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Electric Field

Two parallel ...

Question

Two parallel plates have charges $Q_{1}$ and $Q_{2}$ on them and the capacitance is c. The potential difference between them is

A

\frac{Q_{1} + Q_{2}}{2 C}

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B

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

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C

\frac{Q_{1} + Q_{2}}{C}

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D

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

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Solution

The correct option is B $\frac{Q_{1} - Q_{2}}{2 C}$
$x + y = Q_{1}$
$- x + y = Q_{2}$
So, $x + x = Q_{1} - Q_{2} ⟹ x = \frac{Q_{1} - Q_{2}}{2}$
Potential difference $= \frac{x}{c} = \frac{Q_{1} - Q_{2}}{2 C}$

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