V_{0}}$ then in the absence of the dielectric, the potential would be

$\displaystyle V\left(r\right)=\begin{cases} V_{0} & rR \end{cases} \ \ \ \ \ (1)$

From this we get the field (which is zero inside the sphere of course):

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V_{0}R}{r^{2}}\hat{\mathbf{r}}\quad\left(r>R\right) \ \ \ \ \ (3)$

For a linear dielectric, the polarization is (for ${z<0}$):

 $\displaystyle \mathbf{P}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}\mathbf{E}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}\frac{V_{0}R}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (5)$

The volume bound charge induced by this is

 $\displaystyle \rho_{b}$ $\displaystyle =$ $\displaystyle -\nabla\cdot\mathbf{P}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

(since ${\nabla\cdot\left(\hat{\mathbf{r}}/r^{2}\right)=4\pi\delta_{3}\left(\mathbf{r}\right)}$ and we are considering only points with ${r>R}$ so we avoid the singularity at the origin).

The surface bound charge is ${\sigma_{b}=\mathbf{P}\cdot\hat{\mathbf{n}}}$ and is zero on the plane ${z=0}$ since ${\mathbf{P}\perp\hat{\mathbf{n}}}$. That is, since ${\mathbf{P}}$ is parallel to ${\mathbf{\hat{\mathbf{r}}}}$ and ${\hat{\mathbf{r}}}$ is horizontal across the plane ${z=0}$, all the polarization is parallel to the plane so there is no bound surface charge.

On the lower hemisphere, we consider the surface of the dielectric which has a surface normal pointing towards the origin, which means that ${\hat{\mathbf{n}}=-\hat{\mathbf{r}}}$ so on this surface

 $\displaystyle \sigma_{b}$ $\displaystyle =$ $\displaystyle \mathbf{P}\cdot\hat{\mathbf{n}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\chi_{e}\frac{V_{0}}{R} \ \ \ \ \ (9)$

In order for the sphere to be maintained at a constant potential, the surface charge density on the lower hemisphere would have to increase to compensate for this induced bound charge on the dielectric. This is the same argument as that used when we increase the capacitance of a capacitor by putting dielectric between its plates. Because the dielectric cancels out some of the field between the plates, we need to put more charge on the plates in order to maintain the same potential difference between the plates. In this case, we’re trying to maintain the same potential difference between the sphere and infinity (where ${V=0}$), so we need to increase the charge on the lower hemisphere to maintain it at ${V_{0}}$.

In the absence of dielectric, we can work out the surface charge density on the sphere from the field, since across a surface layer of charge

$\displaystyle \Delta\mathbf{E}_{\perp}=\frac{\sigma}{\epsilon_{0}} \ \ \ \ \ (10)$

Since ${\mathbf{E}=0}$ inside and ${\mathbf{E}=\frac{V_{0}}{R}\hat{\mathbf{r}}}$ outside, we have for the upper hemisphere

$\displaystyle \sigma_{\uparrow}=\frac{\epsilon_{0}V_{0}}{R} \ \ \ \ \ (11)$

In the lower hemisphere, the net surface density must be the same, so

 $\displaystyle \sigma_{\downarrow}$ $\displaystyle =$ $\displaystyle \frac{\epsilon_{0}V_{0}}{R}+\epsilon_{0}\chi_{e}\frac{V_{0}}{R}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon\frac{V_{0}}{R} \ \ \ \ \ (13)$

where ${\epsilon=\epsilon_{0}\left(1+\chi_{e}\right)=\epsilon_{0}\epsilon_{r}}$. That is, the net charge density is the same as if the dielectric was not present at all. Thus by the uniqueness of the potential in systems containing dielectrics, the potential for the half-embedded sphere is the same as that for a sphere without any dielectric.

This argument hinges on the fact that the boundary of the dielectric lies on a surface that is parallel to the field (the plane ${z=0}$ here). Thus we could use the same argument for a sphere embedded in any dielectric in which the surface of the dielectric lay entirely parallel to the field, for example a cone centred at the origin. It would not work for, say, a sphere embedded in a dielectric plane that didn’t split the sphere into two equal hemispheres.

# Uniqueness of potential in dielectrics

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.35.

When we considered electric fields in a vacuum, we found that if we specify the charge distribution in some region ${\mathcal{V}}$, and also specify either the potential or its normal derivative on the boundary of that region, then the potential inside ${\mathcal{V}}$ is unique. Here, we’ll show that uniqueness of the potential also applies to the case where ${\mathcal{V}}$ contains some linear dielectric. The assumptions we make are:

1. The potential ${V}$ is specified on all boundaries of ${\mathcal{V}}$.
2. The free charge distribution ${\rho_{f}}$ is specified everywhere within ${\mathcal{V}}$.
3. The distribution of dielectric within ${\mathcal{V}}$ is fixed, and all dielectric constants are specified.

