A Charge Above a Plane

Introduction

The physics problem that I want to tell you about in this lesson dates back at least 175 years—to the early days when physicists were first beginning to understand the laws of electricity and magnetism.

But the solution that I'm going to describe is not the standard treatment that you may have seen before in textbooks. It's much simpler!

We'll be looking at a classic problem from electrostatics, where a point charge is positioned above a large, conducting sheet of metal. And the question is to determine the total electric field that results.

I'll explain why that kind of problem is usually very hard in just a moment, but what I'll show you in this lesson is a remarkably simple symmetry argument that boils the whole solution down to high-school level physics and a handful of lines of algebra.

By contrast, the approach that textbooks usually take to this problem—what's known as the method of images—relies on some significantly more advanced mathematical machinery involving the theory of boundary value problems and uniqueness theorems and so on.¹

That method is extremely clever and powerful in its own right, and it's very much worth learning. But I also think it can leave students feeling a little bit unsatisfied in terms of the physical intuition that it delivers. I previously shared a short video about the method of images, and in the future I might make a full lesson explaining how it works in detail.

But for now, I want to show you the simpler symmetry argument that any clever high school student could discover for themself, and which really shows, physically, why the solution has to be what it is.

I certainly don't imagine that I'm the first person to think of this argument. For all I know, it's explained step-by-step in one of the many E&M textbooks that are out there. But it's certainly not the standard approach to the problem that textbooks take, and I hope it helps you to develop a deeper understanding of the physics involved.

The Problem Setup

Here again is the setup of the problem. We have a point particle with some positive electric charge $Q$, and we've placed it above a grounded, conducting sheet—which we'll treat as if it's infinitely large.

If the conductor weren't there, then the point charge would simply produce the usual electric field of a stationary particle. The arrows all point radially away from the charge, and they get weaker the farther out that you go. Equivalently, the electric field lines directed along those arrows extend out symmetrically in all directions surrounding the charge.

When we bring the conductor back into the picture, though, things get more complicated. That's because the positive charge is going to attract a whole bunch of negative charges to build up underneath it on the upper surface of the sheet. And those negative charges then produce their own contribution to the electric field, with the net result that instead of radiating out symmetrically, the field lines now bend down toward the sheet, stretching out from our original positive charge to the negative charges below.

The reason problems like this are hard, though, is that we don't know ahead of time how those negative charges are going to distribute themselves. Will they be sharply concentrated in a spike directly under the source charge? Or will they spread out more diffusely like a gentle hill?

Since we don't even know where the charges are, it seems like an impossible task to try to calculate the field that they're going to produce!

And in fact, these kinds of problems can't be solved exactly for conductors of most shapes and sizes, and we usually have to resort to computer simulations to model the results, like for the finite-sized block above. But for some special geometries—like an infinite sheet, as well as a conducting sphere, and a number of others—we can in fact solve for the exact electric field.

First, though, we should back up a step, and better understand some of the fundamental physics of conductors and what happens when they're immersed in an electric field.

The Physics of Conductors

To begin with, let's return to the Coulomb field of a single, stationary charge.

If we write $\vec r$ for the vector pointing from the charge to a given point in space, then the electric field at that point is given by

$$ \vec E(\vec r) = \frac{Q}{4\pi\epsilon_0}\frac{\hat r}{r^2}. $$

Positive charges are pushed along the direction of the field, while negative charges are pulled back the opposite way.

But what happens if we have some electric field and then we place a conducting block of metal inside of it? To keep things simple to begin with, let's suppose that the external field is just a constant pointing uniformly to the right.

The defining property of a conductor is that it has lots of electrons inside of it that aren't tightly bound to their home atoms, so that they're basically free to move around wherever they like. Then if we immerse the conductor in an electric field that's pointing to the right, those negative charges get pulled in the opposite direction to the left, while the positive nuclei that they departed from are left behind on the right. And the key point is that all those displaced charges then produce their own electric field, which points in the opposite direction—from the right to the left.

The net result is that charges will continue to pile up on the sides of the conductor until the electric field that they produce inside exactly cancels the external field that we applied on the outside.

