Question

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

• A. $\frac{Q}{2}$
• B. $\frac{Q}{4}$
• C. $-\frac{Q}{2}$
• D. $-\frac{Q}{4}$

Answer 1

The correct option is C $-\frac{Q}{2}$

Let the charge distribution be as shown in the given figure.


Electric field at a point $\text{P}$ inside leftmost plate due to all six surfaces are:

Due to charge $Q-q$,
$E_1=\frac{Q-q}{2{ϵ}_{0}A}$ (assumed towards right)

Due to charge $+q$,
$E_2=\frac{q}{2{ϵ}_{0}A}$ (towards left)

Due to charge $-q$ (middle plate),
$E_3=\frac{q}{2{ϵ}_{0}A}$ (towards right)

Due to charge $+q$ (middle plate),
$E_4=\frac{q}{2{ϵ}_{0}A}$ (towards left)

Due to charge $-q$ (rightmost plate),
$E_5=\frac{q}{2{ϵ}_{0}A}$ (towards right)

Due to charge $-2Q+q$ (rightmost plate),
$E_6=\frac{-2Q+q}{2{ϵ}_{0}A}$ (assumed towards left)

Electric field at $\text{P}$ should be zero as it is inside the conductor.

So, $\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}+\overrightarrow{E_4}+\overrightarrow{E_5}+\overrightarrow{E_6}=0$

$\Rightarrow \frac{Q-q}{2{ϵ}_{0}A}-\frac{q}{2{ϵ}_{0}A}+\frac{q}{2{ϵ}_{0}A}-\frac{q}{2{ϵ}_{0}A}+\frac{q}{2{ϵ}_{0}A}-\frac{(-2Q+q)}{2{ϵ}_{0}A}=0$

$\Rightarrow (Q-q)-(-2Q+q)=0$

$\Rightarrow 3Q-2q=0\Rightarrow q=\frac{3Q}{2}$

Further, charge on the outer surface of the rightmost plate is $-2Q+q=-2Q+\frac{3Q}{2}=-\frac{Q}{2}$

And the electric field due to the charge on this face will be towards right (assumed direction is wrong)

Hence, option $\left(c\right)$ is correct.