A capacitor of capacitance 9nF having dielectric slab of `epsi_(r)=2.4` dielectric strength 20 MV/m and P.D.=20V calculate area of plates.

Options

A. `2.1xx10^(-4)m^(2)`
B. `4.2xx10^(-4)m^(2)`
C. `1.4xx10^(-4)m^(2)`
D. `2.4xx10^(-4)m^(2)`

Answer

B

Explanation

C=9nF, `epsi_(r)2.4V=20` volt
Dielectric strength=20MV/m
Let separation between plants=d
`E=(v)/(d)`
`20x10^(6)=(20)/(d)`
`d=10^(-6)m`
Now, `C=(in_(0)Ain_(r))/(d)`
`9xx10^(-9)=(8.85xx10^(-12)xxAxx2.4)/(10^(-6))`
`A=(9xx10^(-15))/(8.85xx2.4xx10^(-12))`
`A=0.42xx10^(-3)`
`A=4.2xx10^(-4)m^(2)`