Question

A parallel plate capacitor is to be designed with a voltage rating of $1kV$ using a material of dielectric constant $10$ and dielectric strength ${10^6}V{m^{ - 1}}$. What minimum area of the plates is required to have a capacitance of $88.5pF$ ?

Answer 1

A capacitor is a device to store electrical energy and is also known as a condenser. The parallel plate capacitor is a type of capacitor which is widely used. The capacitance of a parallel plate capacitor is the ratio of charge and voltage and depends on the material, its area, and the distance between the plates.

Complete step by step solution:
Let us first write the given information in the question.
The capacitance of parallel plate capacitor $C = 88.5pF$, the voltage of parallel plate capacitance $V = 1kV = 1000V$, dielectric constant $k = 10$, dielectric strength = ${10^6}V/m$
We have to find the area of the parallel plate capacitor.
The formula to calculate the capacitance of a parallel plate capacitor is given below.
$C = \dfrac{{{\varepsilon _o}kA}}{d}$ …………………….(1)
Here, ${\varepsilon _o}$is the permittivity of free space, $k$is the relative permittivity of the material, A is the area of plates, and $d$ is the distance between the plates.
To find the distance between the plates we use the following formula. $V = Ed \Rightarrow d = \dfrac{V}{E}$
Here, $V$ is the voltage of the capacitor and $E$ is the dielectric strength, and $d$ is the distance between the plates of the capacitor.
Let us put the values in the above formula.
$d = \dfrac{{1000V}}{{{{10}^6}V/m}} = {10^{ - 3}}m$
Now, let us put these values, to calculate the area, in equation (1).
$88.5 \times {10^{ - 12}} = \dfrac{{8.85 \times {{10}^{ - 12}} \times 10 \times A}}{{{{10}^{ - 4}}}}$
Let us rearrange the equation and solve it for $A$ .
$A = {10^{ - 4}}{m^2}$
Therefore, the area of the capacitor to hold the capacitance of $88.5pF$ is ${10^{ - 4}}{m^2}$.

Note:
The energy stored in the capacitor is given by the following formula.
$U = \dfrac{1}{2}C{V^2} = \dfrac{1}{2}QV = \dfrac{1}{2}\dfrac{{{Q^2}}}{C}$
Here, $U$ is the energy, $C$ is the capacitance, $V$ is the voltage, and $Q$ is the charge on the capacitor plates.
For some electrical devices, a large amount of energy is required to start them, in such device’s condenser is used. For example, electric fans.