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)

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
$C_{PQ}=2C$
Now, $C_{PQ}$ and $C$ are in series.
$\therefore C_{XY}=\frac{C\cdot C_{PQ}}{C+C_{PQ}}$
$=\frac{C\cdot 2C}{C+2C}=\frac{2}{3}C$
The capacitance of parallel plate capacitor
$C=\frac{{\epsilon }_{0}A}{d}$
$C_{XY}=\frac{2}{3}\frac{{\epsilon }_{0}A}{d}$