Question

A 2.00-cm-diameter parallel-plate capacitor with a spacing of 0.600 mm is charged to 300 V .
A. What is the total energy stored in the electric field?
B. What is the energy density?

Answer 1

step;1

Given that

diameter D=2 cm

radius r=1 cm=1*10^-2 m

distance d=0.6 mm

potential difference V=300 v

now we find the capacitance

capacitance C=8.85*10^-12*3.14*10^-4/0.6*10^-3=4.632pF

now we find the energy stored

the energy stored E=1/2*4.632*10^-12*300^2=21*10^-8 J

the energy density =21*10^-8/3.14*10^-4*6*10^-4

                           =1.115 J/m^3