The proof follows a similar line of reasoning to that used in the electric field case. As before we’ll suppose that there are two distinct potentials ${V_{1}}$ and ${V_{2}}$ that satisfy the conditions. We’ll also define the displacements due to these potentials as ${\mathbf{D}_{1}}$ and ${\mathbf{D}_{2}}$. Now we consider the difference between these two solutions, so we have ${V_{3}\equiv V_{1}-V_{2}}$ and ${\mathbf{D}_{3}\equiv\mathbf{D}_{1}-\mathbf{D}_{2}}$. Now we look at this volume integral and convert it to a surface integral in the usual way:

 $\displaystyle \int_{\mathcal{V}}\nabla\cdot\left(V_{3}\mathbf{D}_{3}\right)d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle \int_{A}V_{3}\mathbf{D}_{3}\cdot d\mathbf{a} \ \ \ \ \ (1)$

By assumption, on the surface ${A}$, ${V_{3}=V_{1}-V_{2}=0}$ since the potential is specified everywhere on the boundary. Therefore:

$\displaystyle \int_{\mathcal{V}}\nabla\cdot\left(V_{3}\mathbf{D}_{3}\right)d^{3}\mathbf{r}=0 \ \ \ \ \ (2)$

Now we expand the integrand using a standard theorem from vector calculus:

$\displaystyle \nabla\cdot\left(V_{3}\mathbf{D}_{3}\right)=\mathbf{D}_{3}\cdot\nabla V_{3}+V_{3}\nabla\cdot\mathbf{D}_{3} \ \ \ \ \ (3)$

From the formula for the divergence of the displacement:

 $\displaystyle \nabla\cdot\mathbf{D}_{3}$ $\displaystyle =$ $\displaystyle \nabla\cdot\left(\mathbf{D}_{1}-\mathbf{D}_{2}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho_{f}-\rho_{f}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

This follows, since the free charge distribution was fixed by assumption. Therefore we have

$\displaystyle \nabla\cdot\left(V_{3}\mathbf{D}_{3}\right)=\mathbf{D}_{3}\cdot\nabla V_{3} \ \ \ \ \ (7)$

For a linear dielectric, ${\mathbf{D}=\epsilon\mathbf{E}}$ and in general, ${\mathbf{E}=-\nabla V}$, so

 $\displaystyle \nabla\cdot\left(V_{3}\mathbf{D}_{3}\right)$ $\displaystyle =$ $\displaystyle \mathbf{D}_{3}\cdot\nabla V_{3}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon E_{3}^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon\left|\mathbf{E}_{1}-\mathbf{E}_{2}\right|^{2} \ \ \ \ \ (10)$

Thus the volume integral becomes

 $\displaystyle \int_{\mathcal{V}}\nabla\cdot\left(V_{3}\mathbf{D}_{3}\right)d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle -\int_{\mathcal{V}}\epsilon\left|\mathbf{E}_{1}-\mathbf{E}_{2}\right|^{2}d^{3}\mathbf{r}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

(Note that we can’t take ${\epsilon}$ outside the integral since in general it varies over the volume, depending on what dielectrics are present.)

Now the integrand is non-negative everywhere, since the permittivity ${\epsilon\ge\epsilon_{0}}$ so the only way the integral can be zero is if ${\mathbf{E}_{1}=\mathbf{E}_{2}}$ everywhere inside ${\mathcal{V}}$. This means that ${V_{1}-V_{2}=k}$ for some constant ${k}$, but since ${V_{1}=V_{2}}$ on the boundary, we must have ${k=0}$ and ${V_{1}=V_{2}}$ everywhere.

# Point charge in dielectric sphere

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.32.

Yet another example of the bound charge due to polarization in a dielectric. Suppose we have a point charge at the centre of a sphere of linear dielectric (susceptibility ${\chi_{e}}$) of radius ${R}$. We can begin the solution by calculating the displacement, since the system has spherical symmetry. We know that

 $\displaystyle \int_{V}\nabla\cdot\mathbf{D}d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle Q_{f}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{A}\mathbf{D}\cdot d\mathbf{a} \ \ \ \ \ (2)$

where ${Q_{f}}$ is the free charge enclosed by the surface of integration. Here we can take the surface to be a sphere of radius ${r}$ centred at the point charge, which is the only free charge in the problem. Therefore

 $\displaystyle \int_{A}\mathbf{D}\cdot d\mathbf{a}$ $\displaystyle =$ $\displaystyle 4\pi r^{2}D(r)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q\ \ \ \ \ (4)$ $\displaystyle \mathbf{D}\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (5)$

For a linear dielectric we have

$\displaystyle \mathbf{D}=\epsilon_{0}\left(1+\chi_{e}\right)\mathbf{E} \ \ \ \ \ (6)$

so we can get the field

$\displaystyle \mathbf{E}=\begin{cases} \frac{q}{4\pi\epsilon_{0}\left(1+\chi_{e}\right)}\frac{\hat{\mathbf{r}}}{r^{2}} & 0R \end{cases} \ \ \ \ \ (7)$

The polarization, again because we’re dealing with a linear dielectric, is

 $\displaystyle \mathbf{P}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}\mathbf{E}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\chi_{e}q}{4\pi\left(1+\chi_{e}\right)}\frac{\hat{\mathbf{r}}}{r^{2}} \ \ \ \ \ (9)$

The bound charges induced by this polarization are

 $\displaystyle \sigma_{b}$ $\displaystyle \equiv$ $\displaystyle \mathbf{P}\cdot\hat{\mathbf{n}}\ \ \ \ \ (10)$ $\displaystyle \rho_{b}$ $\displaystyle \equiv$ $\displaystyle -\nabla\cdot\mathbf{P} \ \ \ \ \ (11)$

