Boundary-Value Problems in E&M

A charge near a metal plate

Hold a small positive charge a centimetre above a large sheet of grounded metal and let go. It accelerates downward. Nothing in the metal was charged to begin with, and the sheet as a whole is still neutral, so what is pulling?

The charge pulled electrons in the metal toward the spot beneath it. Those electrons pile up into a thin layer of negative surface charge, and that layer attracts the charge. To find the force you need the surface charge density $\sigma$ at every point of the plate. But $\sigma$ arranges itself in response to the total field, and the total field includes the field of $\sigma$ itself. You cannot integrate your way in from the outside. This is a boundary-value problem: you are not told the charge everywhere, you are told a condition on a surface, and you have to build the field that satisfies it.

In the coupled oscillator lessons you found the motion by guessing a form and checking it satisfied the equations. The same move works here, and here it comes with a proof that the guess is the only answer.

Uniqueness is what licenses the guess

Everywhere outside the conductors, the potential obeys Poisson's equation, $\nabla^2 V = -\rho/\varepsilon_0$, which reduces to Laplace's equation $\nabla^2 V = 0$ wherever there is no charge. The uniqueness theorem says: if you specify the charge inside a region and the value of $V$ on every boundary of that region, then at most one potential satisfies both.

Read what that buys you. You are allowed to find the solution any way you like, including inventing fictitious charges out of thin air. If your construction (a) puts the same real charges in the region you care about and (b) produces the correct potential on the boundary, it is the answer. There is nothing left to check. A guess that passes those two tests is a proof.

One immediate consequence is worth having in your pocket. Hollow out an empty cavity inside a conductor held at $0$ V. Inside the cavity there is no charge, so $\nabla^2 V = 0$, and on the whole boundary $V = 0$. The constant function $V = 0$ satisfies both. By uniqueness it is the potential, so $\vec{E} = 0$ inside the cavity, no matter what you do with charges outside. That is electrostatic shielding, proved in three lines.

The image charge for a grounded plane

Put $+q$ at height $d$ above an infinite grounded plane. Grounded means the plane is wired to the earth, free to gain or lose charge, and held at $V = 0$.

Delete the metal. Place $-q$ at the mirror point, a distance $d$ below where the plane was. Now look at any point $P$ on the old plane. It sits at the same distance $r$ from $+q$ and from $-q$, because they are mirror images. The potentials add:

$V_P = \frac{kq}{r} + \frac{k(-q)}{r} = 0$

That holds at every point of the plane. The fake charge sits below the plane, so above the plane the only real charge is still $q$. Both conditions of the uniqueness theorem are met, so above the plane the field of these two point charges is the true field.

The force follows at once. The charges are $2d$ apart, so

$F = \frac{kq^2}{(2d)^2} = \frac{kq^2}{4d^2}$

attractive, pointing at the plane. Note the $2d$: forgetting it and writing $kq^2/d^2$ is the single most common error in this topic.

The induced surface density comes from the field just outside a conductor, $E = \sigma/\varepsilon_0$ pointing along the outward normal. At a point on the plane a horizontal distance $s$ from the foot of the charge, both point charges lie a distance $\sqrt{s^2+d^2}$ away, the horizontal components cancel, and the vertical components add:

$\sigma(s) = -\frac{qd}{2\pi (s^2+d^2)^{3/2}}$

It is negative everywhere, largest in magnitude directly beneath the charge, and it falls off like $1/s^3$ far out. Integrate it over the plane and you get exactly $-q$: the plane pulls up precisely enough charge to terminate every field line leaving $q$.

Example 1: an electron on a copper plate

An electron sits $2.00$ nm above a large grounded copper plate. Find the force on it.

Charge magnitude $e = 1.602\times10^{-19}$ C, $d = 2.00\times10^{-9}$ m.

$F = \frac{ke^2}{4d^2} = \frac{(8.99\times10^{9})(1.602\times10^{-19})^2}{4(2.00\times10^{-9})^2}$

The numerator is $8.99\times10^{9} \times 2.566\times10^{-38} = 2.307\times10^{-28}$. The denominator is $4 \times 4.00\times10^{-18} = 1.60\times10^{-17}$. Dividing, $F = 1.44\times10^{-11}$ N, directed into the plate. This attraction is the image force that makes it take work to pull an electron out of a metal.

Energy: the factor of one half

Here is the trap. Two real charges $+q$ and $-q$ separated by $2d$ have interaction energy $-kq^2/2d$. That is not the energy of the charge-plane system.

Bring the charge in from infinity to height $d$ and add up the work you do against the attraction. At height $z$ the force on you is $kq^2/4z^2$ toward the plane, so you do work

$W = -\int_{\infty}^{d} \frac{kq^2}{4z^2}\,dz = -\frac{kq^2}{4d}$

Half of the two-charge answer. The reason is physical: the image is not an independent object you have to drag around. It moves because the real charge moves, and it costs nothing to move. Only the real charge's trip counts.

