# A simple electric model: A spherical shell of charge

#### Prerequisites

We've now considered the electric fields arising from a number of "simple" distributions of charge — one with no extension or 0 dimensions: the point charge (Coulomb's law), one in 1 dimension (a uniform line charge), and one in 2 dimensions (a uniform sheet of charge). These simple models are useful in a number of ways. First, they show clearly how multiple charges produce dependences on distance that look different from what we see with the basic force law — Coulomb's law. Second, they can serve as approximate models for the fields close to a string or sheet of charge (like a DNA molecule or the surface of a membrane).

So we've done 0, 1, and 2D. Can we go to 3D? Our 1 and 2D models had the property that we could imagine them extending to infinity in all directions. Of course this just means that we are so close to the line or sheet of charge that their edges are too far away to contribute significantly — but it makes the calculations much easier. In 3D if we took a distribution of charge to infinity we would fill all space. While this is interesting (and will be useful when we think about charges inside a medium, like a fluid), to understand what's going on, it will be useful to think about a uniform non-infinite 3D system that has a high symmetry: a spherical distribution of charge.

Since we'll learn that a sphere of charge looks just like a point charge from outside, let's consider the simple model of a thin shell of uniform charge. We'll see that this toy model gives us some interesting insights.

We've been drawing our "point charges" as if they were little spheres, so we might guess that if we are far enough away from a sphere of charge the electric field will look just like Coulomb's law: $E = k_Cq/r^2$ where $r$ is the distance from the center of the sphere. But the result is more interesting and even stronger.

Let's consider not a sphere of charge, but a thin shell of radius $r$ and thickness $dr$ (where $dr << r$). There are two results:

1. Everywhere outside a thin shell of uniform charge, the electric field due to the charge of the shell is exactly the same as the field of a point charge that has the same total charge and is placed at the center of the shell.
2. Everywhere inside a thin shell of uniform charge, the electric field due to the charge of the shell is 0.

These are quite striking results — especially the second one. The fields look something like this:

The field on the left is that due to a point charge at the origin. (Of course there should be an arrow at every point, since the charge creates an E field everywhere, but if we drew them all the space would be filled with red and we couldn't make sense of what it happening.) The field at the right is that due to a thin spherical shell of charge (shown) that has the same total charge as the charge on the left. Outside the shell, everything is exactly the same as in the picture on the left. In fact, if you only had an electric field probe, you couldn't tell whether you had the situation on the left or on the right!

But inside, the result is strikingly different. The point charge's field gets very large as you approach the origin, whereas inside the shell, you see no field at all.

While it is somewhat challenging to prove result 1 (it involves some heavy vector integrals), we can see where result 2 comes from with a very straightforward argument.

Consider a shell with a uniform charge distribution (red in the figure below) and look at the field at a point somewhere inside (the blue dot).  The field felt by the blue dot can be obtained by breaking the shell up into lots of tiny pieces, using Coulomb's law, and integrating the vectors, just like we did for the line of charge and the sheet of charge. Doing this integral directly would be painful. But we can try something simpler. We can use the functional dependence of Coulomb's law to build the result fairly easily.

If we were at the center of the sphere, we would expect to get 0. Every bit of charge on the surface would produce a bit of E field that would be cancelled by the bit of charge on the opposite side that's equally far away. It would be like sitting halfway between two equal positive charges. They would each produce an E field where we were, but the two fields would be equal and opposite and the sum would be 0.

Let's look at the contribution of a bit of the surface to the field felt at the blue dot. From the dot, send out a small narrow cone to the nearer surface. If the surface charge density on the sphere is $σ$, the bit of surface area, $dA_1$, has a charge $Q_1 = σdA_1$. It will produce a field of magnitude

$$dE_1 = \frac{k_C Q_1}{r^2_1} = \frac{k_C \sigma dA_1}{r^2_1}$$

where $r_1$ is the distance to $Q_1$. (We are using "$dA$" instead of "$ΔA$" to indicate that we mean them to be very small so that we could, if we needed, do an integral to add them up. Don't get confused with any distance in the problem.)

If we follow the lines we used to find the charge $Q_1$ backwards to the other side of the shell, we'll find another charged area, $Q_2$. This charge will also produce an E field at the blue dot, but in the opposite direction. It will produce a field of magnitude

$$dE_2 = \frac{k_C Q_2}{r^2_2} = \frac{k_C \sigma dA_2}{r^2_2}$$

where$r_2$ is the distance to $Q_2$.

How do those two E fields compare? If we look at the little cones picking out those two areas, we see that they create similar triangles, so

$$\frac{s_1}{r_1} = \frac{s_2}{r_2}$$

But the areas $dA_1$ and $dA_2$ will be proportional to the square of $s_1$ and $s_2$ since we are really imagining a little cone rather than a triangle. (Imagine, for example, that we had the same triangle in the third dimension so the line representing $s_1$ and $s_2$ are really little squares with sides $s_1$ and $s_2$.) This tells us that

$$\frac{dA_1}{dA_2} = \frac{s_1^2}{s_2^2}.$$

So if we square our similar triangle relation we get:

$$\frac{s_1^2}{r_1^2} = \frac{s_2^2}{r_2^2}$$

But since our areas are proportional to the squares of the $s$'s we can write this as

$$\frac{dA_1}{r_1^2} = \frac{dA_2}{r_2^2}$$

or

$$\frac{Q_1}{r_1^2} = \frac{Q_2}{r_2^2}$$

The bits of charge scale like the square of the distance: the closer one is smaller but closer. The contribution of each of the two bits of charge to the E field has the same magnitude but opposite directions.

So the E fields produced by the two bits of charge will cancel. Since this is true for every bit of charge, the total result has to be 0 and result 2 follows.

Joe Redish  2/16/16

Article 649