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Energy of Capacitor

  • Thread starter U154756
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  • #1
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My physics class is not a calculus based class; therefore, the formulas you see here will not be calculus. My problem states:

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?
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
I get 2.80e-11 J. Everything you said looks correct except for the final answer, so you probably typed an error into your calculator.
 

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