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An air capacitor consists of two identical parallel plates A and B as shown in figure. Plate A is given change Q and plate B is given a charge 3Q . P is the median plane of the capacitor. If $C_{0}$ is the capacitance of the capacitor then

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$V_{P} - V_{A} = \frac{Q}{4 C_{o}}$

$V_{P} - V_{A} = \frac{- Q}{C_{o}}$

$V_{P} - V_{B} = \frac{- Q}{4 C_{o}}$

$V_{P} - V_{A} = \frac{Q}{2 C_{o}}$

answer is B.

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Detailed Solution

The distribution of charges on plates are

E at any point between the plates is

$E = \frac{σ}{2 \in_{0}} + \frac{σ}{2 \in_{0}} = \frac{σ}{\in} = \frac{Q}{A \in_{o}}$

$d v = - E . d r$

$\Rightarrow V_{A} - V_{P} = - \frac{Q}{A \in_{o}} \times \frac{d}{2}$

$∴ V_{P} - V_{A} = \frac{Q}{2 C_{o}}$

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