Question

A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has total charge +2q, and the outer shell has charge +4q.
(a) Calculate the electric field E⃗ (magnitude and direction) in terms of q and the distance r from the common center of the two shells for (i) r < a; (ii) a < r < b; (iii) b < r < c; (iv) c < r < d; (v) r > d. Graph the radial component of E⃗ as a function of r.
(b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell?

Answer 1

Answer:

(a) We should apply Gauss' Law to each region to find the corresponding electric field. In order to use Gauss' Law, we should draw an imaginary spherical surface inside each region. The electric field passes through this imaginary surface is equal to the charge inside in this surface.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The integral in the left-hand side is equal to the area of the imaginary surface.

(i) r < a:

E = 0, since there is no charge inside the imaginary surface.

(ii) a < r < b:

E = 0, since the electric field is always zero within conductors. Another reason for this is that the inner shell has no charge, and the enclosed charge in the Gauss' Law is zero.

(iii) b < r < c:

The imaginary surface is drawn in this region and it contains +2q charge, which is the net charge of the smaller shell. Therefore, Gauss' Law gives

$E4\pi r^2 = \frac{2q}{\epsilon_0}\\E = \frac{q}{2\pi \epsilon_0 r^2}$

(iv) c < r < d:

The enclosed charge inside the imaginary surface in this region is the sum of the total charge of the smaller shell and the inner surface of the larger shell, which are +2q - 2q = 0. Equivalently, the electric field within a conductor is zero always.

E = 0.

(v) r > d:

The imaginary surface is now drawn to the outermost region and contains the total charge of both spheres, which is +6q. Therefore, Gauss' Law reads

$E4\pi r^2 = \frac{6q}{\epsilon_0}\\E = \frac{3q}{2\pi\epsilon_0 r^2}$

(b)

Since both shells are conductors, the charges are distributed around the inner and outer surfaces, so no charge within the conductor.

(i) Since the innermost region (r > a) has no charge, the inner surface of the smaller shell is also neutral.

(ii) Since the total charge of the smaller shell is +2q, and the inner surface is neutral, the outer surface has a charge of +2q.

(iii) The +2q charge on the outer surface of the smaller shell attracts an equal amount of opposite charges on the inner surface of the larger shell. So, the inner surface of the larger shell has a charge of -2q.

(iv) Since the total charge of the larger shell is +4q, and the inner surface contains -2q, then the outer surface has a charge of +6q.