Question

Four metal plates are arranged as shown in the figure. Capacitance between $X$ and $Y$ ( $A\to$ Area of each plate, $d\to$ distance between the plates) is

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

Answer 1

Answer: (C)

From the figure, it is clear that two capacitances will form the equivalent arrangement as shown in figure.

Let $C$ be the capacitance of each capacitor.
The net capacitance of $P$ and $Q$ connected in parallel $C_{PQ}=2C$
Now, $C_{PQ}$ and $C$ are in series.
$\therefore C_{XY}=\frac{C.C_{PQ}}{C+C_{PQ}}$
$=\frac{C.2C}{C+2C}=\frac{2}{3}C$
The capacitance of parallel plate capacitor
$C=\frac{{ϵ}_{0}A}{d}$
$\Rightarrow C_{XY}=\frac{2}{3}\frac{{ϵ}_{0}A}{d}$