Question

Two conducting concentric, hollow spheres A and B have radii a and b respectively, with A inside B.Their common potentials is V. A is now given some charge such that its potential becomes zero. The potential of B will now be :

• A. 0
• B. V(1 - a/b)
• C. V a/b
• D. V b/a

Answer 1

Answer: (B)

V(1 - a/b)

At first, let charges on $A$ and $B$ be $q_A$ and $q_B$ respectively. The radius be $a$ and $b$ respectively.

So, $V=k{q}_A/a+k{q}_B/b$

and, $V=k{q}_A/b+k{q}_B/b$

where $k=1/4\pi {ϵ}_o$

Thus, $k{q}_A/a+k{q}_B/b=k{q}_A/b+k{q}_B/b\Rightarrow q_A=0$

So, $V=k{q}_B/b$

Let $A$ is now given charge $q_A^{{}^{\prime }}$

Now, new potential on spheres are:

$V_A^{{}^{\prime }}=k{q}_A^{{}^{\prime }}/a+k{q}_B/b=0\Rightarrow q_A^{{}^{\prime }}=-q_{B}a/b$

and, $V_B^{{}^{\prime }}=k{q}_A^{{}^{\prime }}/b+k{q}_B/b=k(-q_{B}a/b)/b+k{q}_B/b=k{q}_B(1-a/b)/b=V(1-a/b)$

So, new potential on $B$ is $V(1-a/b)$