Why Does the Equation for Electrical Potential Energy Include a Factor of 1/2?

[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 Feynman 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 Feynman 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?

 

[Aaron Barnard]

The warning you are receiving is because your post contains more than one question. It is generally better to ask one question at a time in order to receive the best answers and to ensure that the answers are relevant to the questions you are asking.

 

[Graeme Robinson]

Electrical potential energy is the energy that is associated with the position of electrically charged particles. It is the energy that is required to move a charged particle from one point to another in an electric field. This energy is dependent on the electric potential, which is a measure of the electric potential energy per unit charge at a given point in space.

The factor of 1/2 in the equation for electrical potential energy is due to the fact that in calculating the energy of a system of charges, we count each pair of charges twice. This is because each charge contributes to the potential energy of every other charge in the system. So, in order to avoid counting the energy of each pair of charges twice, we divide the total energy by 2.

In the equation U = 1/2 ∫ρφdV, the integral is over the entire volume, and the factor of 1/2 takes into account the fact that each pair of charges contributes to the potential energy of the other. When we substitute in φ(1) = ∫ρ(2)/4πε0r12dV2, we are essentially removing one of the charges (ρ(1)) from the integral, which is why it may seem like the factor of 1/2 is not needed. However, this substitution is only valid if ρ(1) is constant, which is not always the case.

In the scenario you described, where the potential from all charges in the second distribution is the same, the factor of 1/2 still applies because each charge in the first distribution contributes to the potential energy of every other charge in the second distribution. So, the total energy would be 1/2 times the potential energy of a single charge in the first distribution.

As for the warning, it is likely just a reminder to check for any errors or typos in your post before submitting it. Hope this helps clarify the concept of electrical potential energy for you!