Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area ('A'/3) and the separation between the plates is âdâ, â2dâ and â3dâ respectively. The equivalent capacity of combination is (â0 = absolute permittivity of free space)

Question

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area ('A'/3) and the separation between the plates is âdâ, â2dâ and â3dâ respectively. The equivalent capacity of combination is (â0 = absolute permittivity of free space) (A) (7â0A/18 d) (B) (11â0A/18 d) (C) (13â0A/18 d) (D) (17â0A/18 d)

Tardigrade Admin · Accepted Answer

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}}{3 d}

Capacitance of second capacitor

C_{2} = \frac{A ϵ _{o}}{6 d}

Capacitance of second capacitor

C_{3} = \frac{A ϵ _{o}}{9 d}

Equivalent capacitance of capacitors connected in aprallel

C_{e q} = C_{1} + C_{2} + C_{3}

∴ C_{e q} = \frac{A ϵ _{o}}{3 d} + \frac{A ϵ _{o}}{6 d} + \frac{A ϵ _{o}}{9 d}

\Rightarrow C_{e q} = \frac{11 A ϵ _{o}}{18 d}

$\frac{7 \in _{0} A}{18 d}$

$\frac{11 \in _{0} A}{18 d}$

$\frac{13 \in _{0} A}{18 d}$

$\frac{17 \in _{0} A}{18 d}$

Q. Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area 3′A′​ and the separation between the plates is ‘d’, ‘2d’ and ‘3d’ respectively. The equivalent capacity of combination is (∈0​= absolute permittivity of free space)

18d7∈0​A​

18d11∈0​A​

18d13∈0​A​

18d17∈0​A​

$\frac{7 \in _{0} A}{18 d}$

$\frac{11 \in _{0} A}{18 d}$

$\frac{13 \in _{0} A}{18 d}$

$\frac{17 \in _{0} A}{18 d}$