The equivalent capacitance between 'P' and 'Q' is [A-area of each plate, d-separation between the plates, $ϵ_{0}$ -permittivity of free space]

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The equivalent capacitance between 'P' and 'Q' is [A-area of each plate, d-separation between the plates, ϵ 0 ​ -permittivity of free space]

TestHub · Accepted Answer

The given arrangement is equivalent to three capacitors connected in parallel between the terminals P and Q, as the potential difference across each section is the same.

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

For the three sections, the area is $A^{'} = \frac{A}{3}$ and the distance between the plates is $d$ .

The individual capacitances are:
$C_{1} = \frac{K _{1} ϵ _{0} ( A /3 )}{d} = \frac{1 \cdot ϵ _{0} A}{3 d} = \frac{ϵ _{0} A}{3 d}$

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

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

Since the capacitors are in parallel, the equivalent capacitance $C_{e q}$ is the sum of the individual capacitances:

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

$C_{e q} = \frac{ϵ _{0} A}{3 d} + \frac{2 ϵ _{0} A}{3 d} + \frac{3 ϵ _{0} A}{3 d}$

$C_{e q} = \frac{ϵ _{0} A}{3 d} (1 + 2 + 3)$

$C_{e q} = \frac{6 ϵ _{0} A}{3 d} = \frac{2 A ϵ _{0}}{d}$

Answer: $\frac{2 A ϵ _{0}}{d}$

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