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Discharging a capacitor

  1. Jun 18, 2012 #1
    Consider the hypothetical situation.
    In a circuit with a capacitor and an ideal cell, potential difference across the capacitor would be the potential difference maintained by the cell. If the ideal cell is removed leaving the only the capacitor in the circuit without any resistance in the circuit, the capacitor would become discharged and finally charge on capacitor would be 0 and thus potential difference would be 0. Now since Capacitance=charge/potential difference, the capacitance would be 0/0? Am i right?
    Or if the cell is real charge would only approach 0 and hence potential difference too would approach 0 and hence C approaches 0/0? Is this correct?
  2. jcsd
  3. Jun 18, 2012 #2


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    Staff: Mentor

    The capacitor discharges only if it is not perfect and can conduct somehow or has other ways to discharge.

    "Capacitance=charge/potential difference" is a relation between values the capacitor can have, not the current values. A better way would be to define "capacitance (at a specific potential difference) is the derivative of the charge with respect to a change of the potential difference".
  4. Jun 18, 2012 #3
    Capacitor does not discharge when it is disconnected after charging.It discharges only when there is another capacitor or any other electronic device(resistor,semiconductor,etc) attached to it.As Q=CV,here V is not zero,it is same as that of the battery that was connected before and Q is the maximum amount of charge that the capacitor can store in it.
    Therefore C=charge stored/potential across the capacitor plates.

    A simple experiment to show you that capacitor does not discharge when disconnected is,
    take a capacitor and connect it to a potential source.Then disconnect it and keep it for some time(even a day).Then connect a small LED to it,and the LED will glow.This shows that capacitor does not discharge when disconnected from any other conductors/semiconductors.
  5. Jun 18, 2012 #4


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    Real capacitor does discharge on its own, but very, very slowly. There is going to be some current between plates even with best of insulators, there is going to be some current due to ions in the air, etc. You can't insulate the plates perfectly.

    But it's kind of irrelevant to the question. Defining capacitance as C=q/V is really talking more about slope rather than any specific value. You'd never take the (0,0) point to measure a slope of the line for the same reason you don't take the 0/0 point here. It doesn't define the value.
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