Question

A parallel plate capacitor has a certain capacitance. When $\dfrac{2}{3}$rd of the distance between the plates is filled with dielectric the capacitance is found to be $2.25$ times the initial capacitance. The dielectric constant of the dielectric is ______
A. 1
B. 3
C. 7
D. 6

Answer 1

The ratio of the amount of electric charge deposited on a conductor to the difference in electric potential is known as capacitance. Self-capacitance and reciprocal capacitance are two closely related concepts of capacitance. Self-capacitance is a property of any material that can be electrically charged.

Complete step by step answer:
A capacitor (also called a condenser) is a two-terminal passive electrical device that stores energy electrostatically in an electric field. Practical capacitors come in a variety of shapes and sizes, but they all have at least two electrical conductors (plates) separated by a dielectric (i.e., insulator).

The Farad (abbreviated to F) is the unit of capacitance and is named after the British scientist Michael Faraday. Capacitance is the electrical property of a capacitor and is the measure of a capacitor's ability to store an electrical charge into its two surfaces.

The capacitance of a parallel plate capacitor with a dielectric between its plates is $C = \kappa {_0}\dfrac{A}{d}$ where the material’s dielectric constant is. Dielectric strength is the highest electric field strength above which an insulating material tends to break down and behave.

When a di-electric slab of thick t and relative permittivity ${\varepsilon _\gamma }$ is added
$C = \dfrac{{{\varepsilon _0}A}}{{d - \left( {t - \dfrac{t}{{{\varepsilon _\gamma }}}} \right)}}$.............(1)
$\Rightarrow C = \dfrac{{A{\varepsilon _0}}}{d}$.............(2)
Substituting the given values we get
$2.25\,C = \dfrac{{A{\varepsilon _0}}}{{d - \left( {\dfrac{{2d}}{3} - \dfrac{{2d}}{{3{\varepsilon _\gamma }}}} \right)}}$
$\Rightarrow 2.25\,C = \dfrac{{A{\varepsilon _0}}}{{d\left( {\dfrac{1}{3} + \dfrac{2}{{3{\varepsilon _r}}}} \right)}}$
From 2 we get
$2.25\,C = \dfrac{C}{{\left( {\dfrac{1}{3} + \dfrac{2}{{3{\varepsilon _r}}}} \right)}}$
$\therefore {\varepsilon _r}= 6$

Hence option D is correct.

Note: The absolute permittivity, also known as permittivity and denoted by the Greek letter $ε$ (epsilon), is a measure of a dielectric's electric polarizability in electromagnetism. A substance with a high permittivity polarises more in response to an applied electric field than one with a low permittivity, allowing it to store more energy. The permittivity of a capacitor is significant in electrostatics since it determines its capacitance.