Reading: Griffiths Chapter 3, sections 1 and 2

You may work together and get help from other students. Your solutions must be written in your own words, without looking at someone else’s solutions while you write them.

Don’t forget the 9 points that we are looking for in your solutions (see Moodle).

You may only use Mathematica on the last one. I don’t actually think you need it on any of the problems, so don’t take that as a sign that I think the last one is harder than the others.

In order to make sure you get your context and meaning/make sense points, next to your answers, please put a “c” with a circle around it for context, and an “m” with a circle around it for meaning.

——–These two problems are from the previous chapter—————–

  1. Griffiths, Chapter 2, Problem 2.43.

  2. Suppose the two cylinders in the previous problem are of length, \(L\), and have nearly the same radii, i.e. \(\frac{(b – a)}{b} << 1\). Further suppose that the inner cylinder is displaced along the axis so that its end is a distance, \(x\), from the end of the outer cylinder. Assume that a charge \(Q\) is placed on the outer cylinder and that a charge \(–Q\) is placed on the inner cylinder. How do you think the charge is distributed on the two conductors? Compute the force on the inner cylinder. Is it sucked in or pushed out of the outer cylinder? From your solution you will see that the analysis breaks down when \(x = L\) and \(x = 0\). Speculate as to why this occurs. Hint: Calculate the energy stored in the capacitor (or work done to bring the pieces together) as a function of \(x\) using the results of problem #1. Recall energy in a capactor is \(\frac{1}{2}QV\) just like \(W = \frac{1}{2}\int \rho V\). Then use \(F = -\frac{dW}{dx}\), i.e. the negative of the derivative with respect to \(x\) of the work is the force required to to increase \(x\). Also, there are a bunch of minus signs floating around. I found it helpful to think about which direction I expected the force to be so that I could check my answer.

  3. A point charge, \(Q\), is placed equidistant between two parallel, grounded, conducting plates separated by a distance, \(D\). Assume the sizes of the plates are much larger than their separation.

    a) What are the strengths and positions of the image charges which will give the potential everywhere between the plates? (There are many, many of them! Also notice that I did not say to find the potential everywhere.)

    b) Find an expression for the surface charge density on either plate at the point adjacent to the charge. (The sum you just found is called Catalan’s Constant.) Hint: Somewhat unexpectedly (at least to me) this problem is easier to do in terms of electric field instead of potential. Remember to get the surface charge density on a conductor you just multiply the field by \(\epsilon_0\), so you really only need the electric field at the point in question.

  4. Example 3.2 on page 128 of Griffiths locates the appropriate image charge for the case of a point charge outside a grounded conducting sphere. What are the magnitudes and positions of the image charges (there are two of them) for the case of a point charge outside an isolated, neutral, conducting sphere? What is the potential of the sphere?

  5. Two “infinite”, parallel, conducting plates separated by a distance, \(s\), are at potentials \(0\) and \(V_0\).

    There is a volume charge density, \(\rho = \rho_0\frac{x}{s}\), between the two plates where \(x\) is the distance from the plate at zero potential.

    a) Use Poisson’s equation to find the potential \(V\) in the region between the plates.

    b) What are the surface charge densities on the plates?

    c) Show that the total charge per unit area (on both plates plus the charge between them) vanishes as it must to ensure that \(E = 0\) inside the conductors.