In the given circuit switch K is open. The charge on the capacitor is C in steady-state is q 1 . Now the key is closed and steady-state charge on C is q 2 . The ratio of charges q 1 / q 2 is -

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Initially when switch is open

Finally when the switch is closed

$I = \frac{E}{3 R}; ∴ V_{A} - V_{B} = I 2 R = E / 3 R \times 2 R = 2 E / 3$

$∴ C h a r g e o n c a p a c i t o r = C V; q_{2} = \frac{C 2 E}{3}$

$∴ \frac{q_{1}}{q_{2}} = \frac{C E}{2 / 3 C E} = 3 : 2$

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