What is the potential energy of an arrangement of eight negative charges on the corners of a cube of side $b$, with a positive charge in the center of the cube? Suppose each negative charge is an electron with charge $−e$, while the central particle carries a double positive charge, $2e$?
This question is an example in Purcell's Electricity and Magnetism. The solution goes as follows: [In this case] there are four different types of pairs. One type involves the center charge, while the other three involve the various edges and diagonals of the cube. Summing over all pairs yields $$ U=\frac{1}{4\pi \epsilon _0}(8\cdot \frac{(-2e^2)}{(\frac{\sqrt{3}}{2})b}+12\cdot \frac{e^2}{b}+12\cdot \frac{e^2}{\sqrt{2}b}+4\cdot \frac{e^2}{\sqrt{3}b}) $$ Then the book introduces the following equation to sum over pairs $$ U=\frac{1}{2}\sum_{j=1}^{N}\sum_{k\neq j}^{}\frac{1}{4\pi \epsilon _{0}}\frac{q_{j}q_{k}}{r_{jk}} $$
I honestly don't understand the purpose of the double sum and how to evaluate it to solve the above question. Can someone explain what these equations mean intuitively and then how to apply the equation to the above example? Thanks. An example of how to solve this question using this equation would be great.
EDIT: Is it that I do the following summation, taking j=1,2,...,N but for every time I take a summation of j=N, {1,2,3,4,5,6,7,8,9} (for the nine charges including the charge of +2e), k is NOT equal to j? In this sense, would I be taking j and summing over k a total of 8 different times for every k, since k isn't equal to j? For the first set where j=1 I would have addend of U, being $$ U_1=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{1}q_{4}}{r_{14}}+\frac{q_{1}q_{5}}{r_{15}}+\frac{q_{1}q_{6}}{r_{16}}+\frac{q_{1}q_{7}}{r_{17}}+\frac{q_{1}q_{8}}{r_{18}}+\frac{q_{1}q_{9}}{r_{19}})) $$ Then the second addend would be $$ U_2=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{2}q_{1}}{r_{21}}+\frac{q_{2}q_{3}}{r_{23}}+\frac{q_{2}q_{4}}{r_{24}}+\frac{q_{2}q_{5}}{r_{25}}+\frac{q_{2}q_{6}}{r_{26}}+\frac{q_{2}q_{7}}{r_{27}}+\frac{q_{2}q_{8}}{r_{28}}+\frac{q_{2}q_{9}}{r_{29}})) $$ Then the third addend would be $$ U_3=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{3}q_{1}}{r_{31}}+\frac{q_{3}q_{2}}{r_{32}}+\frac{q_{3}q_{4}}{r_{34}}+\frac{q_{3}q_{5}}{r_{35}}+\frac{q_{3}q_{6}}{r_{36}}+\frac{q_{3}q_{7}}{r_{37}}+\frac{q_{3}q_{8}}{r_{38}}+\frac{q_{3}q_{9}}{r_{39}})) $$ and so on, making
$$ U_{TOT}=U_1+U_2+U_3+...+U_N, N=9 $$ Is this correct?
