A thin metallic, spherical shell of radius a has a charge . Concentric with it is another thin, metallic, spherical shell of radius b (where b > a) and charge . We have to find the electric field at radial points r where (a) r < a, (b) a < r < b, and (c) r > b. (d) We have to discuss the criterion one would use to determine how the charges are distributed on the inner and outer surfaces of the shells.

Solution:             Click For PDF Version

It is given that a thin spherical conducting shell of radius a has charge . And that another spherical conducting shell of radius b concentric with the shell of radius a has charge .

As the electric field inside a charged conductor has to be zero, there cannot be any charge on the inner surface of the conducting shell of radius a. This follows from the Gauss’ law which states that flux across any closed Gaussian surface within the conducting shell has to be zero as the electric field inside a conductor is zero, and, therefore, there cannot be any enclosed charge.

As the conducting shells are spherical and concentric, the surface charge distribution on the shells will be uniform.

As the electric field within the conducting shell of radius b is zero, an amount of charge on the shell of radius a will induce charge on the inner surface of the shell of radius b. This will add a surplus charge on the outer surface of the shell of radius b. Thus the total amount of charge on the outer surface of the shell of radius b will be .

(a)

Therefore, the electric field for r < a will be zero.

(b)

Electric field in the region a < r < b will be due to the charge distributed uniformly on the shell of radius a. Therefore, it will be

.

(c)

Electric field in the region r > b will be