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Change in energy of a capacitor

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A 2.1 aF capacitor has a net charge of 0.5e (a positive charge, the symbol e is taken as a positive number 1.6 x 10-19 coulomb). What is the energy needed to add one electron (charge -e) to this capacitor?

    2. Relevant equations
    Energy in a Capacitor:
    [itex]U = Q^2 \div 2C[/itex]

    Where U is the energy, Q is the charge and C is the capacitance

    3. The attempt at a solution
    The initial charge is +0.5e and after adding a charge of -e, the final charge would be -0.5e.
    To find the energy needed:
    [itex]\Delta U = U_f - U_i \\
    = \frac{(Q_f)^2 - (Q_i)^2}{2C} \\
    = \frac{(-0.5e)^2 - (0.5e)^2}{2C}
    = 0[/itex]

    Squaring the initial and final charge results in positive e/4 and subtracting these values of equal magnitude gives 0. But having 0 energy change doesn't make sense to me. What am I doing wrong here?

    Thanks in advance for any help
  2. jcsd
  3. Oct 14, 2013 #2


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    Staff: Mentor

    So on its plates this capacitor is storing a charge difference of half an electron?
  4. Oct 14, 2013 #3
    This is indeed the problem as my Professor wrote it.
  5. Oct 14, 2013 #4
    After asking him about it, my Professor responded:

    "The non integer charge can come from polarization of the metal piece by an electric field, that slightly moves the position of many charges."

    If that helps.
  6. Oct 14, 2013 #5

    rude man

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    Homework Helper
    Gold Member

    What is the voltage on the capacitor before and after adding the charge e? What then is the difference in energy?
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