Question

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

• A. $\frac{3}{2} \frac{\in_0 A}{d}$
• B. $\frac{2 \in_0 A}{d}$
• C. $\frac{2}{3} \frac{\in_0 A}{d}$
• D. $\frac{3 \in_0 A}{d}$

Answer 1

Answer: (C)

$\frac{2}{3} \frac{\in_0 A}{d}$

Answer 2

From the figure, it is clear that two capacitances will from 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
$C_{P Q}=2 C$
Now, $C_{P Q}$ and $C$ are in series.
$\therefore C_{X Y} =\frac{C \cdot C_{P Q}}{C+C_{P Q}}$
$=\frac{C \cdot 2 C}{C+2 C}=\frac{2}{3} C$
The capacitance of parallel plate capacitor
$C =\frac{\varepsilon_{0} A}{d}$
$C_{X Y} =\frac{2}{3} \frac{\varepsilon_{0} A}{d}$