Question

Four large identical metallic plates are placed as shown in the figure. Plate $2$ is giren a charge $Q$. All other plates are neutral. Now plates $1$ and $4$ are earthed. Area of each plate is $A$.

Under electrostatic conditions, choose the correct statements

• A. The outer surfaces of plates $\left(1\right)$ and $\left(4\right)$ will not have any charge.
• B. The right side surface of plate $\left(2\right)$ will have a charge $\frac{Q}{3}.$
• C. The right side surface of plate $\left(1\right)$ will have a charge $\frac{-3Q}{4}$
• D. Potential difference between plate $1$ and $2$ is $\frac{3Qd}{4{ϵ}_{0}A}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The outer surfaces of plates $\left(1\right)$ and $\left(4\right)$ will not have any charge.
C The right side surface of plate $\left(1\right)$ will have a charge $\frac{-3Q}{4}$
D Potential difference between plate $1$ and $2$ is $\frac{3Qd}{4{ϵ}_{0}A}$



The opposite faces must have equal and opposite charge.
The outer surface of plate 1 and 4 will have no charge as they are earthed.
$E_1=\frac{Q-q}{{ϵ}_{o}A},E_2=\frac{q}{{ϵ}_{o}A},E_3=\frac{q}{{ϵ}_{o}A}$

Using relation of potential difference between plates
$V_1+E_{1}d-E_2\left(2d\right)-E_{3}d=V_4$
But, $V_1=V_4=0$[As they are earthed]
$\Rightarrow E_1-2E_2-E_3=0$
$\Rightarrow Q-q-2q-q=0\Rightarrow q=\frac{Q}{4}$
The right side surface of plate $\left(1\right)$ will have a charge $=-(Q-q)=\frac{-3Q}{4}$


Potential difference between plate $1$ and $2$ is,
$V=E_{1}d=\frac{3Qd}{4{ϵ}_{o}A}$