Question

A parallel plate capacitor is to be designed with a voltage rating $1kV$ 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 capacitoance of $88.5pF$?

Answer 1

Given that,

Potential rating of a parallel plate capacitor $V=1kV=1000V$

Dielectric constant of a material $\epsilon =10$

Dielectric strength ${10}^{6}V/m$

Now, capacitance of the parallel plate capacitor,

$C=88.5\times {10}^{-12}F$

Distance between the plates is given by,

$d=\frac{V}{E}$

$d=\frac{1000}{{10}^6}$

$d={10}^{-3}m$

Now, capacitance is given by the relation

We know that

$c=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$

$A=\frac{cd}{{\epsilon }_0{\epsilon }_r}$

$A=\frac{88.5\times {10}^{-12}\times {10}^{-3}}{10\times 8.85\times {10}^{-12}}$

$A={10}^{-4}c{m}^2$

Hence, the minimum area is ${10}^{-4}c{m}^2$