Question

A certain substance has a dielectric constant of 3.9 and a dielectric strength of 22 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 6.7 × 10^-2 µ F and to ensure that the capacitor will be able to withstand a potential difference of 3.7 kV?

Answer 1

d = distance between the plates

E = dielectric strength = 22 x 10⁶

V = potetial difference = 3700

d = V/E = 3700/( 22 x 10⁶ )

d = 0.00017 m

capacitance is given as

C = k $\epsilon$ A/d

6.7 x 10^-8 = (3.9) (8.85 x 10^-12) A/(0.00017)

A = 0.33 m²