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Work done to increase the charge on capacitor

  1. Jun 8, 2008 #1
    1. The problem statement, all variables and given/known data

    how much work is done in increasing the charge on a 3 microfarad capacitor from 300 microcoulombs to 600 microcoulombs?

    i am supposed to get work W = 0.045 joules

    2. Relevant equations

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

    change in electric potential V = change in potential energy U/ q where q is charge

    work W = change in potential energy U

    3. The attempt at a solution

    first i calculated the electric potential at 300 microcoulombs using the first eq:

    C = Q/V --> V = QC = (300*10^-6)(3*10^-6) = 9*10^-10 volts

    then calculated the electric potential at 600 microcoulombs using same eq:

    C = Q/V --> V = QC = (600*10^-6)(3*10^-6) = 1.8*10^-9 volts

    since change in electric potential V = change in potential energy U/q ---> change in V = (1.8*10^-9 - 9*10^-10) = 9*10^-10 volts and since ---> change in U = change in V * q ---> i let change in V = 9*10^-10 and q = 3*10^-6 to get:

    change in U = work W = (9*10^-10)(3*10^-6) = 2.7*10^-15 joules

    i know i did something wrong because i am supposed to get work W = 0.045 joules

    what did i do wrong?


    thanks
     
  2. jcsd
  3. Jun 8, 2008 #2
    your equations are wrong: Q = CV not V = QC. (didnt check if the answer is right though with this change)
     
  4. Jun 8, 2008 #3
    i checked my equations again, the C = Q/V is correct, i solved for V incorrectly, originally i had V = QC, when in actuality it should be V = Q/C which changed my numbers to this:

    electric potential for the Q = 3*10^-6, C = 300*10^-6, gave me V = 100 volts

    electric potential for the Q = 3*10^-6, C = 600*10^-6, gave me V = 200 volts

    so the change in electric potential V = 200 - 100 = 100volts, thus change in potential energy U = (change in V)q = 100(3*10^-6) = work W = 3*10^-4 joules

    which is still incorrect, i'm supposed to get 0.045 joules work, what did i do wrong now?

    thanks
     
  5. Jun 8, 2008 #4

    Dick

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    Science Advisor
    Homework Helper

    The energy stored in a capacitor is (1/2)Q^2/C. Why don't you just use that?
     
  6. Jun 8, 2008 #5
    you were correct dick, i didn't even think of using that formula you gave me, i must've forgotten about it. anyways, when i used it, i got the answer i was looking for, work W = 0.045 joules

    thanks again
     
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