Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$ . A metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_{1}$ and $C_{2}$ as shown in the figure. The effective capacitance of the two capacitors is C' and the capacitance of the capacitor initially is C, then $\frac{C ^{'}}{C}$ is :

Question

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Answer 1

Let the gap for capacitor 1 be x and that for capacitor 2 be y. Then,

$C_{1} = \frac{ϵ _{0} A}{x}$ and $C_{2} = \frac{ϵ _{0} A}{y}$
$x + y = d / 2$

Since the two are in series,
$C^{'} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = ϵ_{0} A \frac{\frac{1}{x y}}{\frac{1}{x} + \frac{1}{y}}$

$= \frac{ϵ _{0} A}{x + y} = \frac{2 ϵ _{0} A}{d} = 2 C$

Hence , $\frac{C ^{'}}{C} = 2$

Answer 2

Let the gap for capacitor 1 be x and that for capacitor 2 be y. Then,

C_{1} = \frac{ϵ _{0} A}{x}

and

C_{2} = \frac{ϵ _{0} A}{y}

x + y = d / 2

Since the two are in series,

C^{'} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = ϵ_{0} A \frac{\frac{1}{x y}}{\frac{1}{x} + \frac{1}{y}}

= \frac{ϵ _{0} A}{x + y} = \frac{2 ϵ _{0} A}{d} = 2 C

Hence ,

\frac{C ^{'}}{C} = 2