Question

The dielectric in a capacitor serves two purposes. It increases the capacitance, compared to an otherwise identical capacitor with an air gap, and it increases the maximum potential difference the capacitor can support. If the electric field in a material is sufficiently strong, the material will suddenly become able to conduct, creating a spark. The critical field strength, at which breakdown occurs, is 3.0 MV/m for air, but 60 MV/m for Teflon. A parallel-plate capacitor consists of two square plates 15 cm on a side, spaced 0.40 mm apart with only air between them. What is the maximum energy that can be stored by the capacitor? What is the maximum energy that can be stored if the plates are separated by a 0.40-mm-thick Teflon sheet?

Answer 1

given

the plate area is 0.15*0.15 = 22.5*10^-3 m²

the voltage that can be sustained by 0.4 mm of air dielectric is :

V = 3*10⁶*0.40*10^-3 = 1200 V

(A)

C = *A/d

C = 8.85*10^-12 * 22.5*10^-3 / 0.40*10^-3

C = 497.81*10^-12 F

The energy storedin the capacitor

E = 1/2*C*V²

E = 0.5*497.81*10^-12*(1200)²

E = 358.42*10^-6 J

(B)

V = 60*10⁶ *358.42*10^-6 = 21.50*10³ = 21.50 kV

The relative dielectric constant for teflon is 2.1 .

C = *A/d = 2.1*8.85*10^-12 * 22.5*10^-3 / 0.40*10^-3

C = 1045.40 *10^-12 F

The maximum energy stored is:

E = 1/2*C*V²

E = 0.5*1045.40*10^-12 *(21.50*10³)²

E = 241.61*10^-3 J