Question

A certain substance has a dielectric constant of $2.8$ and dielectric strength of $18\,MV{m^{ - 1}}$ If it’s used as a dielectric material in a parallel plate capacitor, what minimum area should the plates of the capacitor to obtain capacitance of $7 \times {10^{ - 8}}F$ and to ensure that the capacitor will be able to withstand a potential difference of $4\,kV$.

Answer 1

Hint: In order to solve this question, we will first find the magnitude of distance between the plates of parallel plate capacitor with the given electric field and potential difference values and then using these parameters we will solve for the area of plates of capacitor by using the general formula of capacitance of the capacitor.

Formula used:
Capacitance of the capacitor is calculated as
$C = \dfrac{{K{ \in _0}A}}{d}$
where ${ \in _0} = 8.854 \times {10^{ - 12}}{C^2}{m^{ - 2}}{N^{ - 1}}$ is known as permittivity of free space and $K,A,d$ are the dielectric constant, area of plates of capacitor, distance between the two plates of capacitor.

Complete step by step answer:
According to the question we have given that,
$K = 2.8$ Dielectric constant
$\Rightarrow E = 18MV{m^{ - 1}} = 18 \times {10^6}V{m^{ - 1}}$ Electric field inside the capacitor.
$\Rightarrow C = 7 \times {10^{ - 8}}F$ Capacitance of the capacitor.
$\Rightarrow V = 4kV = 4 \times {10^3}V$ Potential difference between the plates of the capacitor.

Let $d$ be the distance between the plates of the capacitor then using electric field, potential difference relation we have,
$d = \dfrac{V}{E}$ On putting the values of parameters we get,
$\Rightarrow d = \dfrac{{4 \times {{10}^3}}}{{18 \times {{10}^6}}}$
$\Rightarrow d = 0.22 \times {10^{ - 3}}m$
Now, let us suppose that A be the area of plates of the capacitor which we need to find the, using the formula $C = \dfrac{{K{ \in _0}A}}{d}$ and putting the values of each parameters we get,
$7 \times {10^{ - 8}} = \dfrac{{2.8 \times 8.854 \times {{10}^{ - 12}} \times A}}{{0.22 \times {{10}^{ - 3}}}}$
$\Rightarrow 0.062 \times 10 = A$
$\therefore A = 0.62\,{m^2}$

Hence, the area of the plates of the capacitor used should be of $A = 0.62\,{m^2}$.

Note:It should be remembered that, the basic unit of conversions which are used as $1kV = {10^3}V$ and $1MV = {10^6}V$ also, capacitor can be made in various forms like spherical capacitors, cylindrical capacitors as per the requirement and each form of capacitor has its unique capacitance formula like a spherical capacitor having radius of r will have a capacitance of $C = 4\pi { \in _0}r$ and the basic concept by which these formulas are derived is such that charge on the capacitor is the product of capacitance and the potential difference across it which is written as $Q = CV.$