At the surface of the sphere, ${\hat{\mathbf{n}}=\hat{\mathbf{r}}}$ and we get

 $\displaystyle \sigma_{b}$ $\displaystyle =$ $\displaystyle \frac{\chi_{e}q}{4\pi\left(1+\chi_{e}\right)R^{2}}\ \ \ \ \ (12)$ $\displaystyle Q_{s}$ $\displaystyle =$ $\displaystyle 4\pi R^{2}\sigma_{b}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\chi_{e}q}{\left(1+\chi_{e}\right)} \ \ \ \ \ (14)$

For the bound charge, we make use of the formula for the three-dimensional delta function:

$\displaystyle \nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r^{2}}\right)=4\pi\delta_{3}(\mathbf{r}) \ \ \ \ \ (15)$

The bound charge is therefore

 $\displaystyle \rho_{b}$ $\displaystyle =$ $\displaystyle -\nabla\cdot\mathbf{P}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\chi_{e}q}{\left(1+\chi_{e}\right)}\delta_{3}(\mathbf{r}) \ \ \ \ \ (17)$

The total volume bound charge is the integral of this over the sphere, which is just

 $\displaystyle Q_{v}$ $\displaystyle =$ $\displaystyle -\int_{V}\frac{\chi_{e}q}{\left(1+\chi_{e}\right)}\delta_{3}(\mathbf{r})d^{3}\mathbf{r}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\chi_{e}q}{\left(1+\chi_{e}\right)} \ \ \ \ \ (19)$

All the volume bound charge is concentrated at the centre, and the total bound charge is ${Q_{v}+Q_{s}=0}$ as required.

# Force on a dielectric

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 4.4.4 & Problem 4.28.

We’ve seen that a conductor carrying some free charge experiences electrostatic pressure due to the interaction of the free charges. A dielectric in an external electric field also experiences a force from that field.

In the case of a parallel plate capacitor filled with dielectric, the field over most of the plates’ area is perpendicular to the plates and thus there is no net force on the dielectric since it is held in place between the plates and the electric force is balanced by the mechanical force exerted by the plates. However, at the edges of the plates, there is a fringing field which curves outwards from the plates themselves, and this tends to pull the dielectric further into the capacitor. If the dielectric fills the capacitor, then the forces at all the edges balance each other and the dielectric experiences no net force. However, if the dielectric only partially fills the space between the plates, there is a net force tending to pull the dielectric further in.

A nice example of this is the following problem. We have two concentric cylindrical tubes. The inner tube has radius ${a}$ and the outer tube has radius ${b}$. The tubes are placed vertically into a dish of dielectric oil with susceptibility ${\chi_{e}}$ and mass density ${\rho}$, and a constant voltage of ${V}$ is applied between the two tubes. The oil will be attracted into the space between the tubes, so it will tend to rise to a height ${h}$.

We’ve already seen that the capacitance per unit length of a coaxial pair of cylinders in a vacuum is

$\displaystyle C=\frac{2\pi\epsilon_{0}}{\ln\left(b/a\right)} \ \ \ \ \ (1)$

If the total length of the cylinder is ${L}$, and the height of the oil is ${h}$, then the portions of the cylinder will have capacitances of

 $\displaystyle C_{oil}$ $\displaystyle =$ $\displaystyle \frac{2\pi\epsilon_{0}\epsilon_{r}}{\ln\left(b/a\right)}h\ \ \ \ \ (2)$ $\displaystyle C_{vac}$ $\displaystyle =$ $\displaystyle \frac{2\pi\epsilon_{0}}{\ln\left(b/a\right)}\left(L-h\right) \ \ \ \ \ (3)$

(Note that the capacitance of the region with the oil is increased by a factor of the dielectric constant ${\epsilon_{r}=1+\chi_{e}}$.) The total capacitance is thus

 $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{2\pi\epsilon_{0}}{\ln\left(b/a\right)}\left[\epsilon_{r}h+L-h\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi\epsilon_{0}}{\ln\left(b/a\right)}\left(\chi_{e}h+L\right) \ \ \ \ \ (5)$

Since we’re holding the voltage constant and the capacitance increases as the oil rises, the amount of charge on the plates must also increase. That is, we must transfer charge from the inner to the outer tube in order to maintain a constant ${V}$. The amount of work required to transfer an amount of charge ${dq}$ through a potential difference ${V}$ is ${Vdq}$. There are thus two sources of work done here: the energy required to make the oil rise between the plates, and the energy required to transfer charge from one tube to the other.

If we choose instead to assume that the charge on the plates rather than the voltage remains constant then no charge is transferred so no work is done by the ${Vdq}$ term, which is now zero. In this case, the voltage would change as the oil rises. However, since the oil would still rise, work is still being done.

