Problem 2.38

A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b, as in Fig. 2.48). The shell carries no net charge. (a) Find the surface charge density σ at R, at a, and at b. (b) Find the potential at the center, using infinity as the reference point. © Now the outer surface is touched to a grounding wire, which drains off charge and lowers its potential to zero (same as at infinity). How do your answers to (a) and (b) change?

(a) The surface charge density at R, a, and b can be found using Gauss’ Law, which states that the electric flux through any closed surface is proportional to the enclosed charge. Since the metal sphere and shell are conductor, the electric field inside is zero and the charge is distributed on the surface.

At R, the surface charge density is simply the total charge on the sphere divided by the surface area of the sphere, which is σR = q/4πR².

At a and b, the surface charge density is the same as at R because the shell is a conductor and any excess charge on the inner surface will distribute itself on the outer surface so that the net charge inside is zero. Therefore, the surface charge density at a and b is also σa = σb =…