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Two concentri...

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 $q_{A} / q_{B} = 1 / 2$
C $| q_{A}^{'} | / | q_{B}^{'} | = 1$
D Potential difference between A and B after earthing becomes V/2
Potential of A:
$2 V = \frac{q_{A}}{R} + \frac{q_{B}}{2 R}$
Potential of B:
$\frac{3 V}{2} = \frac{q_{A}}{2 R} + \frac{q_{B}}{2 R}$
Solving both,
$q_{A} = V R$
$q_{B} = 2 V R$
$\frac{q_{A}}{q_{B}} = \frac{1}{2}$

When shell B is earthed, $V_{B}^{'} = 0$
$\Rightarrow \frac{q_{A}^{'}}{2 R} + \frac{q_{B}^{'}}{2 R} = 0$
$\Rightarrow q_{B}^{'} = - q_{A}^{'}$
$\frac{| q_{A}^{'} |}{| q_{B}^{'} |} = 1$
$V_{A}^{'} = \frac{q_{A}^{'}}{R} + \frac{q_{B}^{'}}{2 R} = \frac{q_{A}^{'}}{2 R}$
Potential difference between A and B: $V_{A}^{'} - V_{B}^{'} = V_{A} - V_{B} = 2 V - \frac{3}{2} V = \frac{1}{2} V$
$\Rightarrow V_{A}^{'} = \frac{1}{2} V$ since $V_{B}^{'} = 0$

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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 q′A and q′B. Then:

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