In a parallel plate capacitor, the separation between the plates is 3 mm with air between them. Now a 1 mm thick layer of a material of dielectric constant 2 is introduced between the plates due to which the capacity increases. In order to bring its capacity of the original value, the separation (in mm ) between the plates must be made-

Question

In a parallel plate capacitor, the separation between the plates is 3 mm with air between them. Now a 1 mm thick layer of a material of dielectric constant 2 is introduced between the plates due to which the capacity increases. In order to bring its capacity of the original value, the separation (in mm ) between the plates must be made- (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C0 =(ϵ0 A/d) Cm =(ϵ0 A/[d prime-t+t / k]) By question Cm =C0 (ϵ0 A/d) =(ϵ0 A/[d'-t+t / k]) d'-1+(1/2)=3 d'=4-(1/2)=(7/2)=3.5 mm

C0​=dϵ0​A​ Cm​=[d′−t+t/k]ϵ0​A​ By question Cm​=C0​ dϵ0​A​=[d′−t+t/k]ϵ0​A​ d′−1+21​=3 d′=4−21​=27​=3.5mm

$C_{0} = \frac{ϵ _{0} A}{d}$
$C_{m} = \frac{ϵ _{0} A}{[ d ^{'} - t + t / k ]}$
By question
$C_{m} = C_{0}$
$\frac{ϵ _{0} A}{d} = \frac{ϵ _{0} A}{[ d ^{'} - t + t / k ]}$
$d^{'} - 1 + \frac{1}{2} = 3$
$d^{'} = 4 - \frac{1}{2} = \frac{7}{2} = 3.5 mm$