Question

A parallel plate capacitor of capacitance $51pF$ is too designed with voltage rating $1kV$ using a material of dielectric constant $200$ and dielectric strength about ${10^7}V{m^{ - 1}}$ . If the electric field is never to exceed $10\% $ of dielectric strength what should be the minimum area of plates?

Answer 1

Hint:In order to solve this question we need to understand the definition of capacitor. Capacitor is an electrical device that is used to store electric energy in form of charges stored on its plates and later used in some circuits as a source of energy. There are three types of capacitors used, one is parallel plate capacitor, second is cylindrical plate capacitor and the last one is spherical plate capacitor.

Complete step by step answer:
According to the question we have, given capacitance $C = 51pF$ dielectric constant $k = 200$ dielectric strength $E = {10^7}V{m^{ - 1}}$ and Voltage$V = 1kV$ Also ${\varepsilon _0} = 8.854 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$.
Since value of electric field permitted to use is $10\% $ so electric field across plates is \[E' = \dfrac{{10}}{{100}} \times {10^7}V{m^{ - 1}}\]
\[E' = {10^6}V{m^{ - 1}}\]
Now let the plate separation be “d” so by using relation $E = \dfrac{V}{d}$ we get
$d = \dfrac{V}{E} = \dfrac{{{{10}^3}V}}{{{{10}^6}V{m^{ - 1}}}}$
$\Rightarrow d = {10^{ - 3}}\,m$
Let plate area be “A” then by using $C = \dfrac{{k{\varepsilon _0}A}}{d}$
Putting values we get
$A = \dfrac{{Cd}}{{k{\varepsilon _0}}}$
$\Rightarrow A = \dfrac{{({{10}^{ - 12}} \times {{10}^{ - 3}})}}{{(200 \times 8.854 \times {{10}^{ - 12}})}}$
$\therefore A = 0.056 \times {10^{ - 5}}{m^2}$

So the minimum plate is $area = 0.056\,\mu {m^2}$.

Note: It should be remembered that dielectric strength is the electric field inside dielectric whereas dielectric is defined as an insulator which gets polarized under the influence of external electric field and thereby generating displacement vector. Also dielectric constant is a measure of resistance shown by insulators in response to their behavior in an external electric field. Capacitance increases if we place a dielectric material between the plates of the capacitor and $1\mu m = {10^{ - 6}}m$.