So the energy of the system is $U = -kq^2/4d$, and the work an external agent does moving the charge from $d_1$ to $d_2$ is $U(d_2) - U(d_1)$.

Example 2: work to double the height

A charge $q = +2.0$ nC is held $5.0$ cm above a grounded plane and moved slowly to $10.0$ cm. Find the work done by the external agent.

First $kq^2 = (8.99\times10^{9})(2.0\times10^{-9})^2 = 3.596\times10^{-8}$ N$\cdot$m$^2$. Then

$U = -\frac{kq^2}{4d} \quad\Rightarrow\quad W = \frac{kq^2}{4}\left(\frac{1}{d_1} - \frac{1}{d_2}\right)$

with $d_1 = 0.050$ m and $d_2 = 0.100$ m. The bracket is $20.0 - 10.0 = 10.0$ m$^{-1}$. So $W = (3.596\times10^{-8}/4)(10.0) = 8.99\times10^{-8}$ J, or $89.9$ nJ. Positive, as it must be: you are pulling against an attraction.

Corners, wedges and plates: more than one image

A single plane needs one image. Bring in a second grounded plane and each image spoils the boundary condition on the other plane, so you reflect again.

Two planes meeting at a right angle, with $q$ at $(a,a)$, need three images: $-q$ at $(-a,a)$, $-q$ at $(a,-a)$, and $+q$ at $(-a,-a)$. Check a point on the vertical wall: the first and third images pair off with the real charge and the second image, and $V = 0$. The same check works on the horizontal wall. Three images, and the series closes.

It closes because $90^\circ$ divides into $360^\circ$ a whole number of times. For a wedge of opening angle $\theta$, the reflections generate $360^\circ/\theta$ charges in total counting the real one, so you need $360^\circ/\theta - 1$ images. A $30^\circ$ wedge needs $12 - 1 = 11$. If $360^\circ/\theta$ is not an integer the images never close up and the method fails outright.

Two parallel grounded plates are the case where the series never closes. The charge reflects in the left plate, that image reflects in the right plate, and so on forever: an infinite ladder of alternating charges spaced by twice the plate separation. The force is then an infinite sum, which converges quickly because distant images fall off like $1/n^2$. The same infinite ladder appears for a conducting sphere near a plane, and it is why the capacitance of that pair is an infinite series rather than a closed formula.

One special case you can do in your head: a charge exactly midway between two grounded plates. Every image on the left has a partner at the mirror distance on the right, so every force cancels in pairs and the net force is zero. It is an unstable equilibrium.

Example 3: charge between two grounded plates

$q = +4.0$ nC sits between grounded plates $12.0$ cm apart, $3.0$ cm from the left plate.

Set up coordinates with the left plate at $x = 0$, the right at $x = L = 0.120$ m, and the charge at $x_0 = 0.030$ m. Reflecting in the left plate puts $-q$ at $-x_0$. Reflecting that in the right plate puts $+q$ at $2L + x_0$, and so on. The full set is $+q$ at $2nL + x_0$ and $-q$ at $2nL - x_0$ for every integer $n$, with $n = 0$ omitted from the first family because that is the real charge.

The nearest few dominate. The $-q$ image at $x = -0.030$ m is $0.060$ m away and pulls left with

$F_1 = \frac{kq^2}{(0.060)^2} = \frac{1.438\times10^{-7}}{3.60\times10^{-3}} = 4.00\times10^{-5}\ \text{N}$

The $-q$ image at $x = 2L - x_0 = 0.210$ m is $0.180$ m away and pulls right with $1.438\times10^{-7}/0.0324 = 4.44\times10^{-6}$ N. The $+q$ image at $x = -2L + x_0 = -0.210$ m is $0.240$ m away and pushes right with $1.438\times10^{-7}/0.0576 = 2.50\times10^{-6}$ N. Already the running total is $4.00 - 0.444 - 0.250 = 3.31\times10^{-5}$ N to the left, and the remaining terms add back about $10\%$. Summing the whole ladder gives $3.66\times10^{-5}$ N, or $36.6$ $\mu$N, toward the near plate.

You will not be asked to sum an infinite series by hand under time pressure. You will be asked to write down the ladder correctly and evaluate enough terms to answer to two figures.

The grounded sphere

The sphere is the other geometry worth memorising. Put $q$ a distance $a$ from the centre of a grounded sphere of radius $R$, with $a > R$. One image still works, but it is neither equal in size nor at the mirror point. Put

$q' = -\frac{R}{a}\,q \qquad \text{at a distance} \qquad b = \frac{R^2}{a}$

from the centre, on the line joining the centre to $q$. Those two numbers are forced: writing the potential at a surface point at angle $\alpha$ from that line and demanding it vanish for all $\alpha$ gives exactly this pair.

