Question

Five conducting parallel plates having area $A$ and separation between them $d$ are placed as shown in figure. Plate number $2$ and $4$ are connected and between points $\text{A}$ and $\text{B}$ a cell of emf $E$ is connected. The charge flow through the cell is

• A. $\frac{2}{3}\frac{{ϵ}_{0}AE}{d}$
• B. $\frac{4{ϵ}_{0}AE}{d}$
• C. $\frac{3}{4}\frac{{ϵ}_{0}AE}{d}$
• D. $\frac{{ϵ}_{0}AE}{d}$

Answer 1

The correct option is A $\frac{2}{3}\frac{{ϵ}_{0}AE}{d}$

Assigning the potentials to be equal for the plates connected by same wire and redrawing the circuit.


The plate facing each other form a capacitor and considering same plate area $A$ and distance $d$, the capacitance will be $C$ for each of them.

$\therefore C=\frac{{ϵ}_{0}A}{d}$

Reduced form of circuit will be as follow


$\therefore C_{eq}=\frac{2C\times C}{2C+C}=\frac{2C}{3}$

The charge flows through call will be,

$Q=C_{eq}\times E$

$\Rightarrow Q=\left(\frac{2C}{3}\right)E=\frac{2EC}{3}$

$\Rightarrow Q=\frac{2}{3}\frac{A{ϵ}_{0}E}{d}$

Hence $\left(b\right)$ is correct option.