The picture above is just a cartoon, of course, but below is a computer simulation of what the field lines actually wind up looking like in this example. Far away from the block, they're the straight, horizontal lines of the constant electric field that we started with. But as they approach the conductor, they bend to meet the positive and negative charges that have built up on its surface, represented by the green and pink gradients.

There are three important general lessons to take away from this example.

1. The electric field inside of a conductor vanishes, regardless of what kind of field we try to apply outside of it.
2. Any remaining net charge, whether it's positive or negative, gets pushed out to the surface of the conductor—the charge density $\rho$ within the interior vanishes. There are still protons and electrons around inside, but they come in equal numbers, such that the total charge inside is zero.
3. The total electric field just outside the conductor has to be perpendicular to the surface. If it weren't—if there were a parallel component to the field—it would again push the charges along the surface until their own field canceled that component out.

And that's exactly what happens in our problem with the infinite sheet: negative charges pile up on the top surface of the conductor, and they deform the electric field produced by $Q$, forcing it to be perpendicular to the surface and vanishing in the interior. In fact, since the plane is infinitely large, it completely screens the electric field above it from penetrating below, so that the field in the entire half-space under the sheet vanishes.

The only other small wrinkle in this problem is that we're assuming the conductor is grounded, which means that we've connected it by a wire to a reservoir like, say, the Earth. The conductor can then draw up whatever negative charge it needs to in order to cancel out the external field from $Q.$ That makes the problem simpler to analyze because there are no residual positive charges to worry about.

But that still leaves us with the question of how to actually compute the field in the upper region above the sheet. The conventional way to tackle these kinds of problems is to first compute a closely related scalar quantity—the electric potential—and then finish solving for the electric field afterward.

Like I mentioned, I might write a separate lesson explaining that approach later on. But for now, we actually already know enough to solve this problem outright using physicists' favorite tool of all: symmetry.

The Solution

As I explained at the beginning, there are two separate contributions to the electric field in this problem.

The first is the ordinary Coulomb field from our original point charge. We know what that is—let's call it $\vec E_Q$. The second contribution is the field produced by the negative induced charges that are spread out on the surface of the sheet. We don't know that part yet, but let's call it $\vec E_\text{induced}$. The total field that we're looking for is their sum:

$$ \vec E = \vec E_Q + \vec E_\text{ind}. $$

The problem, again, is that we don't know how those induced charges are actually distributed, so how can we know what their electric field is going to be?

Here's how we can figure it out. The starting point is the fact I just mentioned: that because we're treating the sheet as if it's infinite, the electric field everywhere underneath it must vanish.

In fact, we could make that argument more rigorous by replacing the thin sheet with a thick slab that fills the entire lower half of the space. Then the electric field inside of the conductor has to vanish, for the reasons we just discussed. The two problems are equivalent, but for the purposes of illustration, it's easier to draw a thin sheet than a thick slab, so that's why I've chosen the former. You'll see both versions of the problem in different textbooks.

Whichever version you choose, the fact that the total electric field vanishes in the lower half space immediately tells us that the induced field under the sheet must be equal-but-opposite to the Coulomb field produced by $Q$:

$$ \vec E_\mathrm{ind} = -\vec E_Q \qquad \text{for $z < 0$}. $$

And therefore, we already know the induced field in the entire lower half of the space. As for figuring it out in the upper half space, that's where the symmetry argument comes into play.

Because the induced charges are confined to the surface of the sheet, the electric field vectors that they produce above and below must be related to each other by a reflection through that plane of symmetry.

For example, the field at a point directly below the center of the sheet points straight upwards along the $z$ axis. The field at its mirror image point directly above is then the reflection of that arrow, pointing straight downward.

Similarly, the field vectors at any other pair of mirror image points above and below the plane will be directed at some angle toward the negatively charged sheet, but always reflected with the opposite $z$ components.

To understand why, consider any little patch of negative charge on the sheet. It produces some contribution to the electric field that's directed along the lines between the points and the patch. What's more, because the two points are the same distance away from the patch, the magnitudes of those two field vectors are identical. The only difference between them is that their $z$ components have flipped signs.

The same goes for every other patch of negative charge making up the sheet, and therefore the total induced field vectors at each pair of mirrored points are indeed related by that same reflection.

