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.
Part A
A parallel-plate capacitor consists of two square plates 16cm on a side, spaced 0.45mm apart with only air between them. What is the maximum energy that can be stored by the capacitor?
Part B
What is the maximum energy that can be stored if the plates are separated by a 0.45-mm-thick Teflon sheet?

Answer 1

Answer: 580 x 10^-3 J

Explanation:

0.6mm is 0.6/1000 = 600*10^-6 m

The plate area is .17*.17 = 28.9*10^-3 m^2

Air:

The voltage that can be sustained by 0.60 mm of air dielectric is:

V = 3.0*10^6* 600*10^-6 = 1800 V

The capacitance is:

C = ε*A/d = 8.854*10^-12 * 28.9*10^-3/600*10^-6 = 426*10^-12 F = 426 pF

The energy stored in a capacitor is:

E = (1/2)*C*V^2 = (1/2)*426*10^-12*(1800)^2 = 691*10^-6 J

Teflon:

The voltage is:

V = 60*10^6* 600*10^-6 = 36*10^3 = 36 kV

According to the listed reference, the relative dielectric constant for teflon is 2.1, this figure multiplies the "ε" of free space.

The capacitance is:

C = ε*A/d = 2.1*8.854*10^-12 * 28.9*10^-3/600*10^-6 = 896*10^-12 F = 896 pF

It would have been easier to note that the capacitance is 2.1 times the air-dielectric case.

The maximum energy stored is:

E = (1/2)*C*V^2 = (1/2)* 896*10^-12* (36*10^3)^2 = 580*10^-3 J