Question

Four metallic plates each with a surface area of one side $A$ are placed at a distance $d$ from each other as shown in figure. Then the capacitance of the system between $X$ and $Y$ is

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

Answer 1

The correct option is D $\frac{3{ϵ}_{0}A}{2d}$

The given circuit can be redrawn as
$C_{eq}=\frac{3C}{2}=\frac{3{ϵ}_{0}A}{2d}$.