Electrostatics 5 Question 33

35. A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2 C$ is similarly charged to a potential difference $2 V$. 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

$(1995,2 \mathrm{M})$

(a) zero

(b) $\frac{3}{2} C V^{2}$

(c) $\frac{25}{6} C V^{2}$

(d) $\frac{9}{2} C V^{2}$

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Answer:

Correct Answer: 35. (b)

Solution:

  1. The diagramatic representation of given problem is shown in figure.

The net charge shared between the two capacitors is

$$ Q^{\prime}=Q_{2}-Q_{1}=4 C V-C V=3 C V $$

The two capacitors will have the same potential, say $V^{\prime}$.

The net capacitance of the parallel combination of the two capacitors will be

$$ C^{\prime}=C_{1}+C_{2}=C+2 C=3 C $$

The potential difference across the capacitors will be

$$ V^{\prime}=\frac{Q^{\prime}}{C^{\prime}}=\frac{3 C V}{3 C}=V $$

The electrostatic energy of the capacitors will be

$$ U^{\prime}=\frac{1}{2} C^{\prime} V^{\prime 2}=\frac{1}{2}(3 C) V^{2}=\frac{3}{2} C V^{2} $$



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