Since the energy stored in a capacitor is ${W=\frac{1}{2}CV^{2}}$, if ${V}$ is constant, the change in energy if we move the oil a distance ${dh}$ is

$\displaystyle dW=\frac{1}{2}V^{2}\frac{dC}{dh}dh \ \ \ \ \ (6)$

Since the force we apply to move the oil is in opposition to the electrical force ${F}$ we get

 $\displaystyle F$ $\displaystyle =$ $\displaystyle -\frac{dW}{dh}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}V^{2}\frac{dC}{dh} \ \ \ \ \ (8)$

However, in the constant voltage problem, we also have to move a charge ${dq}$ from one plate to the other, requiring work ${Vdq}$. This adds another term onto the force, which is, since ${q=CV}$:

 $\displaystyle F_{q}$ $\displaystyle =$ $\displaystyle V\frac{dq}{dh}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle V^{2}\frac{dC}{dh} \ \ \ \ \ (10)$

This results in a total force of

 $\displaystyle F_{t}$ $\displaystyle =$ $\displaystyle F+F_{q}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}V^{2}\frac{dC}{dh} \ \ \ \ \ (12)$

It’s worth looking at this in a bit more detail. The capacitance ${C}$ increases as the dielectric constant ${\epsilon_{r}}$ increases, so if we keep the charge ${Q}$ on the capacitor constant as the oil rises (and thus increases the dielectric constant over a greater range, increasing ${C}$) then since ${Q=CV}$, ${V}$ must decrease. The energy stored in the capacitor is ${W=\frac{1}{2}CV^{2}=\frac{Q^{2}}{2C}}$ therefore also decreases. This energy goes into the work done to raise the oil.

However, if we keep ${V}$ constant as the oil rises, then from ${Q=CV}$, ${Q}$ must increase as ${C}$ increases. The energy stored in the capacitor is ${W=\frac{1}{2}CV^{2}}$ and thus it increases. The extra energy is provided by an external battery which adds charge to the capacitor in order to maintain a constant voltage, as well as provide the energy to raise the oil. This is the extra ${Vdq}$ term.

Returning to the problem of determining the height of the oil, the electrical force on the oil is balanced by the gravitational force, so we get

 $\displaystyle \frac{1}{2}V^{2}\frac{dC}{dh}$ $\displaystyle =$ $\displaystyle mg\ \ \ \ \ (13)$ $\displaystyle \frac{1}{2}V^{2}\frac{2\pi\epsilon_{0}\chi_{e}}{\ln\left(b/a\right)}$ $\displaystyle =$ $\displaystyle \pi\left(b^{2}-a^{2}\right)\rho gh\ \ \ \ \ (14)$ $\displaystyle h$ $\displaystyle =$ $\displaystyle \frac{\epsilon_{0}\chi_{e}V^{2}}{\rho g\left(b^{2}-a^{2}\right)\ln\left(b/a\right)} \ \ \ \ \ (15)$

# Energy of conducting sphere in a dielectric shell

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 4.4.3 & Problem 4.26.

We’ve seen that the energy in a system containing dielectrics can be written as

$\displaystyle W=\frac{1}{2}\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r} \ \ \ \ \ (1)$

This is the energy required to place the free charges, and includes the energy needed to polarize the dielectric.

As a simple example of this formula, suppose we have a spherical conductor of radius ${a}$ that has a free charge ${Q}$ on it, and we surround this conductor with a spherical shell of linear dielectric that extends from ${r=a}$ to ${r=b}$. The free charge on the conductor will polarize the dielectric, resulting in surface charges on the inner and outer surfaces of the dielectric.

We can find the displacement ${\mathbf{D}}$ from its relation to the free charge. That is,

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

where ${Q_{f}}$ is the free charge (excluding the bound charge) that is contained within the surface of integration.

In this case, ${Q_{f}=Q}$ and because the system has spherical symmetry, we can take the surface of integration to be a sphere. Since the enclosed free charge is ${Q}$ for any surface with ${r>a}$, we get in this region:

 $\displaystyle 4\pi r^{2}D$ $\displaystyle =$ $\displaystyle Q\ \ \ \ \ (3)$ $\displaystyle \mathbf{D}$ $\displaystyle =$ $\displaystyle \frac{Q}{4\pi r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (4)$

For a linear dielectric ${\mathbf{D}=\epsilon\mathbf{E}=\epsilon_{0}\left(1+\chi_{e}\right)\mathbf{E}}$. We therefore have for the electric field:

$\displaystyle \mathbf{E}=\begin{cases} \frac{Q}{4\pi\epsilon_{0}\left(1+\chi_{e}\right)r^{2}}\hat{\mathbf{r}} & ab \end{cases} \ \ \ \ \ (5)$

The energy is then

 $\displaystyle W$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi Q^{2}}{2\left(4\pi\right)^{2}\epsilon_{0}}\left[\int_{a}^{b}\frac{r^{2}dr}{\left(1+\chi_{e}\right)r^{4}}+\int_{b}^{\infty}\frac{r^{2}dr}{r^{4}}\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Q^{2}}{8\pi\epsilon_{0}}\left[\frac{1}{1+\chi_{e}}\left(\frac{1}{a}-\frac{1}{b}\right)+\frac{1}{b}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Q^{2}}{8\pi\epsilon_{0}\left(1+\chi_{e}\right)}\left(\frac{1}{a}+\frac{\chi_{e}}{b}\right) \ \ \ \ \ (9)$

If we remove the dielectric, this is the same as setting ${\chi_{e}=0}$, so the energy stored in the field of a charged conducting sphere is

$\displaystyle W=\frac{Q^{2}}{8\pi\epsilon_{0}a} \ \ \ \ \ (10)$

# Energy in a dielectric

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 4.4.3 & Problem 4.27.

We now have a look at the analogous formula when we have free charges embedded in a dielectric. To find the work required to assemble only the free charges (excluding the bound charges), we start with the ideas developed originally for finding the energy required to add a point charge to an existing configuration of charges. Assuming that the potential is zero at infinity and we bring in the charge ${q}$ from there to point ${\mathbf{r}}$, the energy required to do this is ${W=qV(\mathbf{r})}$.

