Question

Initially the spheres $\text{A}$ and $\text{B}$ are at potential $V_A$ and $V_B$. The potential of sphere $\text{A}$ when sphere $\text{B}$ is earthed is

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

Answer 1

The correct option is A $V_A-V_B$

Before grounding:
Suppose charge on $\text{A}$ is $q_A$ and $\text{B}$ is $q_B$.

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

$V_B=\frac{k{q}_A}{r_B}+\frac{k{q}_B}{r_B}$

Now, potential difference of shell is given by

$V_A-V_B=k{q}_A[\frac{1}{r_A}-\frac{1}{r_B}]$

We can see that potential difference $(V_A-V_B)$ depends only on charge at $\text{A}$ $(i.e.q_A)$.

After Grounding:
When sphere $\text{B}$ is grounded, charge on $\text{A}$ remains the same, therefore, potential difference remain the same.

Let $V_A^{{}^{\prime }}$ and $V_B^{{}^{\prime }}$ be the new potential.

Then, $V_A^{{}^{\prime }}-V_B{}^{\prime }$ = $V_A-V_B$

$V_B{}^{\prime }=0\because \text{B is grounded}$

$\Rightarrow V_A^{{}^{\prime }}=V_A-V_B$

Why this question ? Concept - sphere $\text{A}$ is isolated, therefore, grounding sphere $\text{B}$ will not change the charge on $\text{A}$.