First slide

Capacitance

Question

Four metallic plates of each with a surface area of one side A, are placed at a distance d from each other. The alternate plates are connected to points A and B as shown in fig. The equivalent capacitance of the system will be

Easy

A

$ε_{0} A / d$
B

$2 ε_{0} A / d$
C

$3 ε_{0} A / d$
D

$4 ε_{0} A / d$

Solution

The four plates are alternatively connected and form three capacitors in parallel. The capacity of each capacitor is ( $ε_{0} A / d$ ). Hence, the net capacitance between A and B is given by
$C_{AB} = C + C + C = (\frac{3 ε_{0} A}{d})$

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