Question

The following arrangement consists of five identical metal plates parallel to each other. Area of each plate is $A$ and separation between the successive plates is $d$. The capacitance between $\text{P}$ and $\text{Q}$ is

• A. $\frac{5{ϵ}_{0}A}{d}$
• B. $\frac{7}{3}\frac{{ϵ}_{0}A}{d}$
• C. $\frac{4}{3}\frac{{ϵ}_{0}A}{d}$
• D. $\frac{5}{3}\frac{{ϵ}_{0}A}{d}$

Answer 1

The correct option is D $\frac{5}{3}\frac{{ϵ}_{0}A}{d}$

Let us assign the same potential to all points connected with the same conducting wire. Also, the plate attached to same wire will be at the same potential as that of wire.


Reduced circuit will be like this


So, net equivalent capacitance of the circuit will be

$C_{net}=\left[\frac{C\times 2C}{C+2C}\right]+C$

$\Rightarrow C_{net}=\frac{2C}{3}+C=\frac{5C}{3}$

$\because C=\frac{A{ϵ}_o}{d}$

$\therefore C_{net}=\frac{5A{ϵ}_o}{3d}$

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