Question

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

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

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 $C_{P Q}=2C$
Now, $C_{P Q}$ and $C$ are in series.
$\therefore \, \, C_{X Y}=\frac{C . C_{P Q}}{C + C_{P Q}}$
$=\frac{C . 2 C}{C + 2 C}=\frac{2}{3}C$
The capacitance of parallel plate capacitor
$C=\frac{\epsilon _{0} A}{d}$
$\Rightarrow \, \, C_{X Y}=\frac{2}{3}\frac{\epsilon _{0} A}{d}$