Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area A/3 and separation between the plates is d, 2d and 3d respectively. The equivalent capacity of the combinat

Question

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area A/3 and separation between the plates is d, 2d and 3d respectively. The equivalent capacity of the combination is ( ϵ 0 ​ -permittivity of free space)

TestHub · Accepted Answer

The capacitance of a parallel plate capacitor is given by $C = \frac{ϵ _{0} A ^{'}}{d ^{'}}$

For the three capacitors, the capacitances are:

$C_{1} = \frac{ϵ _{0} ( A /3 )}{d} = \frac{ϵ _{0} A}{3 d}$

$C_{2} = \frac{ϵ _{0} ( A /3 )}{2 d} = \frac{ϵ _{0} A}{6 d}$

$C_{3} = \frac{ϵ _{0} ( A /3 )}{3 d} = \frac{ϵ _{0} A}{9 d}$

Since the capacitors are connected in parallel, the equivalent capacitance is the sum of individual capacitances:

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

$C_{e q} = \frac{ϵ _{0} A}{3 d} + \frac{ϵ _{0} A}{6 d} + \frac{ϵ _{0} A}{9 d}$

$C_{e q} = \frac{ϵ _{0} A}{d} (\frac{1}{3} + \frac{1}{6} + \frac{1}{9})$

$C_{e q} = \frac{ϵ _{0} A}{d} (\frac{6 + 3 + 2}{18}) = \frac{11 ϵ _{0} A}{18 d}$

Answer: $\frac{11 ϵ _{0} A}{18 d}$

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