Question 2.61: What is the minimum-energy configuration for a system of N e...

Suppose the n point charges are evenly spaced around the circle, with the jth particle at angle j(2π/n). According to Eq. 2.42, the energy of the configuration is

W=\frac{1}{2} \sum_{i=1}^{n} q_{i} V\left( r _{i}\right)                 (2.42)

W_{n}=n \frac{1}{2} q V ,

where V is the potential due to the (n – 1) other charges, at charge # n (on the x axis).

V=\frac{1}{4 \pi \epsilon_{0}} q \sum_{j=1}^{n-1} \frac{1}{ᴫ_{j}}, \quadᴫ_{j}=2 R \sin \left(\frac{j \pi}{n}\right)

(see the figure). So

W_{n}=\frac{q^{2}}{4 \pi \epsilon_{0} R} \frac{n}{4} \sum_{j=1}^{n-1} \frac{1}{\sin (j \pi / n)}=\frac{q^{2}}{4 \pi \epsilon_{0} R} \Omega_{n} .

Mathematica says

\Omega_{10}=\frac{10}{4} \sum_{j=1}^{9} \frac{1}{\sin (j \pi / 10)}=38.6245

\Omega_{11}=\frac{11}{4} \sum_{j=1}^{10} \frac{1}{\sin (j \pi / 11)}=48.5757

\Omega_{12}=\frac{12}{4} \sum_{j=1}^{11} \frac{1}{\sin (j \pi / 12)}=59.8074

If (n-1) charges are on the circle (energy \Omega_{n-1} q^{2} / 4 \pi \epsilon_{0} R ), and the nth is at the center, the total energy is 

W_{n}=\left[\Omega_{n-1}+(n-1)\right] \frac{q^{2}}{4 \pi \epsilon_{0} R} .

For

n=11: \quad \Omega_{10}+10=38.6245+10=48.6245>\Omega_{11}

n=12: \quad \Omega_{11}+11=48.5757+11=59.5757<\Omega_{12}

Thus a lower energy is achieved for 11 charges if they are all at the rim, but for 12 it is better to put one at the center.