Question

A capacitor has a capacitance of 50pF, which increases to 175pF with a dielectric material between its plates. What is the dielectric constant of the material?

Answer 1

Hint: A capacitor is a device that stores electrical energy in the form of an electric field. It is made up of two conductors that are separated by a dielectric. The ability of a capacitor to store an electric charge is known as capacitance.

Complete step by step answer:
Dielectric material can be an insulator, because a dielectric has poor conductivity to electrical charge. It is also known as relative permittivity. The dielectric constant of a substance is defined as the ratio of the permittivity of the substance to the permittivity of the free space. It is denoted by the symbol kappa $\kappa$. It is expressed in units Farad per meter.
It is given that the capacitor has a capacitance of $50pF$ but when a dielectric material is insulated between the plates, the capacitance increases to $175pF$.
If C is the capacitance of the capacitor when its capacitance was $50pF$.
Let C’ is the capacitance when its capacitance increases to $175pF$.
According to the given condition
Capacitance of new capacitor = $\kappa$(Capacitance of the capacitor earlier)
$C^{\prime }=\kappa C$
$\Rightarrow \kappa ={\displaystyle \frac{C^{\prime }}{C}}$
On substituting the corresponding values of capacitance,
$\Rightarrow \kappa ={\displaystyle \frac{175}{50}}$
On simplification,
$\Rightarrow \kappa =3.5$

Therefore, the dielectric constant of the material will be $3.5$.

Note:
It is to be noted that the capacitance of a capacitor can be increased if the dielectric constant is inserted between the plates. By increasing the capacitance, the ability of the capacitor to store opposite charges on the plates will increase. Dielectric constant is basically a number that has no dimensions. An insulator is used as a dielectric as it fills the space between the plates but if a metal is used then the field inside the metal will be zero. Therefore there will be no charge stored.