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A simple electric model: a sphere of charge

Page history last edited by Joe Redish 4 years, 11 months ago

Working Content > Electric Field > Motivating simple electric models

 

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. 

 

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 = kcq/r2 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 << ). 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.

 

This is quite a striking result -- 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. This would be painful. But we can try something simpler. 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, dA1, has a charge Q1 = σdA1. It will produce a field of magnitude

 

dE1 = kCQ1/r12 = kCσ dA1/r12 

 

where r1 is the distance to Q1. (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 Q1 backwards to the other side of the shell, we'll find another charged area Q2. This charge will also produce an E field at the blue dot, but in the opposite direction. It will produce a field of magnitude

 

dE2 = kCQ2/r22 = kCσ dA2/r22 

 

where r2 is the distance to Q2.

 

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

 

s2/r2 = s1/r1

 

But the areas dA1 and dA2 will be proportional to the square of s1 and s2 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 s1 and s2 are really squares with those dimensions.) This tells us that

 

dA1/dA2 = s12/s22.

 

So if we square our similar triangle relation we get:

 

 

s22/r22 = s12/r12

 

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

 

dA2/r22 = dA1/r12

or

Q2/r22 = Q1/r12.

 

So the E fields produced by the two bits of charge on the opposite side of the shell will have the same magnitude, be in the opposite directions, and cancel. Since this is true for every bit of charge, the total result has to be 0.

 

 

 

Joe Redish  2/16/16

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