A parallel plate capacitor of capacitance is charged to a potential . The energy stored in the capacitor is . [2023]

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Solution

Q.
A parallel plate capacitor of capacitance $2 F$ is charged to a potential $V$ . The energy stored in the capacitor is $E_{1}$ . The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_{2}$ . The ratio $E_{2} / E_{1}$ is [2023]

1 2 : 1

2 1 : 2

3 1 : 4

4 2 : 3

Ans.
(2)

Initially,

$Q_{1} = C V = (2) V$

$E_{1} = \frac{1}{2} C V^{2} = \frac{1}{2} (2) V^{2} = V^{2}$

Finally,

Charge on each capacitor,

$Q_{2} = \frac{Q_{1}}{2} = \frac{2 V}{2} = V$

$E_{2} = 2 (\frac{1}{2} \frac{Q_{2}^{2}}{C}) = \frac{V^{2}}{2}$

$∴$ $\frac{E_{2}}{E_{1}} = \frac{1}{2}$

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