Three things follow. The total induced charge on a grounded sphere is $q'$ itself, since all the field lines from $q$ that land on the sphere must terminate on that much charge. The force is the attraction between $q$ and $q'$, separated by $a - b$:

$F = \frac{kqq'}{(a-b)^2} = -\frac{kq^2 R a}{(a^2-R^2)^2}$

And the energy, halved again for the same reason as before, is $U = -kq^2R/\left[2(a^2-R^2)\right]$.

Two variants come up constantly:

• Isolated and neutral. The sphere cannot take on net charge $q'$, so add a second image $+\frac{R}{a}q$ at the centre. A charge at the centre gives the same potential at every surface point, so the surface is still an equipotential (just not at zero), and the net image charge is now zero.
• Held at potential $V_0$. Put a charge $Q_c = RV_0/k$ at the centre instead. That alone would put the sphere at $V_0$, and the original image keeps the surface uniform.

The same algebra covers a charge inside a grounded spherical cavity of radius $R$ at distance $s < R$: the image is $-Rq/s$ at $R^2/s$, now outside the cavity. The force magnitude is $kq^2Rs/(R^2-s^2)^2$, pulling the charge toward the nearest wall.

Example 4: an isolated neutral sphere

$q = +2.0$ nC sits $15.0$ cm from the centre of an isolated uncharged sphere of radius $5.0$ cm. Find the force.

Image one: $q' = -(0.05/0.15)(2.0) = -0.667$ nC at $b = 0.0025/0.15 = 0.0167$ m. Image two: $+0.667$ nC at the centre.

Attraction to image one, over a separation $a - b = 0.150 - 0.0167 = 0.1333$ m:

$F_1 = \frac{(8.99\times10^{9})(2.0\times10^{-9})(0.667\times10^{-9})}{(0.1333)^2} = \frac{1.199\times10^{-8}}{0.01778} = 6.74\times10^{-7}\ \text{N}$

Repulsion from image two, at the full $0.150$ m:

$F_2 = \frac{1.199\times10^{-8}}{0.0225} = 5.33\times10^{-7}\ \text{N}$

Net $6.74\times10^{-7} - 5.33\times10^{-7} = 1.42\times10^{-7}$ N, or $142$ nN, still attractive. A neutral conductor always attracts a nearby charge.

Lines, wires and dielectrics

The same mirror works for a line charge. A wire with linear density $\lambda$ a height $h$ above a grounded plane images to $-\lambda$ at depth $h$. Using the line-charge field $E = 2k\lambda/r$, the attraction per unit length is

$\frac{F}{L} = \frac{2k\lambda^2}{2h} = \frac{k\lambda^2}{h}$

Because the potential of the pair is $V = 2k\lambda\ln(r_-/r_+)$, the wire surface sits at $V = 2k\lambda \ln(2h/a)$ relative to the plane for a wire of radius $a \ll h$, so the capacitance per unit length is

$\frac{C}{L} = \frac{\lambda}{V} = \frac{2\pi\varepsilon_0}{\ln(2h/a)}$

Two identical wires of radius $a$ with axes a distance $D$ apart are the same problem cut in half: the midplane between them is at zero potential and can be replaced by a grounded plane, so each wire has $C/L$ to that plane with $2h = D$, and the two in series give $C/L = \pi\varepsilon_0/\ln(D/a)$.

Finally, a charge $q$ in vacuum a height $d$ above a semi-infinite dielectric with relative permittivity $\varepsilon_r$. The dielectric is not a conductor and the boundary condition is different, but the construction survives with a weakened image:

$q' = -q\,\frac{\varepsilon_r - 1}{\varepsilon_r + 1}$

at the mirror point. When $\varepsilon_r \to \infty$ you recover the perfect conductor, $q' = -q$. The bound surface charge density follows the same shape as $\sigma$ did for the conductor, with $q$ replaced by $q'$.

Recognising a boundary-value problem

Reach for images when a problem has all three of these features.

A conductor, or a dielectric surface, with a symmetric shape. Plane, sphere, cylinder, or a corner built from planes. Anything else and the images will not close.

A stated boundary condition rather than a stated charge distribution. "Grounded", "held at 200 V", "isolated and neutral". The word grounded is the loudest possible signal: it means $V = 0$ on that surface, which is exactly what an image is built to enforce.

A question about force, energy, induced charge, or capacitance on a charge that sits outside the conductor. Gauss's law cannot touch these, because the induced charge is not uniform and there is no surface on which $E$ is constant.

Then run the checklist. Identify the region you care about. Put images only outside it. Verify $V$ on the boundary. Use the full formula for force, and half of it for energy. And before you write the number down, ask whether the sign makes sense: a grounded or neutral conductor always pulls a charge in, never pushes it away.

These tools turn a geometry into an algebra problem. Next you will do the same for spacetime, replacing awkward velocity transformations with a single object that carries its own bookkeeping, in four-vectors and the relativistic Doppler effect.