In the continuous case, if we bring in a small amount ${\Delta\rho_{f}}$ of free charge, then the energy required to do this is

$\displaystyle \Delta W=\int\left[\Delta\rho_{f}\left(\mathbf{r}\right)\right]V\left(\mathbf{r}\right)d^{3}\mathbf{r} \ \ \ \ \ (1)$

(Note that it is incorrect to start with the formula ${W=\frac{1}{2}\int\rho\left(\mathbf{r}\right)V\left(\mathbf{r}\right)d^{3}\mathbf{r}}$, since that formula was derived for a situation where we considered all the charges (not just the free charges) already in place, so that the potential in the formula is from the final distribution of charges. In our case, we’re essentially retracing the derivation used originally for the case of adding a point charge to an existing distribution and building up the formula from there.)

Since ${\nabla\cdot\mathbf{D}=\rho_{f}}$ (see here), we can write this as

$\displaystyle \Delta W=\int\nabla\cdot\left[\Delta\mathbf{D}\right]V\left(\mathbf{r}\right)d^{3}\mathbf{r} \ \ \ \ \ (2)$

Using the vector calculus identity

$\displaystyle \nabla\cdot\left(\mathbf{A}V\right)=V\nabla\cdot\mathbf{A}+\mathbf{A}\cdot\nabla V \ \ \ \ \ (3)$

and the relation between field and potential ${\mathbf{E}=-\nabla V}$, we get

$\displaystyle \Delta W=\int\nabla\cdot\left[V\Delta\mathbf{D}\right]d^{3}\mathbf{r}+\int\Delta\mathbf{D}\cdot\mathbf{E}d^{3}\mathbf{r} \ \ \ \ \ (4)$

Using the divergence theorem, we can convert the first integral into a surface integral and let the surface go to infinity, making the usual assumption that the quantity ${V\Delta\mathbf{D}}$ goes to zero faster than ${1/r^{2}}$, so that the surface integral also goes to zero. We therefore end up with

$\displaystyle \Delta W=\int\Delta\mathbf{D}\cdot\mathbf{E}d^{3}\mathbf{r} \ \ \ \ \ (5)$

Now for a linear dielectric, ${\mathbf{D}=\epsilon_{0}\epsilon_{r}\mathbf{E}}$, so we can write, for incremental changes

 $\displaystyle \Delta\left(\mathbf{D}\cdot\mathbf{E}\right)$ $\displaystyle =$ $\displaystyle \epsilon_{0}\epsilon_{r}\Delta\left(\mathbf{E}\cdot\mathbf{E}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\epsilon_{r}\Delta E^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\epsilon_{0}\epsilon_{r}E\Delta E\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\mathbf{E}\cdot\Delta\mathbf{D} \ \ \ \ \ (9)$

So if we pull the physicist’s trick of interchanging differentials and integrals, we can write

$\displaystyle \Delta W=\frac{1}{2}\Delta\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r} \ \ \ \ \ (10)$

So in general, the work done to assemble free charge in a dielectric is

$\displaystyle W=\frac{1}{2}\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r} \ \ \ \ \ (11)$

Earlier we saw that the energy of a static charge distribution could be written in terms of the electric field:

$\displaystyle W=\frac{\epsilon_{0}}{2}\int E^{2}d^{3}\mathbf{r} \ \ \ \ \ (12)$

If we compare these two formulas they appear to be incompatible, since if ${\mathbf{D}=\epsilon_{0}\epsilon_{r}\mathbf{E}}$, the first integral becomes

$\displaystyle W=\frac{\epsilon_{0}\epsilon_{r}}{2}\int E^{2}d^{3}\mathbf{r} \ \ \ \ \ (13)$

so that there’s a factor of ${\epsilon_{r}}$ in the first integral that is absent from the second.

The resolution of this dilemma is that we derived the dielectric formula for free charge only. In the absence of a dielectric, all charge is free charge, and ${\epsilon_{r}=1}$ so the two formulas become the same. However, when we have dielectric present, bringing in free charge will cause a polarization of the dielectric, which requires energy, since we are essentially pulling apart the positive and negative bound charges in the dielectric’s atoms. Thus the energy required to place a free charge next to a dielectric is greater (by the factor ${\epsilon_{r}}$) than that required to place the same charge without the dielectric being there.

As an example of the difference between the two formulas, consider again the problem of the uniformly polarized sphere with polarization ${\mathbf{P}}$. We found that the field inside the sphere was uniform:

$\displaystyle \mathbf{E}_{r

Outside the sphere, the potential is

$\displaystyle V_{r>R}=\frac{R^{3}}{3\epsilon_{0}r^{2}}P\cos\theta \ \ \ \ \ (15)$

This is the same as the potential of a pure dipole with dipole moment

$\displaystyle \mathbf{p}=\frac{4}{3}\pi R^{3}\mathbf{P} \ \ \ \ \ (16)$

Thus the field outside the sphere is the same as that of a pure dipole, so we get

$\displaystyle \mathbf{E}_{r>R}=\frac{p}{4\pi\epsilon_{0}r^{3}}\left(2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\theta}\right) \ \ \ \ \ (17)$

Outside the sphere, the energy is the same no matter which formula we use since there is no dielectric here. We therefore have

