Q2.61P - Introduction to Electrodynamics

Question

What is the minimum-energy configuration for a system of N equal

point charges placed on or inside a circle of radius R?  Because the charge on

a conductor goes to the surface, you might think the N charges would arrange

themselves (uniformly) around the circumference. Show (to the contrary) that for

N = 12 it is better to place 11 on the circumference and one at the center. How about for N = 11 (is the energy lower if you put all 11 around the circumference, or if you put 10 on the circumference and one at the center)? [Hint: Do it numerically-you'll need at least 4 significant digits. Express all energies as multiples of $\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}$ ]

Step-by-Step Solution

Answer

Answer

The energy of the configuration having n charges of magnitude q equally spread in a circle of radius R is $\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{n}}{\mathbf{4}\mathbf{s}\mathbf{i}\mathbf{n}\left({\displaystyle \frac{j\pi }{n}}\right)}$. The energy of the configuration having n - 1 charges of magnitude q equally spread in a circle of radius R and one charge at the center is $\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{4}\mathbf{sin}\mathbf{\left(}{\displaystyle \frac{\mathrm{j\pi }}{n-1}}\mathbf{\right)}}\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}$. For n = 12 the former configuration has lower energy but for n = 11 the latter configuration has lower energy. 

Step 1: Given data

There are N equal point charges placed on or inside a circle of radius R.

Step 2: Potential of a point charge

The potential from a charge q at a distance R is

$\mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$ 

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$ is the permittivity of free space.

Step 3: Energy of charge configuration

When n charges are equally spread on a circle of radius R, distance of the j-th charge from one reference charge is

${\mathit{r}}_{\mathbf{j}}\mathbf{=}\mathbf{2}\mathit{R}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\left(\frac{j\pi }{n}\right)$ 

From equation (1), the potential on the reference charge from the remaining  charges is  

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{j}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{1}}{\mathbf{2}\mathbf{R}\mathbf{s}\mathbf{i}\mathbf{n}\left({\displaystyle \frac{j\pi }{n}}\right)} \end{gathered}$$ 

All the charges have this same potential. The net energy of the configuration is

${\mathit{E}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathit{q}\mathit{V}$ 

The half factor is to avoid double counting. Substitute the value in the above equation and get

$$\begin{gathered} {\mathit{E}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathit{q}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{1}}{\mathbf{2}\mathbf{R}\mathbf{s}\mathbf{i}\mathbf{n}\left({\displaystyle \frac{j\pi }{n}}\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{n}}{\mathbf{4}\mathbf{s}\mathbf{i}\mathbf{n}\left({\displaystyle \frac{j\pi }{n}}\right)} \end{gathered}$$ 

If  n - 1 charges are kept at the circle and one is kept at the center, the net energy of the configuration is

${\mathit{E}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{=}\mathbf{1}}\frac{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{4}\mathbf{ }\mathbf{s}\mathbf{i}\mathbf{n}\left({\displaystyle \frac{j\pi }{n-1}}\right)}\mathbf{+}\frac{\left(n-1\right){\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}$ 

The values for  n = 10, 11, 12 are obtained from Mathematica as follows

$$\begin{gathered} {\mathit{E}}_{\mathbf{\circ }}\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{10}\mathbf{)}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\times }\mathbf{38}\mathbf{.}\mathbf{6245} \\ {\mathit{E}}_{\mathbf{\circ }}\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{11}\mathbf{)}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\times }\mathbf{48}\mathbf{.}\mathbf{5757} \\ {\mathit{E}}_{\mathbf{\circ }}\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{12}\mathbf{)}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\times }\mathbf{59}\mathbf{.}\mathbf{8074} \\ {\mathit{E}}_{\mathbf{i}}\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{11}\mathbf{)}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\times }\mathbf{48}\mathbf{.}\mathbf{6245} \\ {\mathit{E}}_{\mathbf{i}}\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{12}\mathbf{)}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\times }\mathbf{59}\mathbf{.}\mathbf{5757} \end{gathered}$$

Thus, for  n = 12, the configuration with 11 charges at the circle and one charge at the center has lower energy but for  n = 11 the configuration with all charges at the center has lower energy.