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Electrical Potential Energy?

  1. Jun 11, 2004 #1
    In Feynmann's Lectures on Physics, many of you will have come across the chapter on electrostatic energy. I wonder if someone could explain to me why we have the factor 1/2 in the equation:
    [tex]
    U = \frac{1}{2} \int \rho \phi dV
    [/tex]
    where [tex] \rho [/tex] is the charge density of one distribution of charges, and [tex] \phi [/tex] is the potential due to another. This is derived from
    [tex]
    U = \frac{1}{2} \iint \frac{\rho(1) \rho(2)}{4 \pi \epsilon_0 r_{12}} dV_1 dV_2
    [/tex]
    and by substituting in
    [tex]
    \phi(1) = \int \frac{\rho(2)}{4 \pi \epsilon_0 r_{12}}dV_2
    [/tex]
    whence the 1's disappear since there are no 2's around.
    Feynmann explains it by saying that in this expression we count each pair of charges twice. I can understand this in the double integral, where each charge density is integrated over both volumes, but in doing the substitution Feynmann has removed [tex] \rho(1) [/tex] out of the inner integral. It seems to me that the argument for the factor 1/2 in the first place rested on the fact that one couldn't do that.
    Put another way, if we take my first expression that I have trouble with, and imagine a situation where the second charge distribution is arranged so that the potential from all its charges on the first is the same. Then [tex] \phi [/tex] would be constant and
    [tex]
    U = \frac{\phi}{2} \int \rho dV = \frac{\phi(1) q(1)}{2}
    [/tex]
    which Feynmann himself says is only half the total electrostatic energy of the system. Clarification would be greatly appreciated.

    p.s. What's this warning thing I have? What have I done?
     
    Last edited: Jun 12, 2004
  2. jcsd
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