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Question

Two concentric shells have radii R and 2R charges $q_{A}$ and $q_{B}$ and potentials 2V and (3/2) V respectively. Now shell B is earthed and let charges on them become $q_{A}$ ' and $q_{B}$ '. Then :

q A / q B = 1 / 2

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q A^{'} / q B^{'} = 1

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potential of A after earthing becomes (3/2)V

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potential difference between A and B after earthing becomes V/2

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Solution

The correct options are
A $q A / q B = 1 / 2$
D potential difference between A and B after earthing becomes V/2
Before grounding,
$2 V = k q_{A} / R + k q_{B} / 2 R$ , where $k = 1 / 4 π ϵ_{o}$
and, $3 V / 2 = k q_{A} / 2 R + k q_{B} / 2 R$
$\Rightarrow q_{A} / q_{B} = 1 / 2$
and $V = k q_{A} / R - (1)$

After grounding of $B$ , let charges on $A$ and $B$ be $q_{A}^{^{'}}$ and $q_{B}^{^{'}}$ respectively.
As $A$ is isolated, $q_{A}^{^{'}} = q_{A}$
so option A is correct.

Now, new potential on spheres are:
$V_{A}^{^{'}} = k q_{A}^{^{'}} / R + k q_{B}^{^{'}} / 2 R = k q_{A} / R + k q_{B}^{^{'}} / 2 R$
and, $V_{B}^{^{'}} = k q_{A}^{^{'}} / 2 R + k q_{B}^{^{'}} / 2 R = k q_{A} / 2 R + k q_{B}^{^{'}} / 2 R = 0 \Rightarrow$
$q_{B}^{^{'}} = - q_{A} = - q_{A}^{^{'}}$
$\Rightarrow q_{A}^{^{'}} / q_{B}^{^{'}} = - 1$
so option B is incorrect.

So, $V_{A}^{^{'}} = k q_{A} / R - k q_{A} / 2 R = k q_{A} / 2 R = V / 2$ using $(1)$
so option C is incorrect
$\Rightarrow V_{A}^{^{'}} - V_{B}^{^{'}} = V / 2$
so option D is correct.

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Two concentric shells have radii R and 2R charges qA and qB and potentials 2V and (3/2) V respectively. Now shell B is earthed and let charges on them become qA' and qB'. Then :

Two concentric shells have radii R and 2R charges $q_{A}$ and $q_{B}$ and potentials 2V and (3/2) V respectively. Now shell B is earthed and let charges on them become $q_{A}$ ' and $q_{B}$ '. Then :