 $\displaystyle W_{r>R}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\epsilon_{0}\int E^{2}d^{3}\mathbf{r}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p^{2}\epsilon_{0}}{2\left(4\pi\epsilon_{0}\right)^{2}}2\pi\int_{R}^{\infty}\int_{0}^{\pi}\frac{r^{2}\sin\theta}{r^{6}}\left(4\cos^{2}\theta+\sin^{2}\theta\right)d\theta dr\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{12\pi\epsilon_{0}R^{3}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi P^{2}R^{3}}{27\epsilon_{0}} \ \ \ \ \ (21)$

Inside the sphere, since the field is constant, we have, using the total charge formula

 $\displaystyle W_{r $\displaystyle =$ $\displaystyle \frac{\epsilon_{0}}{2}\int E^{2}d^{3}\mathbf{r}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi P^{2}R^{3}}{27\epsilon_{0}} \ \ \ \ \ (23)$

The total energy calculated this way is thus

 $\displaystyle W_{tot}$ $\displaystyle =$ $\displaystyle \frac{4\pi P^{2}R^{3}}{27\epsilon_{0}}+\frac{2\pi P^{2}R^{3}}{27\epsilon_{0}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi P^{2}R^{3}}{9\epsilon_{0}} \ \ \ \ \ (25)$

Calculated using the free charge formula, we use the definition of displacement as ${\mathbf{D}=\epsilon_{0}\mathbf{E}+\mathbf{P}=-\mathbf{P}/3+\mathbf{P}=2\mathbf{P}/3}$. Then

 $\displaystyle W_{r $\displaystyle =$ $\displaystyle \frac{1}{2}\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4\pi P^{2}R^{3}}{27\epsilon_{0}} \ \ \ \ \ (27)$

The total energy in this case thus comes out to ${W_{tot}=0}$, which is because there is no free charge in the problem. However, the derivation above also relied on the dielectric being linear, which isn’t the case in this problem since we have a frozen in polarization with no external electric field. Thus the second result doesn’t really mean much anyway.

# Point charge embedded in dielectric plane

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.25.

The following is a problem that combines the theory of linear dielectrics with the method of images. Suppose we divide 3-d space into two parts, separated by the plane ${z=0}$. The region ${z<0}$ is filled with a dielectric with constant ${\epsilon_{r}}$, while the region ${z>0}$ is filled with a dielectric with a different constant ${\epsilon_{r}'}$. We also place a point charge ${q}$ on the ${z}$ axis at ${z=d}$.

This problem is similar to that of a point charge above a conducting plane that served as the introductory example for the method of images. However, in that case, we were able to replace the conductor by a single point charge ${-q}$ at location ${z=-d}$ and then show that the resulting potential was valid for ${z>0}$. (Inside a grounded conductor, ${V=0}$.) Here, the situation isn’t quite as simple.

The point charge will polarize both dielectrics with the result that there will be bound surface charge where the dielectrics meet. We call this surface charge ${\sigma_{b}}$ and ${\sigma_{b}'}$ for ${z<0}$ and ${z>0}$ respectively. The surface charge is related to the polarization by

$\displaystyle \sigma_{b}=\mathbf{P}\cdot\hat{\mathbf{n}} \ \ \ \ \ (1)$

and for a linear dielectric

$\displaystyle \mathbf{P}=\epsilon_{0}\chi_{e}\mathbf{E} \ \ \ \ \ (2)$

so at the boundary between the dielectrics, we get

 $\displaystyle \sigma_{b}$ $\displaystyle =$ $\displaystyle \mathbf{P}\cdot\hat{\mathbf{n}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}E_{z}\ \ \ \ \ (4)$ $\displaystyle \sigma_{b}'$ $\displaystyle =$ $\displaystyle \mathbf{P}'\cdot\hat{\mathbf{n}'}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\chi_{e}'E_{z}' \ \ \ \ \ (6)$

We need to be careful with the directions of the various vectors here. Suppose that ${q}$ is positive. Then the induced polarization will be such that negative charges are attracted to ${q}$, so that ${\sigma_{b}}$ will consist of negative charge and ${\sigma_{b}'}$ will consist of positive charge. That is, the polarization vector points towards ${q}$ in both cases.

In the calculation of ${\mathbf{P}\cdot\hat{\mathbf{n}}}$ the normal vector to the plane points upwards but in the calculation of ${\mathbf{P}'\cdot\hat{\mathbf{n}}'}$ the normal vector points downwards, while ${\mathbf{P}'}$ still points in the same direction as ${\mathbf{P}}$. That is the reason for the minus sign in the result for ${\sigma_{b}'}$.

The electric field is discontinuous across a layer of surface charge, with the discontinuity given by

$\displaystyle E_{z(above)}-E_{z(below)}=\frac{\sigma}{\epsilon_{0}} \ \ \ \ \ (7)$

The fields due to the surface charges in this problem are therefore

 $\displaystyle E_{z}'$ $\displaystyle =$ $\displaystyle E_{z(above)}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma_{b}+\sigma_{b}'}{2\epsilon_{0}}\ \ \ \ \ (9)$ $\displaystyle E_{z}$ $\displaystyle =$ $\displaystyle E_{z(below)}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\sigma_{b}+\sigma_{b}'}{2\epsilon_{0}} \ \ \ \ \ (11)$

The perpendicular component of the field due to the point charge is, at ${z=0}$:

 $\displaystyle E_{z(q)}$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi\epsilon'}\frac{q}{r^{2}+d^{2}}\cos\theta\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle -\frac{1}{4\pi\epsilon'}\frac{qd}{\left(r^{2}+d^{2}\right)^{3/2}} \ \ \ \ \ (13)$

Here ${\epsilon'=\epsilon_{0}\epsilon_{r}}$ and reflects the fact that the electric field is reduced by the factor of ${\epsilon_{r}}$ inside a dielectric. The variable ${r}$ is the distance from the origin to a point on the ${z=0}$ plane, as usual. The angle ${\theta}$ is that between the ${z}$ axis and a line from ${q}$ to the point on the plane, so that ${\cos\theta=d/\sqrt{d^{2}+r^{2}}}$. This formula is valid on both sides of the boundary.

