Question

What is the effect on the capacitance of a parallel plate capacitor when a dielectric slab is filled between the plate of capacitor?

Answer 1

Answer:

New capacitance will be Dielectric constant times the old capacitance value

Explanation:

Let $E_0=\dfrac{Q}{A\epsilon_0}$ be the electric field before the dielectric is added  when the dielectric slab is filled between the plates the New electric field will be  $E=\dfrac{Q}{A\epsilon_0 k}$ where k is dielectric constant.

Now the potential difference the plates will be

$v=Ed\\v=\dfrac{Qd}{\epsilon_0 A k}$

Now the new capacitance will be

$C'=\dfrac{Q}{V}=\dfrac{kQA\epsilon_0}{Qd}=kC_0$

where $C_0$ is initial capacitance value.