Question

Four large identical metal plates are placed as shown in the figure. Plate $2$ is given a charge $Q$. All other plates are neutral. Now Plate $1$ and $4$ are earthed. Area of each plates is $A$. The charge appearing on the right side of plate $3$ is

• A. $Q$
• B. $\frac{Q}{2}$
• C. $\frac{Q}{8}$
• D. $\frac{Q}{4}$

Answer 1

The correct option is D $\frac{Q}{4}$

Let us assume $x$ be the charge on the left side of plate $2$.

Now, due to induction , charge distribution of all plates is as shown in the figure.

Charge on the right face of the plate $2$ be $(Q-x)$.

Since, the left face of the plate $3$ is opposite to the right face of plate $2$ the charge on this face will be $-(Q-x)$.

Similarly charges on the other faces can be determined for the plates $3$ and $4$ as shown in figure.


Assume, charges on the left face of plate $1$ as $y$.

Since, the left face of the plate $1$ and right face of the plate $4$ are earthed, so the potential for both the faces will be same.i.e.,
$V_1=V_4\Rightarrow V_4-V_1=0$

Hence, the charges on the right face of plate $4$ will be equal to the charge on the left face of plate $1$ which is $y$

Let,

$V_2-V_1=\frac{-xd}{A{\epsilon }_o}$, is the potential difference across the plates $1$ and $2$.

$V_3-V_2=\frac{(Q-x)2d}{A{\epsilon }_o}$, is the potential difference across the plates $2$ and $3$.

$V_4-V_3=\frac{(Q-x)d}{A{\epsilon }_o}$, is the potential difference across plate $3$ and $4$.

Substituting the values we get,

$\frac{-xd}{A{\epsilon }_o}+\frac{(Q-x)2d}{A{\epsilon }_o}+\frac{(Q-x)d}{A{\epsilon }_o}=0$

$\Rightarrow -xd+Q\left(2d\right)-x\left(2d\right)+Qd-xd=0$

$\Rightarrow 3Q=4x$

$\Rightarrow x=\frac{3Q}{4}$

Since the charge appearing on the right side of plate $3$ is $(Q-x)$

$\therefore (Q-x)=Q-\frac{3Q}{4}=\frac{Q}{4}$

Hence, option (b) is the correct answer.