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Does the presence of a charge change the amount of energy stored in a capacitor?

  1. Sep 9, 2011 #1
    Imagine that we have a charged parallel plate capacitor.

    I hold a charge somewhere between the plates.

    Is the electrostatic energy stored by the capacitor modified in any way?

    Here is my thinking:

    The total field energy between the plates is the integral of the field energy density given by:

    Energy = Integral (eps_0 / 2) * |E|^2

    where |E| is the total field at each point.

    E = E_capacitor + E_charge

    |E|^2 = E.E = (E_cap + E_charge) . (E_cap + E_charge)

    |E|^2 = |E_cap|^2 + |E_charge|^2 + 2 E_charge.E_cap

    Because of the spherical symmetry of E_charge I believe the integral of E_charge . E_cap is zero. Thus there is no mutual electrostatic energy between the capacitor and the charge.

    The total electrostatic energy of the system is just the sum of the energy of the capacitor plus the energy of the charge.

    Thus the amount of energy stored in the capacitor alone has not changed.

    Is this right?
     
  2. jcsd
  3. Sep 9, 2011 #2
    If you place a totally free charge in between two charge plates, it will accelerate to one of the plates and discharge on the plate it hits. If you try to hold the charge still by placing it on a conductor; say a small metal sphere held in place between the plates by insulating strings; than the metal sphere is a conductor and must be considered part of the system of conductors that make up a capacitor. If you use a dielectric sphere instead, than the dielectric material will polarize and tend to weaken the fields, allowing more charge to be stored. Anyway to construct it, you are going to change the system.
     
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