Question

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about ${10}^{7}V{m}^{-1}$. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength.What minimum area of the plates is required to have a capacitance of 50 pF ?

Answer 1

$\text{Step 1 : Electric field inside capacitor}$

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field intensity,

$E=10\%$ of ${10}^{7}V/m$

$={10}^{6}V/m$

Given $V=1kV=1000V$

We know that, $V=E\times d$

$\therefore$Distance between the plates $d=\frac{V}{E}$$=\frac{1000}{{10}^6}={10}^{-3}m$

$\text{Step 2: Capacitance}$

Given, Capacitance $C=50pF$

Also, $C=\frac{KA{\epsilon }_o}{d}$

Dielectric constant of material, $K=3$

$\therefore A=\frac{Cd}{{\epsilon }_{o}K}=\frac{50\times {10}^{-12}\times {10}^{-3}}{8.85\times {10}^{-12}\times 3}\approx 19c{m}^2$