And since we already know the induced field below the plane, we can construct the field above by reflecting each arrow to its mirror image. Having done so, we've now figured out the full electric field sourced by the induced charges, without ever even knowing where exactly they all were in the first place!

Now, by adding that contribution to the Coulomb field from our original charge, we can draw the total field $\vec E = \vec E_Q + \vec E_\mathrm{ind}$ everywhere, as pictured below. As expected, the field vanishes below the sheet, and above the field bends downward from $Q$ toward the negative charges induced on the plane, meeting the surface at a right angle.

Alternatively, we can draw the field lines directed along those arrows. Here's what they look like:

It's very helpful to see what the field looks like, but of course we can also work through the few lines of algebra that are required to write down the explicit formula. With $xy$ coordinates set up on the surface of the plane and $z$ running perpendicular to it, the Coulomb field produced by $Q$, whose location is $(x = 0, y = 0, z = d)$, can be expanded as

$$ \vec E_Q(x,y,z) = \frac{Q}{4\pi \epsilon_0} \frac{x \hat x + y \hat y + (z-d) \hat z}{(x^2 + y^2 + (z-d)^2)^{3/2}}. $$

Alternatively, if we define $s = \sqrt{x^2+y^2}$ to be the distance from the $z$ axis, then we can write the above equation more simply in cylindrical coordinates as

$$ \vec E_Q(s,z) = \frac{Q}{4\pi \epsilon_0} \frac{s \hat s + (z-d) \hat z}{(s^2 + (z-d)^2)^{3/2}}, $$

where $s \hat s = x \hat x + y \hat y$.

As for the induced field, below the plane it's equal-but-opposite to $\vec E_Q$:

$$ \vec E_\mathrm{ind}(s,z) = -\frac{Q}{4\pi \epsilon_0} \frac{s \hat s + (z-d) \hat z}{(s^2 + (z-d)^2)^{3/2}} \qquad \text{for } z < 0. $$

Above the plane, the symmetry argument said that the induced field at a point $(x,y,z)$ is given by whatever the field was at the mirror image point $(x,y,-z)$ below the plane, but with its $z$ component flipped. Thus, compared to the equation above, we flip the sign of the $z$ coordinate along with the sign of the $z$ component in order to obtain

$$ \vec E_\mathrm{ind}(s,z) = -\frac{Q}{4\pi \epsilon_0} \frac{s \hat s \textcolor{red}- (\textcolor{red}-z-d) \hat z}{(s^2 + (\textcolor{red}-z-d)^2)^{3/2}} \qquad \text{for } z > 0. $$

All together, then, the induced field is given by

$$ \vec E_\mathrm{ind}(s,z) = \begin{cases} \displaystyle -\frac{Q}{4\pi \epsilon_0} \frac{s \hat s + (z-d) \hat z}{(s^2 + (z-d)^2)^{3/2}} & z < 0\\[2mm] \displaystyle -\frac{Q}{4\pi \epsilon_0} \frac{s \hat s + (z+d) \hat z}{(s^2 + (z+d)^2)^{3/2}} & z > 0. \end{cases} $$

Adding it to our formula for $\vec E_Q$ from above in order to obtain the total $\vec E = \vec E_Q + \vec E_\mathrm{ind}$, we obtain

$$ \vec E(s,z) = \begin{cases} \displaystyle \vec 0 & z < 0\\[2mm] \displaystyle \frac{Q}{4\pi \epsilon_0} \left[ \frac{s \hat s+ (z-d) \hat z}{(s^2 + (z-d)^2)^{3/2}}- \frac{s \hat s+ (z+d) \hat z}{(s^2 + (z+d)^2)^{3/2}} \right] & z > 0. \end{cases} $$

The first term for $z > 0$ is the contribution from $\vec E_Q$, and the second is the contribution from the induced charges. This is the algebraic form of the total field that I showed you above, and it's the standard solution that you'll find in textbooks.

Speaking of the induced charges, now that we know the field, we can work backward to figure out where exactly all those negative charges actually are.

The Surface Charge

To work out the surface charge, we need to know one more property about conductors.

We've already discussed that, in a conductor, any net charge to be found will always be located on the surface, that the electric field inside the conductor vanishes, and that just outside the surface the field must point in the perpendicular direction.

