Two identical metal plates are given positive charges Q₁ and Q₂ (<Q₁) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is

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$\frac{(Q_{1} + Q_{2})}{(2 C)}$

$\frac{(Q_{1} + Q_{2})}{C}$

$\frac{(Q_{1} - Q_{2})}{C}$

$\frac{(Q_{1} - Q_{2})}{(2 C)}$

answer is D.

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Detailed Solution

The charges on plates are shown in fig.

Here $q = \frac{Q_{1} - Q_{2}}{2}$
$Now C = \frac{q}{V} or V = \frac{q}{C}$
$∴$ Potential difference between the plates
$V = \frac{q}{C} = \frac{Q_{1} - Q_{2}}{2 C}$

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