A $12 p F$ capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ? If another capacitor of $6 p F$ is connected across the combination, find the charge stored and potential difference across each capacitor.

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Here, $C_{1} = 12 p F = 12 \times 10^{- 12} F$ .
V = 50 volt.
Electrostaic energy stored $U = \frac{1}{2} C_{1} V^{2}$
$= \frac{1}{2} \times 12 \times 10^{- 12} {(50)}^{2}$
$U = 1.5 \times 10^{- 8} J$
Now, $C_{2} = 6 p F = 6 \times 10^{- 12} F$
$\frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} = \frac{1}{4}$
$C_{s} = 4 p F = 4 \times 10^{- 12} F$
Charge on each condenser,
$q_{1} = q_{2} = C_{s} V = 4 \times 10^{- 12} \times 50$
$= 2 \times 10^{- 10} C$
$V_{1} = \frac{q_{1}}{C_{1}} = \frac{2 \times 10^{- 10}}{12 \times 10^{- 12}} = \frac{50}{3}$ volt
$V_{2} = \frac{q_{2}}{C_{2}} = \frac{2 \times 10^{- 10}}{6 \times 10^{- 12}} = \frac{100}{3}$ volt

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Updated on:21/07/2023

Class 12 PHYSICS ELECTRIC CHARGES AND FIELDS

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