- #1

U154756

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A parallel-plate capacitor has 3.56 cm^2 plates that are separated by 6.08 mm with air between them. The permittivity of a vacuum is 8.85419e-12 C^2/Nm^2. If a 10.4 V battery is connected to this capacitor, how much energy dies it store. Answer in units of J.

Capacitance between parallel plates:

C = (epsilon * A)/d (epsilon is the permittivity of a vacuum.)

Once you have capacitance, you can figure the stored energy using:

U = 1/2CV^2

A = 3.56 cm^2 = 3.56e-4 m^2

d = 6.08 mm = .00608 m

Epsilon = 8.85419e-12 C^2/Nm^2

V = 10.4

Plugging the above numbers into the formula:

U = (1/2)((epsilon * A)/d)(V^2)

I get 9.9811e-15 J, but the answer is wrong. Am I missing something here or just flat out working the problem wrong?