A parallel-plate capacitor is connected to a resistanceless circuit with a battery until the&nb

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2. Electric Potential and Capacitance

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A parallel-plate capacitor is connected to a resistanceless circuit with a battery until the capacitor is fully charged. The battery is then disconnected from the circuit and the plates of the capacitor are moved to half of their original separation using insulated gloves. Let $V_{new}$ be the potential difference across the capacitor plates when the plates have moved. Let $V_{old}$ be the potential difference across the capacitor plates when they were connected to the battery $\frac{V_{new}}{V_{old}}=$......

A

$0.25$

B

$0.5$

C

$1$

D

$2$

Solution

$\mathrm{C}_{\mathrm{new}}=2 \mathrm{C}_{\mathrm{old}}$

$\mathrm{V}_{\mathrm{new}}=\frac{1}{2} \mathrm{V}_{\mathrm{old}}$

Standard 12

Physics

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