Question

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness 3d/4, where d is the distance between plates. How is the capacitor changed when the slab is inserted between the plates?

Answer 1

Hint: The description of condenser and the properties of dielectric medium and dielectric constant must be recalled for answering this form of questions.

Complete step-by-step answer:
A condenser is a tool used to store charges. The amount of charge Q, which a condenser can store depends on two major factors, the voltage applied and the physical characteristics of the condenser, such as its thickness.

Capacity C is the volume of charge that is deposited per volt, or $C = \dfrac{Q}{V}$. A parallel plate capacitor’s capacitance is given by ${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$.

When separating the plates by air or free space, ${\varepsilon _0}$ is called the free-space permittivity. The capacitance of a parallel plate condenser with a dielectric between its plates is ${C_0} = k\dfrac{{A{\varepsilon _0}}}{d}$ here $k$ material is dielectric constant. The average strength of the electric field in which an insulating material starts to break down and conduct is called dielectric strength.

Now for our question, we have a slab of material of dielectric constant $k$ which has the same area as that of plates of a parallel plate capacitor but has thickness $\dfrac{{3d}}{4}$, where d is the distance between plates then how the capacitor changes when the slab is inserted between the plates.

Here we have

${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$
$C' = \dfrac{{A{\varepsilon _0}}}{{d - t + \dfrac{t}{k}}}$
When we put $t = \dfrac{{3d}}{4}$
$C' = \dfrac{{A{\varepsilon _0}}}{d} \cdot \dfrac{{4k}}{{3 + k}}$
$C' = C\dfrac{{4k}}{{3 + k}}$

Hence the capacitor changes $C' = C\dfrac{{4k}}{{3 + k}}$ when the slab is inserted between the plates.

Note: We can see the importance of the dielectric constant in this type of question. Therefore, we must also consider the dielectric condenser equation which is ${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$. It will help to solve these types of questions in a very short time and very easily.