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?

Answers (1)

Given

Voltage rating in designing capacitorV=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

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

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

₹ 4999/-
Buy Now
Knockout NEET May 2021

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

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

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

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions