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Parallel Plate Capacitance

  1. Feb 13, 2009 #1
    A capacitor with capacitance of C = 8.5±1 μF has a potential difference of V = 4.5±0.4 Volts. What are the charge on the capacitor and the error in this value, in μC?

    To find charge I just use Q = C*V correct? And how do I calculate the error?

    Thank you!
     
  2. jcsd
  3. Feb 13, 2009 #2

    rock.freak667

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    For s=ab
    [tex]\frac{\delta s }{s}= \frac{\delta a}{a} + \frac{\delta b}{b}[/tex]

    Where [itex]\delta s[/itex] is the error in s and similiarly for a and b.
     
  4. Feb 13, 2009 #3

    Delphi51

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    Another way to say that is that for multiplying and dividing, the % errors in quantities are added.
     
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