A capacitor is made of a flat plate of area A | vedclass

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(D) The capacitor can be modeled as two capacitors connected in parallel,each having an area of $A/2$.
For the first part,the separation between the

Vedclass · Accepted Answer

$\frac{\epsilon_0 A}{2 d}\left[\frac{2 d+b}{d+b}\right]$

Vedclass · Answer

(D) The capacitor can be modeled as two capacitors connected in parallel,each having an area of $A/2$.
For the first part,the separation between the plates is $d$. Thus,$C_1 = \frac{\epsilon_0 (A/2)}{d} = \frac{\epsilon_0 A}{2d}$.
For the second part,the separation between the plates is $d+b$. Thus,$C_2 = \frac{\epsilon_0 (A/2)}{d+b} = \frac{\epsilon_0 A}{2(d+b)}$.
Since they are in parallel,the total capacitance is $C = C_1 + C_2$.
$C = \frac{\epsilon_0 A}{2d} + \frac{\epsilon_0 A}{2(d+b)} = \frac{\epsilon_0 A}{2} \left[ \frac{1}{d} + \frac{1}{d+b} \right]$.
$C = \frac{\epsilon_0 A}{2} \left[ \frac{d+b+d}{d(d+b)} \right] = \frac{\epsilon_0 A}{2} \left[ \frac{2d+b}{d(d+b)} \right]$.
This simplifies to $C = \frac{\epsilon_0 A}{2d} \left[ \frac{2d+b}{d+b} \right]$.

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