We now have enough information to write equations for the two surface charges. We get

 $\displaystyle \sigma_{b}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}E_{z}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}\left[-\frac{1}{4\pi\epsilon'}\frac{qd}{\left(r^{2}+d^{2}\right)^{3/2}}-\frac{\sigma_{b}+\sigma_{b}'}{2\epsilon_{0}}\right]\ \ \ \ \ (15)$ $\displaystyle \sigma_{b}'$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\chi_{e}'E_{z}'\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}'\left[\frac{1}{4\pi\epsilon'}\frac{qd}{\left(r^{2}+d^{2}\right)^{3/2}}-\frac{\sigma_{b}+\sigma_{b}'}{2\epsilon_{0}}\right] \ \ \ \ \ (17)$

As a check at this stage, we observe that if we set ${\chi_{e}=\chi_{e}'}$ (that is, we make the same dielectric fill all space, thus eliminating the boundary), and then add these two equations together, we get

$\displaystyle \sigma_{b}+\sigma_{b}'=-\chi_{e}\left(\sigma_{b}+\sigma_{b}'\right) \ \ \ \ \ (18)$

Since ${\chi_{e}\ne0}$ in general, this means that ${\sigma_{b}+\sigma_{b}'=0}$ in this case. This makes sense, since with no boundary between the dielectrics, we would expect there to be no net surface charge.

Returning to the general case, we can solve these two simultaneous equations (by hand or using Maple), and get, using the relation ${\epsilon'=\epsilon_{0}\left(1+\chi_{e}'\right)}$

 $\displaystyle \sigma_{b}$ $\displaystyle =$ $\displaystyle -\frac{\chi_{e}qd}{2\pi\left(2+\chi_{e}+\chi_{e}'\right)\left(r^{2}+d^{2}\right)^{3/2}}\ \ \ \ \ (19)$ $\displaystyle \sigma_{b}'$ $\displaystyle =$ $\displaystyle \frac{\chi_{e}'qd}{2\pi\left(2+\chi_{e}+\chi_{e}'\right)\left(r^{2}+d^{2}\right)^{3/2}}\frac{\left(1+\chi_{e}\right)}{\left(1+\chi_{e}'\right)} \ \ \ \ \ (20)$

Again, we note that if ${\chi_{e}=\chi_{e}'}$, ${\sigma_{b}=-\sigma_{b}'}$.

In general:

$\displaystyle \sigma_{b}+\sigma_{b}'=\frac{\left(\chi_{e}'-\chi_{e}\right)qd}{2\pi\left(2+\chi_{e}+\chi_{e}'\right)\left(1+\chi_{e}'\right)\left(r^{2}+d^{2}\right)^{3/2}} \ \ \ \ \ (21)$

At this stage, if we want to find the potential, we might try a direct integration, since

$\displaystyle 4\pi\epsilon_{0}V(\mathbf{r})=\frac{q}{\left|\mathbf{r}-\mathbf{d}\right|}+\int\frac{\sigma_{b}(\mathbf{r}')+\sigma_{b}'\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}dx'dy' \ \ \ \ \ (22)$

where ${\mathbf{r}}$ is the point at which we wish to find the potential, and the integration variable ${\mathbf{r}'}$ ranges over the plane ${z=0}$. However, I couldn’t find any way of doing this integral (Maple just got stuck) ${V}$.

I’ve never liked the method of images since it’s largely informed guesswork, but anyway…

If we work out the total bound surface charge we get

 $\displaystyle Q$ $\displaystyle =$ $\displaystyle 2\pi\int_{0}^{\infty}\left(\sigma_{b}+\sigma_{b}'\right)rdr\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\chi_{e}'-\chi_{e}\right)q}{\left(1+\chi_{e}'\right)\left(2+\chi_{e}+\chi_{e}'\right)} \ \ \ \ \ (24)$

We pause here to look at a few limiting cases. If ${\chi_{e}'=0}$, so that the upper region becomes a vacuum, we get

$\displaystyle Q=\frac{-\chi_{e}}{2+\chi_{e}}q \ \ \ \ \ (25)$

which is the result given in Griffiths’s book for that case. Further, if we let ${\chi_{e}\rightarrow\infty}$, we get ${Q=-q}$, which is the result for a point charge next to a conducting plane. Also, if we let ${\chi_{e}'\rightarrow\infty}$, we get ${Q=0}$, since in that case we’ve embedded ${q}$ inside a conductor, which shields it completely, so there is no induced charge at the boundary.

