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Four large identical metallic plates are placed as shown in the figure. Plate-2 is given a charge Q. All other plates are neutral. Now plate-1 and plate 4 are earthed. Area of each plate is A. Then charge on right surface of plate 3 is.

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$\frac{Q}{4}$

$\frac{Q}{3}$

$\frac{Q}{2}$

$- \frac{Q}{3}$

answer is C.

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Detailed Solution

$V_{1} = V_{4} = 0$

$\frac{Q - q}{A ε_{0}} d - \frac{q}{A ε_{0}} 2 d - \frac{q}{A ε_{0}} d = 0$

$q = \frac{Q}{4}$

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