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A capacitor of capacitance C=900 pF is charged fully by 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:

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$3 .25 \times 10^{- 6} J$

$4 .5 \times 10^{- 6} J$

$2 .25 \times 10^{- 6} J$

$1 .5 \times 10^{- 6} J$

answer is B.

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Detailed Solution

$after joining their common potential will be \frac{c_{1} v_{1} + c_{2} (0)}{c_{1} + c_{2}}$

$\frac{9 \times 100}{18} = 50 V$

$and combine capacitance = 2 c = 18 pF$

$energy = \frac{1}{2} 18 \times 10^{- 10} {(50)}^{2}$

$= 9 \times 25 \times 10^{- 8} = 2.25 \times 10^{- 6} J$

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