A capacitor of capacitance C = 900 is charged fully by a 100 battery as shown in figure . Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 as shown in figure . The electrostatic energy stored

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Solution

Q.
A capacitor of capacitance C = 900 $pF$ is charged fully by a 100 $V$ battery $B$ as shown in figure $(a)$ . Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 $pF$ as shown in figure $(b)$ . The electrostatic energy stored by the system $(b)$ is [2022]

1 $4.5 \times 10^{- 6}$ J

2 $3.25 \times 10^{- 6}$ J

3 $2.25 \times 10^{- 6}$ J

4 $1.5 \times 10^{- 6}$ J

Ans.
(3)

Here, $C_{1} = C_{2} = 900$ $pF$

$V_{1} = 100$ $V, V_{2} = 0$ $V$

when $C_{1}$ and $C_{2}$ are joined, their common potential is given as,

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{900 \times 100 + 900 \times 0}{1800}$

$V = 50$ $V$

Electrostatic energy stored in system (b) is

$U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

$U = \frac{1}{2} \times 1800 \times 10^{- 12} \times 50 \times 50$

$U = 2.25 \times 10^{- 6}$ J

According to figure (b), charge on both the plate of capacitor is positive which is not possible.

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