Question

If you want to store 0.0125 J of electrical energy at 240 volts, what capacitance do you need? If you have a dielectric you can use with K=3 and can be made as thin as 200 micrometers, what plate area do you need?

Answer 1

apply energy across capaccitor U = 0.5 cv^2

so

Capacitance C = 2U/v^2

C = 2* 0.0125/(240*240)

C = 434.02 nF

--------------------------------

apply C = KeoA/d

so Area A = Cd/Keo

A = 434.02 e-9   * 2e-6/(3*8.85e-12)

A = 3.26 *10^-2 m^2

Answer 2

C = 2U/V²

U = 0.0125 J

V = 240 V

that gives C = 4.34 x 10^-7f

b) we have

$C =k\epsilon _{o}A/d$

$A = Cd/k\epsilon _{o}$

A = 3.27 m²