# Electrical Potential Energy?

1. Jun 11, 2004

### speeding electron

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:
$$U = \frac{1}{2} \int \rho \phi dV$$
where $$\rho$$ is the charge density of one distribution of charges, and $$\phi$$ is the potential due to another. This is derived from
$$U = \frac{1}{2} \iint \frac{\rho(1) \rho(2)}{4 \pi \epsilon_0 r_{12}} dV_1 dV_2$$
and by substituting in
$$\phi(1) = \int \frac{\rho(2)}{4 \pi \epsilon_0 r_{12}}dV_2$$
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 $$\rho(1)$$ 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 $$\phi$$ would be constant and
$$U = \frac{\phi}{2} \int \rho dV = \frac{\phi(1) q(1)}{2}$$
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