Initially the spheres $A$ and $B$ are at potential $V_{A}$ and $V_{B}$ . The potential of sphere $A$ when sphere $B$ is earthed is

V_{A} + V_{B}

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V_{B} - V_{A}

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V_{A}

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V_{A} - V_{B}

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Solution

The correct option is D $V_{A} - V_{B}$
Before grounding:
Suppose charge on $A$ is $q_{A}$ and $B$ 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 $A$ $(i . e . q_{A})$ .

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

Let $V_{A}^{^{'}}$ and $V_{B}^{^{'}}$ be the new potential.

Then, $V_{A}^{^{'}} - V_{B}^{'}$ = $V_{A} - V_{B}$

$V_{B}^{'} = 0 ∵ B is grounded$

$\Rightarrow V_{A}^{^{'}} = V_{A} - V_{B}$

Why this question ?

Concept - sphere $A$ is isolated, therefore, grounding sphere $B$ will not change the charge on $A$ .

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