The divergence theorem? Gauss’ law? If you’ve had college physics, at least the second is familiar to you. Now those two things are really fascinating, and I’m going to put a math-lite layman’s explanation in a future post. But hey, it’s spring semester finals time and that means all you people in a second-semester physics class are going to have at least a problem or two about this. Shall we practice one, straight from the final I just finished grading? Of course!
A nonconducting sphere of radius R has a uniform charge density ?. At the center of the sphere is a point particle with charge Q. What is the electric field E at points r < R?
This one’s pretty easy, but a quite large number of students missed it at least partially. Try it for yourself, then check the answer below.
We have spherical symmetry, and that immediately suggests we can use a sphere as our Gaussian surface. So draw the imaginary Gaussian sphere with radius r inside the actual charged sphere, and write down Gauss’ law.
Now we know E is going to be emerging perpindicularly to our Gaussian surface, thus the dot product is just E da. And E is constant over the surface, so we just have
But the integral is now just the total area of the sphere: 4 pi r^2. Thus,
And the charge enclosed is the point charge Q plus however much of the solid sphere’s charge we enclose is. And that latter quantity is just the volume of the Gaussian sphere times the charge density ?. Substituting,
Divide to solve for E and we’re done!
If you wanted, you could go ahead and divide this out. But as it is, it’s a complete solution which will give you E at any point inside the sphere.
Note two more things. First, the radius of the sphere R is a red herring: we don’t need it to find the field inside. Second, that wouldn’t be true if r > R, because we’d no longer be including more and more charge in the Gaussian sphere as r increased. Then the total charge enclosed would just be Q + ?(4/3 ? R^3).