Question

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{{}^{\prime }{A}^{\prime }}{3}$ and the separation between the plates is 'd', '2d' and '3d' respectively. The equivalent capacity of combinations is (${\epsilon }_0$ = absolute permittivity of free space)

• A. $\frac{7{\epsilon }_{0}A}{18d}$
• B. $\frac{11{\epsilon }_{0}A}{18d}$
• C. $\frac{13{\epsilon }_{0}A}{18d}$
• D. $\frac{17{\epsilon }_{0}A}{18d}$

Answer 1

The correct option is B $\frac{11{\epsilon }_{0}A}{18d}$

Capacitance of parallel plate capacitor $C=\frac{A{ϵ}_o}{d}$

where $A$ is the area of each plate and $d$ is the separation betwee the plates.

Thus capacitance of first capacitor $C_1=\frac{A{ϵ}_o}{3d}$

Capacitance of second capacitor $C_2=\frac{A{ϵ}_o}{6d}$

Capacitance of second capacitor $C_3=\frac{A{ϵ}_o}{9d}$

Equivalent capacitance of capacitors connected in aprallel $C_{eq}=C_1+C_2+C_3$

$\therefore$ $C_{eq}=\frac{A{ϵ}_o}{3d}$ $+\frac{A{ϵ}_o}{6d}$ $+\frac{A{ϵ}_o}{9d}$

$⟹$ $C_{eq}=\frac{11A{ϵ}_o}{18d}$