Question

Four metallic plates are placed as shown in the figure. Plate $2$ is given a charge $Q$ whereas all other plates are uncharged. Plates $1$ and $4$ are joined together. All plates have same area. The charge appearing on the right side of plate $3$ is

• A. zero
• B. $+\frac{Q}{4}$
• C. $-\frac{3Q}{4}$
• D. $\frac{+Q}{2}$

Answer 1

The correct option is B $+\frac{Q}{4}$

Let the charges on all the plates be as shown in the figure.


Since left face of the plate $1$ and right face of the plate $4$ are connected by wire , the net charge on the both the surfaces should be equal .

Let $q_b=q$ then from conservation of charge principle,
$q_c=-q$ ; $q_d=Q+q$ ; $q_e=-(Q+q)$ ; $q_f=+(Q+q)$ and $q_g=-(Q+q)$


We know that potential due plate placed in electric field $E$ can be calculated by
$$V=Ed=\frac{\sigma }{{\epsilon }_0}d=\frac{qd}{{\epsilon }_{0}A}$$ where, $\sigma$ is charge density, $\sigma =\frac{q}{A}$

Writing potential from plate $1$ to plate $4$
$V_1-\frac{qd}{{\epsilon }_{0}A}-$ $\frac{(Q+q)2d}{{\epsilon }_{0}A}$$-\frac{(Q+q)d}{{\epsilon }_{0}A}$=$V_4$

As, the plate $1$ and $4$ are connected so, there potential must be equal. i.e., $V_1=V_4$

$\Rightarrow V_1-\frac{qd}{{\epsilon }_{0}A}-\frac{(Q+q)2d}{{\epsilon }_{0}A}-\frac{(Q+q)d}{{\epsilon }_{0}A}=V_1$

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

$\therefore q=-\frac{3}{4}Q$

Charge on right side of plates $3$
$q_f=Q+q$

$\Rightarrow q_f=Q+\frac{-3Q}{4}$

$\therefore q_f=\frac{+Q}{4}$

Hence, (b) is correct answer.