Question

A parallel-plate capacitor has 2.00 cm2 plates that are separated by 5.00 mm with air between them. If a 12.0 V battery is connected to this capacitor, how much energy does it store?

Answer 1

energy stored

E = (1/2)C * V^2

C = (ε_oA/d)

ε_o= 8.85 * 10^-12 C^2/Nm^2

A = 2.00 cm^2 = 2.00 * (10^-2)^2 m^2 = 2.00 * 10^-4 m^2

d = 5.00 mm = 5.00 * 10^-3 m

V = 12.0 V

E=(1/2)* [ 8.85*10^(-12) * 2.00*10^(-4)/5*10^-3]*12^2 = 2.5488 x 10^(-11)

Check the calcs again! It should be ok but go over it just in case.

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