Answer to the Question 05/01

DIPOLE IN A SHELL

The question was:



A dipole of size p has been placed inside a conducting grounded spherical shell or radius R at a distance a from the center of the shell (a < R). The dipole points in radial direction. (The dimensions of the dipole are small enough, so it can be treated as an ideal dipole.) Find the electrical potential inside the shell.

(7/2001) The problem has been solved by Gerome Gosset (16/5/2001) (e-mail jerome.gosset1@libertysurf.fr).

Answer: The potential is a sum of potentials created by the dipole p described in the problem, and an image dipole P=p(R/a)3 located at distance l=R2/a from the center, and a charge (monopole) Q=pR/a2 located at the same position as the image dipole. (p and P have the same direction.)

The solution:



This question can be solved using the "method of images", i.e. by considering a potential created by charges inside the shell and imaginary charges outside the shell. (Since we are only interested in solution for the potential inside the shell, the charges outside the shell can be added at will and adjusted in such a way that correct boundary conditions are satisfied at the boundary of the shell.)

It is quite difficult, to find proper images for a dipole sitting inside the shell. Therefore we shall start with a simpler problem. If a charge q (red) is placed at distance a from the center of the sphere then (as is well known from elementary textbooks), we can find an image charge q' (green) at distance l from the center of the shell such that potential created by both charges vanishes on the shell. To determine the values of q' and l it is enough to write down the potential at two points of the shell (the closest, and the most remote to the charge):

q/(R-a)+q'/(l-R)=0;
q/(R+a)+q'/(l+R)=0;

By solving these equations we immediately find that:

q'=-qR/a;
l=R2/a

By substituting these values into a general expression for a potential created by two charges, we can easily verify that the potential vanishes everywhere on the spherical shell.

Now consider two charges, q and -q, at positions a and a-d, inside the shell when d is small and qd=p. The images of these two charges will be q', mentioned above and q"=qR/(a-d) located at l"=R2/(a-d).

Now take the limit of vanishing d, diverging q, while their product (p) is fixed. First we notice that the sum q'+q" does not vanish in that limit but rather approaches Q=pR/a2. Thus we have an image charge Q at position l. In addition, the dipole moment of the images (=q"(l"-l)) approaches P=p(R/a)3. Thus the image of the dipole p consists of a dipole and a charge located at distance l from the center.


P.S. (10/05) Kirk T. McDonald from Princeton (e-mail mcdonald@puphed.princeton.edu suggested (22/10/2001) a generalized version of the above problem, with dipole pointing in arbitrary direction. The generalized problem and its solution are presented in the follwoing pdf file.
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