A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

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$\frac{9}{2} {CV}^{2}$

$\frac{3}{2} {CV}^{2}$

$\frac{35}{6} {CV}^{2}$

answer is B.

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Detailed Solution

$Net charge Q = Q_{2} - Q_{1} potential is V_{l}$

$∴ V_{1} = (\frac{C_{0}}{C + C_{0}}) V_{0}$

$Similarly after nth operation ; E = 1 / 2 C^{1} V^{2}$

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