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Capacitor and work

  • Thread starter scholio
  • Start date
  • #1
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Homework Statement



a 20 microfarad capacitor is charges to 50 volts. how much work would it require to add an additional charge of 0.001 coulombs to each plate?

Homework Equations



capacitance C = Q/V where Q is charge, V is electric potential

electric potential energy U = (1/2)Q^2/C

work W = deltU = change in electric potential energy = U_f - U_i

The Attempt at a Solution



i am not sure about the wording of this problem, when they say add 0.001 coulombs to EACH plate, do i quadruple the 0.001 charge for Q rather than only double it in the electric potential energy equation.

i tried only doubling the Q, but firstly:

solve for charge with C = 20*10^-6 farads, V = 50 volts use C = Q/V, solve for Q = 0.001 coulombs

solve for U_i , so Q = 0.001, C = 20*10^-6 --> U_i = 0.025 joules

solve for U_f, so Q = 0.002, C = 20*10^-6 --> U_f = 0.10 joules

solve for work W = U_f = U_i = 0.10 - 0.025 = 0.075 joules

did solve it correctly, did i make the correct assumption (double instead of quadruple)?

thanks
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
In a charged capacitor two plates have equal and opposite charges. If you add equal amount of charges to both plates, on one plate charges will increase and in the other plate charges will decrease.
 
  • #3
alphysicist
Homework Helper
2,238
1
Hi scholio,

Your answer looks correct to me. The language in the problem seems sloppy, but it should mean that the capacitor charge is increasing by 0.001 C.
 

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