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Why does a charged capacitor provide current to a circuit?

  1. Jan 2, 2013 #1
    I have a conceptual doubt concerning capacitors.

    Suppose that I connect the terminals of a resistor to a charged capacitor, so that current will flow through the resistor.

    Usually, in calculations involving capacitors, the electric field outside the capacitor is taken to be zero, because it is negligible (even though in a real parallel-plate capacitor, the external electric field is not zero, because the plates are not infinite).

    My doubt is: if the external electric field of the capacitor is negligible, why does the capacitor cause current to flow through the resistor?

    I guess that, if the external electric field of the capacitor was exactly zero, then no current would flow through the resistor, even though the capacitor is charged. Is this reasoning correct?
  2. jcsd
  3. Jan 2, 2013 #2


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    Yes, you are correct. The external field cannot be exactly zero. It is only exactly zero for a capacitor with infinite plates. For a real capacitor, keep in mind that the integral of electric field along any path gives you a potential difference. So if you have a potential difference across a capacitor, wire leading around a capacitor will certainly have electric field in it. That's the source of electromotive force resulting in the current.
  4. Jan 2, 2013 #3


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    Staff: Mentor

    Also note that there is no way to construct a circuit with an ideal, infinite-plate capacitor. How would the connecting wire get from one side to the other? :uhh:
  5. Jan 2, 2013 #4
    Thank you for the complete explanation.

    I have one more question about the external electric field of a capacitor: Let the potential difference across the plates of the capacitor be V. So, if we calculate the integral of the electric field along a path outside the capacitor from one plate to the other, we must also get (in absolute value) V, is this correct?

    Since the electric field outside the capacitor is taken to be negligible in most calculations, does it mean that the external electric field very near each plate of the capacitor must be very strong, in order for the integral of the electric field along the external path to sum up to V?
  6. Jan 2, 2013 #5


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    The external field is negligible only as compared to internal field. And that has to do with the fact that path along external field line is much greater than path along internal field lines.
  7. Jan 2, 2013 #6
    OK, I get it. This is because, for a capacitor with a relatively large capacitance, the plate separation is taken to be small. Right?
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