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Properties of Charges

Two plates ar...

Question

Two plates (area = $5$ ) charged to $+ q_{1}$ and $+ q_{2} (q_{2} < q_{1})$ are brought closer to form a capacitor of capacitance $C$ . The potential difference across the plates is :

A

\frac{q_{1} - q_{2}}{2 C}

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B

\frac{q_{1} - q_{2}}{C}

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C

\frac{q_{1} - q_{2}}{4 C}

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D

\frac{2 (q_{1} - q_{2})}{C}

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Solution

The correct option is A $\frac{q_{1} - q_{2}}{2 C}$

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