# A simple electric model: A sheet of charge

#### Prerequisites

In our quest for simple models of distributed charges that produce electric fields that can be simply described, the infinite flat sheet of charge is one of the most useful. Anytime we have a (reasonably) smooth surface with charges spread over it (reasonably) uniformly, the field near to it (but not TOO near so you notice the individuality of each charge) looks a lot like the field near an infinite smooth flat sheet of charge. It's like flat-earth gravity. If you're close enough to the surface, you don't notice the curve. When we're close to a cell membrane, this approximation is pretty reasonable.

It's not quite as easy to do the actual integral as it is for the line charge and it's not as easy to draw a picture since you really need to look in 3D to see the difference between a line and a plane. Let's do the same thing we did in our line charge example, start with finite distributions and make them bigger and bigger.

## A model of an infinite uniform sheet of charge: surface charge density

An infinite sheet of charge sounds cumbersome — and difficult to think about — so let's imagine a finite set first. Looking down from the top, consider having an $L \times L$ square (an area of $L^2$) uniformly spread with a charge of $Q$. If we double the dimensions we now have a $(2L) \times (2L)$ square or four squares. If each is identical to the original, we now have a charge of $4Q$ spread over an area of $4L^2$. Similarly if we go to $(3L) \times (3L)$ we have 9 copies of the original square: a charge of $9Q$ spread over an area of $9L^2$. In each case, the ratio of the charge to the area is a *density of surface charge* — the amount of charge per unit area — of $Q/L^2$. We typically use the symbol $σ$ (sigma) to represent this combination. Our situation (looking down on the sheets) is shown in the figure below.

The *surface charge density*, the charge per unit area, is always $σ$ even as we let the area of the sheet become very large — so large we don't know the actual total area or charge so we model them by an infinite sheet.

## What does the field near an infinite (very large) sheet look like?

For a line charge we could use the symmetry of the line — there was identical charge to the left as there was to the right so the field wouldn't know which way to point. It would have to point straight up. We have a similar result for the plane. If we imagine ourselves some distance above an absolutely flat uniform plane (think flying over a very flat state like Nebraska, for example), we have nothing to choose one direction over another. If somebody standing in front of you gave an argument why the field where you are should point to the left, another person standing behind you giving the identical argument would conclude that the field where you are should point to the right. As a result, the only possibility is that it points straight up. Since every point is the same — for an infinite plane there is always as much plane in any direction as there is in any other — we conclude that the field has to point perpendicular at the plane from every point. Something like the figure shown at the right.

## Why does it look that way?

Just like any distribution of charge, we can think of the field due to the infinite plane as being due to all the little bits of charge in the plane, each contributing to the field at the point we are considering through Coulomb's law for the E field:

$$d\overrightarrow{E} = \frac{k_C(dq)}{r^2} \hat{r} $$

We have considered a tiny bit of area on the plane, $dA$. It will have a tiny bit of charge, $dq = σdA$. This will therefore produce a tiny bit of E field, which we write as $dE$. We now have to add up all the tiny bits of area. This will of course wind up being an integral, which we will not actually do here, but we will see how thinking about it helps us understand what's happening. The picture below shows geometrically how we add up the effects of each little bit of charge on the plane at a particular point — in this case, at the little pink circle a distance *d* up from the plane. The little bit of (blue) charge on the left, labeled $dq$, produces a little blue arrow of E field (at the pink spot) on the right. Similarly, the little bit of (blue charge on the right, labeled dq', produces a little blue arrow of E field (at the pink spot) on the left. Clearly the horizontal components of these two blue arrows produce an E field that points directly upward. As we run around the (gray) ring, we'll get a bunch of these that will add up to the red arrow, $E$.

Now we have to add all the rings.

One of the things you can easily imagine is as the ring gets larger and larger, (the radius, $R$, of the ring grows) the little blue arrows get smaller for two reasons:

- the bits of charge are now farther away ($r$ is bigger) and
- there is more cancellation because the little blue arrows get more and more horizontal.

