Five plates are arranged as shown in the figure and connected across a battery of emf V. The separation between each plate is d and surface area of each plate is A.

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Equivalent capacity between P and Q is $\frac{ε_{0} A}{5 d}$

Charge on plate 1 is $\frac{ε_{0} AV}{5 d}$

Equivalent capacity between P and Q is $\frac{3 ε_{0} A}{5 d}$

Charge on plate 3 is $\frac{3 ε_{0} AV}{5 d}$

answer is B, C, D.

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Detailed Solution

$C = \frac{{Aε}_{0}}{d} \frac{1}{C_{eq}} = \frac{1}{C} + \frac{2}{3 C} = \frac{5}{3 C}$ Option (B) is correct. Option (A) is wrong.
Question Image

Total charge, $Q = {(C_{e q})}_{P Q} V = \frac{3 A ε_{0} V}{5 d}$
In parallel combination, charge is proportional to capacitance.
Charge on plate 1, $\frac{C / 2}{3 C / 2} \times Q = \frac{Q}{3} = \frac{1}{3} \frac{3 A ε_{0} V}{5 d} = \frac{A ε_{0} V}{5 d}$

Option (C) is correct.
Charge on plate 3 in lower branch $= \frac{A ε_{0} V}{5 d}$
Charge on plate 3 in upper branch $= Q - \frac{A ε_{0} V}{5 d} = \frac{2 A ε_{0} V}{5 d}$
Total charge on plate 3 $= Q = \frac{3 A ε_{0} V}{5 d}$ Option (D) is correct.