A parallel plate capacitor of capacitance 2 F is charged to a potential V. The energy stored in the capacitor is E₁. The capacitor is now connected to another uncharged identical capacitor in a parallel combination. The energy stored in the combination is E₂. The ratio $E_{2} / E_{1}$ is :

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2 : 1

2 : 3

1 : 4

1 : 2

answer is D.

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Detailed Solution

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

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