# 2.33 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 $10^{-7}Vm^{-1}$. (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?

Given

Voltage rating in designing capacitor$V=1kV=1000V$

The dielectric constant of the material $K=\epsilon _r=3$

Dielectric strength of material = $10^7V/m$

Safety Condition:

$E=\frac{10}{100}*10^7=10^6V/m$

The capacitance of the plate $C=50pF$

Now, As we know,

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

$d=\frac{V}{E}=\frac{10^3}{10^6}=10^{-3}m$

Now,

$C=\frac{\varepsilon _0\varepsilon_rA }{d}$

$A=\frac{Cd}{\epsilon_0 \epsilon_r }=\frac{50*10^{-12}*10^{-3}}{8.85*10^{-12}*3}=1.98*10^{-3}m^2$

Hence the minimum required area is $1.98*10^{-3}m^2$

## Related Chapters

### Preparation Products

##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
##### Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
##### Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-