Question

Four metallic plate each with a surface area of A are placed at a distance d from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between a and b is-

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

Answer 1

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

The potential difference between the plates, $V=\frac{\sigma d}{{ϵ}_0}=\frac{Qd}{A{ϵ}_0}$
So, the capacitance of the each parallel plate capacitor, $C=\frac{Q}{V}=\frac{A{ϵ}_0}{d}$
The equivalent circuit diagram is shown below.
The equivalent capacitance of the circuit,$C_{eq}=\frac{C\times C}{C+C}+C=\frac{C}{2}+C=\frac{3}{2}C=\frac{3A{ϵ}_0}{2d}$
Ans:$\left(D\right)$