So we conclude:

- At any point the E field above the plane points up, perpendicular to the plane (as long as the charge density is positive)
- Although there is an infinite amount of total charge, if you get too far away, the effect cancels. In fact it is only a circle of radius R ~ 10d that provides almost all (90%) of the E field. (To get this factor, you actually have to do the integral.)

[Doing the actual integral is a task in Calc 3 — vector calculus. But using the physics, that is, the geometry and the symmetry of the charge, we reduce it to a straightforward Calc 1 integral. If you want to see the details, check out the page E field from a sheet of charge (technical)]

## How does it depend on stuff?

We can use dimensional analysis to see how the E field above the plane is going to depend on σ (Coulombs/m^{2}) and d (meters). It really can't depend on anything else, since we don't have a total area or charge if the plane is infinite. (Perhaps a better statement is: since we only see the effect of the charge within a radius of about $10d$ around the point directly beneath us, we can't tell how much total charge there is, as long as we are not near to an edge.)

we know that what we are doing is adding up contributions to the E field. The magnitude of the E field from a point charge looks like

$$E = \frac{k_Cq}{r^2} $$

the Coulomb constant, times a charge, divided by a length square. Our result from adding a lot of these up will always have the same structure dimensionally. There's always a $k_C$ and it's messy dimensionally so let's factor it out and look at the dimension of $E/k_C$. This is just a charge over a distance squared, or, in dimensional notation:

$$\bigg[\frac{E}{k_C}\bigg] = \bigg[\frac{q}{r^2}\bigg] = \frac{\mathrm{Q}}{\mathrm{L^2}}$$

That is, $E/k_C$ has dimensions of charge divided by length squared. So for a line charge we have to have this form as well. But for an infinite plane charge we don't have a charge to work with. The charge is infinite! We therefore have to work with $σ$. This is a charge per unit area so it has dimensions Q/L^{2}. We have no room * *left for any dependence on $d$! This tells us that the only combination we can make with the correct dimensions from this parameter set is $σ$.

Therefore we must conclude that the $E$ from the sheet of charge is proportional to ($\propto$) $k_C σ$ just by dimensional analysis alone. Actually doing the integral shows that there is actually a factor of $2π$ (as you might guess, this comes from doing the integral over the ring), so near a plane the E field is given by

$$\frac{E}{k_C} \propto \sigma \quad E = 2\pi k_C\sigma$$

Basically, we know an E field looks like a charge divided by two lengths (dimensionally). Since our charge already comes with two lengths, we don't have any room left for $d$ anywhere!

This is quite a remarkable result! How can the field near an infinite plane of charge not depend on the distance from the plane? What about Coulomb's law and all that "$1/r^2$" stuff? The answer is quite sensible from the detail remarked above. The entire area of the plane doesn't contribute significantly to the E field; only a circle right below the point of a radius $\sim 10d$. As we go up to higher values of $d$, we are indeed farther away so that each contribution to the field falls like $1/d^2$. But the are that is contributing is proportional to $d^2$ as well. So as we go up, the relevant area — and therefore the amount of charge that matters — grows like $d^2$. These two changes have the same functional dependence and act in opposite directions, so they cancel, leaving a result that is constant. At least until the edge of our growing circle reaches the edge of the actual plane (which is not really infinite in size). At that point the amount of relevant charge stops growing. Eventually, as $s^2$ gets large enough the E field will start falling off and the plane that looked infinite when we were close to it will look like a point charge when we are far enough away!

This analysis not only gives us the result of what the field looks like close to a plane of charge, it tells us when that approximation will begin to fail.

Note also that the field on the other side of the plane will appear the same — perpendicular to the plane pointing away (as long as σ is positive) and constant, independent of the distance from the plane.

Joe Redish 2/15/12

#### Follow-on

Last Modified: May 24, 2019