Three identical metal plates with large surface area are kept plate parallel to each other as shown in fig. The leftmost plate is given charge Q, the rightmost given charge $- 2 Q$ and the middle one remains neutral. Then the charge appearing on outer surface of the rightmost plate is

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$- \frac{3 Q}{2}$

$+ \frac{Q}{2}$

$\frac{3 Q}{2}$

$- \frac{Q}{2}$

answer is B.

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Detailed Solution

Distribution of charge on different surface plates has been shown

Since inside a conductor, E = 0

Take a point P on the leftmost plate. The electric field is

$E = \frac{Q - q}{2 A ε_{0}} - \frac{q}{2 A ε_{0}} + \frac{q}{2 A ε_{0}} - \frac{q}{2 A ε_{0}} + \frac{q}{2 A ε_{0}} - \frac{(q - 2 Q)}{2 A ε_{0}} = 0$

So, $Q - q - q + 2 Q = 0$
or $2 q = 3 Q \Rightarrow q = \frac{3 Q}{2}$
The charge appearing on the outer surface of the rightmost plate is
$q - 2 Q = \frac{3 Q}{2} - 2 Q = - \frac{Q}{2}$