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Hard Capacitor problem

  1. Feb 19, 2004 #1
    Although I figure you will say easy :P

    What we know:

    • Area of plates: 5 meters squared
    • When the plates are separated .004 m, the voltage increases 100V

    What we don't know:

    • The original plate separation
    • The original voltage
    • The capacitance

    Find the charge on the positive plate....it is some sort of simultaneous equations, but mine are leading me in circles. Please help!
     
  2. jcsd
  3. Feb 20, 2004 #2

    NateTG

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    The voltage in a capacitor is:
    [tex]V=\frac{Q}{C}[/tex]
    and
    [tex]C=\frac{\epsilon_0 A}{d}[/tex]

    so you've got
    [tex]V=\frac{Qd}{\epsilon_0A}[/tex]
    so
    [tex]V_0-V_1=\frac{Qd_0}{\epsilon_0A}-\frac{Qd_1}{\epsilon_0A}[/tex]
    The change in voltage is 100, the area is 5, and [tex]\epsilon_0[/tex] is a constant. It doesn't look so bad to me.
     
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