Getting back to the method of images, we have to find the potential in the two regions. For ${z>0}$, we have a shielded charge with effective charge ${q/\epsilon_{r}'}$ at ${z=d}$ and we want to replace the surface charge with a point image in the region ${z<0}$. What should be the amount of this image charge? In the case of the conducting plane, the image charge was the same amount as the total surface charge, and at location ${z=-d}$, so we can try that. In this case, we get

$\displaystyle V_{above}=\frac{1}{4\pi\epsilon_{0}}\left[\frac{q}{\epsilon_{r}'\sqrt{r^{2}+\left(z-d\right)^{2}}}+\frac{Q}{\sqrt{r^{2}+\left(z+d\right)^{2}}}\right] \ \ \ \ \ (26)$

For ${z<0}$, we still see the original charge at ${z=+d}$, and we can try replacing the surface charge by a point charge the same distance beyond the plane as we did when calculating the image for the ${z>0}$ region. That is, we place an image charge of size ${Q}$ at ${z=+d}$ (so it coincides with the original point charge). In this case, the potential is

$\displaystyle V_{below}=\frac{1}{4\pi\epsilon_{0}}\frac{q/\epsilon_{r}+Q}{\sqrt{r^{2}+\left(z-d\right)^{2}}} \ \ \ \ \ (27)$

We can check that this potential satisfies the required boundary conditions. First, it must be continuous at ${z=0}$, which it obviously is:

 $\displaystyle V_{above}(z$ $\displaystyle =$ $\displaystyle 0)=\frac{1}{4\pi\epsilon_{0}}\frac{q/\epsilon_{r}'+Q}{\sqrt{r^{2}+z^{2}}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle V_{below}(z=0) \ \ \ \ \ (29)$

Second, we should be able to derive the discontinuity in the field above. We have

 $\displaystyle \left.\frac{\partial V_{above}}{\partial z}\right|_{z=0}$ $\displaystyle =$ $\displaystyle \frac{d}{4\pi\epsilon_{0}\left(r^{2}+d^{2}\right)^{3/2}}\left[\frac{q}{\epsilon_{r}'}-Q\right]\ \ \ \ \ (30)$ $\displaystyle \left.\frac{\partial V_{below}}{\partial z}\right|_{z=0}$ $\displaystyle =$ $\displaystyle \frac{d}{4\pi\epsilon_{0}\left(r^{2}+d^{2}\right)^{3/2}}\left[\frac{q}{\epsilon_{r}'}+Q\right] \ \ \ \ \ (31)$

Since ${E_{z}=-\partial V/\partial z}$ we get

 $\displaystyle E_{z(above)}-E_{z(below)}$ $\displaystyle =$ $\displaystyle \frac{Qd}{2\pi\epsilon_{0}\left(r^{2}+d^{2}\right)^{3/2}}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\chi_{e}'-\chi_{e}\right)qd}{2\pi\epsilon_{0}\left(r^{2}+d^{2}\right)^{3/2}\left(1+\chi_{e}'\right)\left(2+\chi_{e}+\chi_{e}'\right)}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma_{b}+\sigma_{b}'}{\epsilon_{0}} \ \ \ \ \ (34)$

So this checks out as well.

# 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 sphere in uniform electric field

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.23.

We can work out the field inside a sphere of dielectric placed in a uniform electric field in the same way we tackled the cylinder in the last post (for those interested, this is done in Griffiths’s book). However, we can also use a method of successive approximations to get the answer.

The starting point is to assume that the uniform exterior field ${\mathbf{E}_{0}}$ covers all space, including the interior of the sphere. We then know that this uniform field will cause the sphere to become polarized. Since we’re dealing with a linear dielectric, this polarization is

$\displaystyle \mathbf{P}_{0}=\epsilon_{0}\chi_{e}\mathbf{E}_{0} \ \ \ \ \ (1)$

However, a uniformly polarized sphere produces its own electric field, which we need to add onto the original uniform field. We worked out the field inside a uniformly polarized sphere earlier, so we can quote that result. We get

 $\displaystyle \mathbf{E}_{1}$ $\displaystyle =$ $\displaystyle -\frac{1}{3\epsilon_{0}}\mathbf{P}_{0}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\chi_{e}}{3}\mathbf{E}_{0} \ \ \ \ \ (3)$

This new field produces more polarization, so we get

 $\displaystyle \mathbf{P}_{1}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\chi_{e}\mathbf{E}_{1}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{\chi_{e}^{2}}{3}\mathbf{E}_{0} \ \ \ \ \ (5)$

Another round gives the next correction to the field:

$\displaystyle \mathbf{E}_{2}=\frac{\chi_{e}^{2}}{3^{2}}\mathbf{E}_{0} \ \ \ \ \ (6)$

It’s fairly obvious that the general pattern is

$\displaystyle \mathbf{E}_{n}=\left(-1\right)^{n}\left(\frac{\chi_{e}}{3}\right)^{n}\mathbf{E}_{0} \ \ \ \ \ (7)$

so the total field is the sum of all these increments:

$\displaystyle \mathbf{E}=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(\frac{\chi_{e}}{3}\right)^{n}\mathbf{E}_{0} \ \ \ \ \ (8)$

The sum is a geometric series, so we can sum it using the standard formula and we get

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{1}{1+\chi_{e}/3}\mathbf{E}_{0}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2+\epsilon_{r}}\mathbf{E}_{0} \ \ \ \ \ (10)$

where ${\epsilon_{r}=1+\chi_{e}}$ is the dielectric constant. This agrees with the result obtained using Legendre polynomials, as given in Griffiths.