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Formula for capacitance is C=Q/V

  1. Oct 17, 2005 #1
    The problem entails the following:

    Two conductors having net charges of +14.0 µC and -14.0 µC have a potential difference of 14.0 V.

    (a) Determine the capacitance of the system

    I know that the formula for capacitance is C=Q/V, but do I need to convert units, or add the two conductors together, or what?

    So far the answers I have come up with are 1E-5 F and 2E-6 F, and they are both wrong...:cry:
     
  2. jcsd
  3. Oct 17, 2005 #2
    C= Q/V

    Q's units will be coulombs
    V's units will be volts
    C's answer will be farads

    1 coulomb/ 1 volt = 1 farad so basically how you have explained the problem it would seem that you just need to plug and chug the answer in to the equation.

    it seems that how your numbers are though that it would come out
    to just + and - 1 farad wouldnt it? since 14/14 and -14/14. I hope this helps.. im looking out of my book to try and help you. speaking of which.. what is the answer supposed to be since you apparently know that yours is wrong and know the correct answer?
     
  4. Oct 17, 2005 #3
    Thx for the help, I found out that it was to the wrong power of 10 (shoulda been 1E-6 instead of 1E-5)
     
  5. Oct 17, 2005 #4
    when a capacitor is charged the values for charge upon each plate (surface) are equal in magnitude but opposite in sign. The total charge on the system is 14 uC. The total potential stored is 14V.

    thus total capacitance = Q/Vab= 14uC/14V= (14*10^-6)C/(14*10^0)V=1uF.

    I think that you lost a decimal place somewhere. The proposal was 14uC not 140*10^-5 C.

    micro is an abbreviation for 10^-6, all exponents in engineering notation are multiples of 3. All the results of exponential calculation in engineering notation should result in numbers which are a multiple of three.

    14uC= 14 *10^6 C
     
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