In a parallel plate capacitor the separation between the plates is 3 mm with air between them. Now, a dielectric 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 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 dielectric 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 between the plates must be made (A) 1.5 mm (B) 2.5 mm (C) 4 mm (D) 6 mm

Tardigrade Admin · Accepted Answer

Given, initial separation d=3 mm Let new separation between the plates is d prime. ∴ (C prime d prime/k)=C d Given, C prime=C ⇒ d prime=k d ⇒ d prime=(2)(3)=6 mm

1.5 mm

2.5 mm

4 mm

6 mm

Given, initial separation d=3mm Let new separation between the plates is d′. ∴kC′d′​=Cd Given, C′=C ⇒d′=kd ⇒d′=(2)(3)=6mm

Given, initial separation $d = 3 mm$
Let new separation between the plates is $d^{'}$ .
$∴ \frac{C ^{'} d ^{'}}{k} = C d$
Given, $C^{'} = C$
$\Rightarrow d^{'} = k d$
$\Rightarrow d^{'} = (2) (3) = 6 mm$