Question

There are two conducting hollow spherical shells. One has inner radius a and outer radius b. Other has inner radius c and outer radius d. Inner shell (of radius a and b) has total charge +2q and outer shell has charge +4q.

• A.

• B.

• C.

• D.

Answer 1

Answer: (A)

1

Answer 2

The electric field in different regions is calculated using Gauss's Law and the properties of conductors.

1. For $0 \le r < a$: Inside the inner hollow shell, $Q_{enclosed} = 0$, so $E = 0$.
2. For $a \le r < b$: Inside the conductor of the inner shell, $E = 0$. The charge on the inner surface of the inner shell is 0, and the charge on the outer surface of the inner shell is +2q.
3. For $b \le r < c$: Between the shells, $Q_{enclosed} = +2q$. $E = \frac{1}{4\pi \epsilon_0} \frac{2q}{r^2} = \frac{2kq}{r^2}$.
4. For $c \le r < d$: Inside the conductor of the outer shell, $E = 0$. The charge on the inner surface of the outer shell is -2q, and the charge on the outer surface of the outer shell is +6q.
5. For $r \ge d$: Outside the outer shell, $Q_{enclosed} = +2q + +4q = +6q$. $E = \frac{1}{4\pi \epsilon_0} \frac{6q}{r^2} = \frac{6kq}{r^2}$.

So, the electric field as a function of r is: $E(r) = 0$ for $0 \le r < b$ $E(r) = \frac{2kq}{r^2}$ for $b \le r < c$ $E(r) = 0$ for $c \le r < d$ $E(r) = \frac{6kq}{r^2}$ for $r \ge d$

The graph of E vs r will show:

• E=0 from r=0 up to r=b.
• At r=b, E jumps to $\frac{2kq}{b^2}$.
• From r=b to r=c, E decreases as $1/r^2$.
• At r=c, E drops to 0.
• From r=c to r=d, E=0.
• At r=d, E jumps to $\frac{6kq}{d^2}$.
• For r>d, E decreases as $1/r^2$.

Comparing this behavior with the given options, only the first option shows the correct regions where E is zero and non-zero, and the correct qualitative behavior of E in the non-zero regions.