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Five identical plates each of area A are joined as shown in the figure. The distance between the plates is $d$ . The plates are connected to a potential difference of $V$ volts. The charge on plates 1 and 4 will be

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$\frac{ε_{0} A V}{d}, \frac{2 ε_{0} A V}{d}$

$\frac{ε_{0} A V}{d}, \frac{- 2 ε_{0} A V}{d}$

$\frac{- ε_{0} A V}{d}, \frac{- 2 ε_{0} A V}{d}$

answer is B.

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Detailed Solution

The given circuit can be redrawn as follows. All capacitors are identical and each having capacitance $C = \frac{ε_{0} A}{d}$

Charge on each capacitor = Charge on each plate
$= \frac{ε_{0} A}{d} V$

Therefore charge on plate 4 is

$= \frac{2 ε_{0} A}{d} V$

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