Question

(17)

A 5.90 μF , parallel-plate, air capacitor has a plate separation of 3.90 mm and is charged to a potential difference of 390 V

A) Calculate the energy density in the region between the plates.

Answer 1

given are

C=5.90uF=5.90*10^-6 F

V=390V

d=3.90mm=3.90*10^-3m

U = 1/2 CV^2

u(density)=total energy/volume of plates

=1/2CV^2/Ad

area of plate

c=eoA/d

so A=Cd/eo

u=1/2CV^2/(cd/eo)d

=eo/2 (v^2/d^2)

=8.854*10^-12/2(390^2/(3.90*10^-3)^2)

=4.427*10^-12*10000*10^6

=4.427*10^-2

=0.0427J/m^3