A dielectric slab of dielectric constant K is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge q' on the surface of slab is given by

Question

A dielectric slab of dielectric constant K is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge q' on the surface of slab is given by (A) q'=q-(q/K) (B) q'=-q+(q/K) (C) q'=q[(1/K)+1] (D) q'=-q(1+(1/K))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b93ace6bfb230c9f140260a066444915-.png" /> q'=-q(1-(1/K))

Q. A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$ . The induced charge $q^{'}$ on the surface of slab is given by

$q^{'} = q - \frac{q}{K}$

$q^{'} = - q + \frac{q}{K}$

$q^{'} = q [\frac{1}{K} + 1]$

$q^{'} = - q (1 + \frac{1}{K})$

$q^{'} = - q (1 - \frac{1}{K})$

Q. A dielectric slab of dielectric constant K is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge q′ on the surface of slab is given by

q′=q−Kq​

q′=−q+Kq​

q′=q[K1​+1]

q′=−q(1+K1​)