A system consists of two concentric thin conducting spherical shells. The inner shell has a radius $

Question

A system consists of two concentric thin conducting spherical shells. The inner shell has a radius $a$ and carries a positive charge $q_1$. The outer shell has a radius $b$ (where $b > a$). An unknown charge $q_2$ is deposited on the outer shell such that the electric potential of the inner shell becomes identically zero. Which of the following statements is/are CORRECT?

Professor PC Thomas Classes · Answer

A is Correct: The potential of the inner shell is $V_a = \frac{q_1}{4\pi\epsilon_0 a} + \frac{q_2}{4\pi\epsilon_0 b}$. For $V_a = 0$, we have $q_2 = -q_1 \frac{b}{a}$.
 B is Correct: The potential of the outer shell is $V_b = \frac{q_1 + q_2}{4\pi\epsilon_0 b}$. Substituting $q_2$, we get $V_b = \frac{q_1}{4\pi\epsilon_0 b} - \frac{q_1}{4\pi\epsilon_0 a} = \frac{q_1}{4\pi\epsilon_0 b} \left(1 - \frac{b}{a}\right)$.
C is Incorrect: By Gauss's Law, the electric field for $a < r < b$ is purely determined by the enclosed charge $q_1$, so $E = \frac{q_1}{4\pi\epsilon_0 r^2}$.
D is Correct: The potential at distance $r$ is $V(r) = \frac{q_1}{4\pi\epsilon_0 r} + \frac{q_2}{4\pi\epsilon_0 b}$. Substituting $q_2 = -q_1 \frac{b}{a}$, we get $V(r) = \frac{q_1}{4\pi\epsilon_0}\left(\frac{1}{r} - \frac{1}{a}\right)$.