Question

A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18 MV/m. It is to be used in a parallel plate capacitor that has a capacitance of 7.0 10^-2 microF and that can withstand a potential difference of 4.0 kV. What minimum dimensions (d & A) must the capacitor have?

Answer 1

Given that:

dielectric strength of certain substance = 18 MV/m = 18*10^6 V/m

Potential difference which capacitor can wtihstand = 4.0 kV = 4.0*10^3 V

Nos We know that:

Dielectric strength = Potential difference/Plate separation

So,

E = V/d

d = V/E = 4.0*10^3/(18*10^6)

d = 0.222*10^-3 m = 0.222 mm

Now Capacitance of a capacitor is given by:

C = k*e0*A/d

k = dielectric constant of substance = 2.8

C = Capacitance = 7.0*10^-2 microF = 7.0*10^-8 F

So,

A = Area of plates = ?

A = C*d/(k*e0)

A = 7.0*10^-8*0.222*10^-3/(2.8*8.854*10^-12)

A = 0.627 m^2 = Area of plates

Let me know if you've any query.