And in fact, there's a very simple relationship between that perpendicular electric field and the amount of charge per unit area $\sigma$ located at that point on the surface.

To derive it, we'll turn to Gauss's law, which I've told you a bit about in the last couple of lessons. Given any collection of charges, it says that if we surround them by a surface and then integrate the electric field over the surface, it will measure for us the total amount of charge contained inside.

$$ \int_\mathrm S \vec E \cdot \mathrm d\vec a = \frac{Q_\mathrm{in}}{\epsilon_0}. $$

To apply it in this case, let's choose the surface to be a very thin box of cross-sectional area $A$ straddling the surface of the conductor. Then the amount of charge contained inside of it is that area $A$ times the charge per area $\sigma$ at that location:

$$Q_\mathrm{in} = \sigma A.$$

Meanwhile, since the electric field vanishes inside the conductor and is perpendicular to it just outside, the only contribution to the integral comes from the top surface of the box, which gives $E_\perp A$. Gauss's law therefore implies that

$$ E_\perp A = \frac{\sigma A}{\epsilon_0}. $$

Canceling out the $A$'s, we learn that the electric field just outside of any conductor is simply given by its surface charge density $\sigma$ at that point, up to a constant:

$$ E_\perp = \frac{\sigma}{\epsilon_0} . $$

Flipping that logic around, if we know the field just outside the conductor, then we can compute the charge density using $\sigma = \epsilon_0 E_\perp.$

In the case of our infinite sheet, we've already computed the electric field, and evaluating it just above the surface of the $xy$ plane gives

$$ \vec E(s, z = 0^+) = \frac{Q}{4\pi \epsilon_0} \left[ \frac{-2d \hat z}{(s^2 + d^2)^{3/2}} \right] . $$

The surface charge density on the sheet is therefore

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

As expected, the induced charge is negative, and I've plotted its magnitude over the $xy$ plane below. It's strongest in the middle, right underneath the original positive charge, and falls off the farther away you get from the $z$ axis as $1/s^3$.

Now that we know exactly where all the charges are, we could go back and re-solve the whole problem for the electric field. But of course, we'd get the same answer that we already found by a much simpler argument!

Looking Ahead: The Method of Images

As I mentioned at the beginning, the approach we've taken here is not how you'll typically see this problem solved in textbooks. Instead, the standard treatments use another, very clever trick called the method of images to arrive at the same answer. And in fact, the solution that we've obtained for the electric field gives a hint as to where that other method comes from.

It might have struck you, looking at the picture of the electric field above the sheet, that you've probably seen a very similar-looking field before. In your intro physics classes, I'm guessing you were made to sketch pictures of the fields produced by different arrangements of charged particles, like the field of a pair of equal-but-opposite charges below.

Now, take your hand and cover up the bottom half of that picture; what remains in the top half looks identical to the field lines that we just obtained for the charge sitting above the infinite conducting sheet!

In fact, they don't just look identical—they are identical. The exact formula that we derived earlier for the electric field above the sheet,

$$ \vec E = \frac{Q}{4\pi \epsilon_0} \left[ \frac{s \hat s+ (z-d) \hat z}{(s^2 + (z-d)^2)^{3/2}}- \frac{s \hat s+ (z+d) \hat z}{(s^2 + (z+d)^2)^{3/2}} \right] \qquad \text{for } z > 0, $$

is simply the sum of two point-charge fields: that of a point charge $Q$ located on the $z$ axis at $z = d$, along with a second point charge $-Q$ located at $z = -d.$

This connection is no mere coincidence, and it's the key insight behind the method of images. Instead of fretting over how the induced charges on the surface of the conductor are distributed, we could simply have erased the conductor altogether and replaced it with a single mirror image charge $-Q$ located on the opposite side of the plane!

Fully understanding why that trick works, though, requires quite a bit of more advanced math than we've had need of so far. I may share a follow-up lesson explaining the method of images in more detail; let me know if you'd like to see it!

1. The image method solution originally dates back, as far as I know, to William Thomson (aka "Lord Kelvin") in 1848. The relevant section of his paper On the Mathematical Theory of Electricity in Equilibrium is available here. ↩