Electric Potential Energy of a System of Point Charges — Question 2

Question 2

Half the charge-weighted source potentials counts every interaction once

N fixed point charges occupy distinct positions. Define $V_i=\sum_{j\ne i}kq_j/r_{ij}$, excluding the potential of charge i itself.

1. Derive the formula $U=\tfrac12\sum_iq_iV_i$.

2. Apply it to three equal charges q at the vertices of an equilateral triangle of side a, and identify the error from omitting the factor one half.

Question 2 - Solution

Half the charge-weighted source potentials counts every interaction once

(a) Method and calculation

$$\sum_iq_iV_i=\sum_i\sum_{j\ne i}\frac{kq_iq_j}{r_{ij}}=2\sum_{i<j}\frac{kq_iq_j}{r_{ij}}.$$$$U=\sum_{i<j}\frac{kq_iq_j}{r_{ij}}=\tfrac12\sum_iq_iV_i.$$

(b) Interpretation and further calculation

$$V_i=2kq/a\Rightarrow U=\tfrac12(3q)(2kq/a)=3kq^2/a.$$$$\sum_iq_iV_i=6kq^2/a=2U.$$

Physical check

Point-charge self-potentials are singular and are not part of this finite mutual-interaction energy.