Question

Initially the spheres A and B are at potential $V_A$ and $V_B$ respectively. Now sphere B is earthed by closing the switch. The potential of A will now become :

• A. $V_A-V_B$
• B. 0
• C. $V_B$
• D. $V_A$

Answer 1

Answer: (A)

Before grounding, let charges on $A$ and $B$ be $q_A$ and $q_B$ respectively, and radius be $r_A$ and $r_B$ respectively.

So, $V_A=k{q}_A/r_A+k{q}_B/r_B$

and, $V_B=k{q}_A/r_B+k{q}_B/r_B$

where $k=1/4\pi {ϵ}_o$

After grounding of $B$, let charges on $A$ and $B$ be $q_A^{{}^{\prime }}$ and $q_B^{{}^{\prime }}$ respectively.

As $A$ is isolated, $q_A^{{}^{\prime }}=q_A$

Now, new potential on spheres are:

$V_A^{{}^{\prime }}=k{q}_A^{{}^{\prime }}/r_A+k{q}_B^{{}^{\prime }}/r_B=k{q}_A/r_A+k{q}_B^{{}^{\prime }}/r_B$

and, $V_B^{{}^{\prime }}=k{q}_A^{{}^{\prime }}/r_B+k{q}_B^{{}^{\prime }}/r_B=k{q}_A/r_B+k{q}_B^{{}^{\prime }}/r_B=0\Rightarrow q_B^{{}^{\prime }}=-q_A$

So, $V_A^{{}^{\prime }}=k{q}_A/r_A-k{q}_A/r_B=k{q}_A/r_A+k{q}_B/r_B-(k{q}_A/r_B+k{q}_B/r_B)=V_A-V_B$