Question

The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$ . The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $'y'$ from one plate as $K = \lambda \sec \left( {\dfrac{{\pi y}}{{2d}}} \right)$ , where $\lambda$ is a dimensionless constant The capacitance of this capacitor is
A. $\dfrac{{\pi {\varepsilon _o}\lambda A}}{{2d}}$
B. $\dfrac{{\pi {\varepsilon _o}\lambda A}}{d}$
C. $\dfrac{{2\pi {\varepsilon _o}\lambda A}}{d}$
D. None

Answer 1

One needs to do definite integration in here. All the capacitors are connected in series. Therefore write the relation for capacitance for any position in between the plates as a function of distance between the plates in proper format and integrate with proper limits

Complete step by step answer:
Let us assume a small capacitor with separation $dy$ at a distance of $y$ from one plate. The area of the plates will be $A$ and the dielectric constant will be $K = \lambda \sec \left( {\dfrac{{\pi y}}{{2d}}} \right)$
Therefore the capacitance can be written as
$C = \dfrac{{K{\varepsilon _o}A}}{d} \\\ \Rightarrow dC = \dfrac{{\lambda \sec \left( {\dfrac{{\pi y}}{{2d}}} \right){\varepsilon _o}A}}{{dy}} \\\$
These small capacitors are connected in series. In a series combination of capacitors the reciprocal of the total capacitance is the sum of reciprocals of the individual capacitance. It is written as
$\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + {\dfrac{1}{C}_3}...$
Or in the case of integrals it is written as

$$\dfrac{1}{C} = \int\limits_0^d {\dfrac{1}{{dC}}} \\\ \Rightarrow\dfrac{1}{C} = \int\limits_0^d {\dfrac{1}{{\dfrac{{\lambda \sec \left( {\dfrac{{\pi y}}{{2d}}} \right){\varepsilon _o}A}}{{dy}}}}} \\\ \Rightarrow\dfrac{1}{C} = \int\limits_0^d {\dfrac{{cos\left( {\dfrac{{\pi y}}{{2d}}} \right)}}{{\lambda {\varepsilon _o}A}}dy} \\\ \Rightarrow\dfrac{1}{C} = \dfrac{1}{{\lambda {\varepsilon _o}A}}\int\limits_0^d {cos\left( {\dfrac{{\pi y}}{{2d}}} \right)dy} \\\ \Rightarrow\dfrac{1}{C} = \dfrac{1}{{\lambda {\varepsilon _o}A}}\left[ {\dfrac{{\sin \left( {\dfrac{{\pi y}}{{2d}}} \right)}}{{\dfrac{\pi }{{2d}}}}} \right]_0^d \\\ \Rightarrow\dfrac{1}{C} = \dfrac{{2d}}{{\lambda {\varepsilon _o}A\pi }} \\\ \therefore C = \dfrac{{\lambda {\varepsilon _o}A\pi }}{{2d}} \\\$$

Hence the total capacitance of the capacitor is $C = \dfrac{{\lambda {\varepsilon _o}A\pi }}{{2d}}$.

Hence option A is the correct answer.

Additional information:
A capacitor is a two-terminal electrical device that possesses the ability to store energy in the form of an electric charge. It consists of two electrical conductors that are separated by a distance. The space between the conductors may be filled by vacuum or with an insulating material known as a dielectric. The ability of the capacitor to store charges is known as capacitance.

Capacitors store energy by holding apart pairs of opposite charges. The simplest design for a capacitor is a parallel plate, which consists of two metal plates with a gap between them. But, there are different types of capacitors manufactured in many forms, styles, lengths, girths, and many materials.

Note: These types of questions are very common. One needs good knowledge of integration. Here also the main catch is to write the expression correctly. Also one should be careful with the type of combination of capacitors whether it is series or in parallel. The limits to be put should also be taken care of.