Question

A material (Teflon) of dielectric constant 2.1 is inserted between the inner wire and the outer cylinder of the capacitor in problem #4, which was 18.6 pF.

a) What will the capacitance be now? 39.1 pF

b) How much energy will be stored in this capacitor (with the dielectric) with the 30 volts applied to it? 17.6 nJ .

*How to set up part b and solve?*

Answer 1

here,

the dielectric constant of Teflon , K = 2.1

the initial capacitance , C0 = 18.6 pF

a)

the capacitance of capacitor , C = K * C0

C = 2.1 * 18.6 pF

C = 39.1 pF

b)

V = 30 V

the energy will be stored in this capacitor , U = 0.5 * C * V^2

U = 0.5 * 39.1 * 10^-12 * 30^2 J

U = 17577 * 10^-12 J

U = 17.6 * 10^-9 J = 17.6 nJ