A slab of copper of thickness $\frac{d}{2}$ is introduced between the plates of a parallel plate capacitor where d is the separation between its two plates. If the capacitance of the capacitor without copper slab is C and with copper slab is C' then $\frac{C^{'}}{C}$ is;-

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Solution

From above figure
$C = \frac{A ϵ_{0}}{d}$

$C^{'} = \frac{A ϵ_{0}}{d - \frac{d}{2}}$

$= \frac{2 A ϵ_{0}}{d}$

$C^{'} = 2 C$

So
$= \frac{c^{'}}{c} = \frac{1}{2}$

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A slab of copper of thickness d2 is introduced between the plates of a parallel plate capacitor where d is the separation between its two plates. If the capacitance of the capacitor without copper slab is C and with copper slab is C' then C′C is;-

From above figureC=Aϵ0dC′=Aϵ0d−d2=2Aϵ0dC′=2CSo =c′c=12

A slab of copper of thickness $\frac{d}{2}$ is introduced between the plates of a parallel plate capacitor where d is the separation between its two plates. If the capacitance of the capacitor without copper slab is C and with copper slab is C' then $\frac{C^{'}}{C}$ is;-

From above figure
$C = \frac{A ϵ_{0}}{d}$

$C^{'} = \frac{A ϵ_{0}}{d - \frac{d}{2}}$

$= \frac{2 A ϵ_{0}}{d}$

$C^{'} = 2 C$

So
$= \frac{c^{'}}{c} = \frac{1}{2}$