# Delta otential energy between two charge configurations

Chapter 24, Question 61

Given two configurations, ##C_1##, ##C_2## of ##N## point charges each, determine the smallest value of ##N## s.t. ##V_1>V_2##.

##C_1##:

##N## point charges are uniformly distributed on a ring s.t. the distance between adjacent electrons is constant

##C_2##:

##N-1## point charges are uniformly distributed on a ring s.t. the distance between adjacent electrons is constant and one charge is placed in the center of the ring.

Approach I

1. If we consider a gaussian surface inside the ring, ##E=0##. We know that the voltage at the center of the ring is $$V_\text{center}=\frac{Ne}{r}$$ and furthermore, because $V=\int E\circ ds$, $$V_\text{inside} = V_\text{center}$$
2. From this previous result, $$U_1 = eV_\text{center} = \frac{N(N-1)e^2}{r}$$
3. ##C_2##
the configuration potential without the center electron is $$(1/2)(N-1)(N-2)\frac{e}{r}$$ The center electron adds ##(N-1)\frac{e}{r}## yielding $$U_2 = (1/2)(N-1)(N-2)\frac{e}{r} + (N-1)\frac{e}{r}$$
4. Let ##k = \frac{e}{r}##, and, setting ##(N^2-N)k = (N^2-3N+2)k+(2N-2)k##
$$\implies 0=0$$

Approach II

Let ##N## charges be arranged along a circle with radius ##R##. The position of an arbitrary particle at an angle ##\theta## relative to the positive direction of the x-axis is ##\vec{P}(\theta) = (R\cos\theta, R\sin\theta)##. Pick one charge located at angle $\theta = \theta_i$ and another particle located at ##\theta = \theta_j## relative to the positive direction of the x-axis. The distance between the two particles ##\vec{r}## is

\begin{align} \vec{r} &= \|\vec{P}(\theta_j) - \vec{P}(\theta_i)\| \\ &= \|(R\cos\theta_j, R\sin\theta_j) - (R\cos\theta_i, R\sin\theta_i) \\ &= \|R(\cos\theta_j-\cos\theta_i, \sin\theta_j - \sin\theta_j)\|\\ &= R\|(\cos\theta_j-\cos\theta_i, \sin\theta_j - \sin\theta_j)\| \\ &= R \sqrt{ \cos^2\theta_j -2 \cos\theta_j\cos\theta_i + \cos^2\theta_i \\ + \sin^2\theta_j -2 \sin\theta_j \sin\theta_i + \sin^2\theta_i } \\ &= R\sqrt{ 1-\sin^2\theta_j +1-\sin^2\theta_i + \sin^2\theta_j + \sin^2\theta_i \\ -2(\cos\theta_j\cos\theta_i+ \sin\theta_j \sin\theta_i) }\\ &= R\sqrt{ 2-2(\cos\theta_j \cos\theta_i + \sin\theta_j \sin\theta_i) }\\ &=R\sqrt{2(1-\cos\theta_j \cos\theta_i - \sin\theta_j \sin\theta_i) } \\ & = R\sqrt{2(1-\cos(\theta_j + \theta_i))}\\ \therefore \vec{r} &= R\sqrt{2(1-\cos(\theta_j + \theta_i))} \end{align}

Where

$$\theta_k = k\Delta \theta \\ \Delta\theta = \frac{2\pi}{N}$$

We get

\begin{aligned} \vec{r} =R\sqrt{2}\sqrt{1-\cos\left((i+j)\frac{2\pi}{N}\right)} \end{aligned}

With this expression for ##\vec{r}##, we can write the net potential energy for particle ##j## along the circle with the equation assuming all particles have charges ##q##

\begin{align} U_\text{particle i} & = \sum_{j=1, i \ne j}^{N} { \frac{q^2}{4\pi\epsilon_0R\sqrt{2(1-\cos(\theta_i + \theta_j))}}}\\ & = \frac{q^2}{4\pi\epsilon_0R\sqrt{2}}\sum_{j=1, i \ne j}^{N} { \frac{1}{\sqrt{1-\cos(\theta_i + \theta_j)}}} \\ & = \frac{q^2}{4\pi\epsilon_0R\sqrt{2}}\sum_{j=2}^{N} { \frac{1}{\sqrt{1-\cos(\theta_i + \theta_j)}}} \end{align}

we set ##j_\text{initial} =2## which is equivalent to the conditions ##j=1, j\ne i##

For concision, let

$$L = \frac{q^2}{4\pi\epsilon_0R}$$

Then net potential energy can be expressed as

\begin{align} U(n) & = \frac{1}{2}\sum_i^{n}U_i \\ & = \frac{L}{2} \sum_{i=1, j=2}^{n,n} { \frac{1}{\sqrt{2-2\cos(\Delta \theta(i+j-2))}} } \\ &= \frac{nq^2}{8\pi\epsilon_0R} \sum_{j=1}^{n-1}{ \frac{1}{\sqrt{2}\sqrt{1-\cos(j\Delta\theta))}} }\\ &= \frac{nq^2}{8\pi\epsilon_0R} \sum_{j=1}^{n-1}{ \frac{1}{\sqrt{2}\sqrt{1-\cos(j\frac{2\pi}{N}))}} }\\ &= \frac{nq^2}{8\pi\epsilon_0R} \sum_{j=1}^{n-1}{ \frac{1}{4\sin(\frac{j\pi}{N})} }\\ &= \frac{nq^2}{32\pi\epsilon_0R} \sum_{j=1}^{n-1}{ \csc\left(j\frac{\pi}{N}\right) } \\ \end{align}

For two configurations with ##N## charges we define the potential energies:

$$U_1 = U(N) \\ U_2 = U(N-1) + (N-1)L$$

where the second term in the definition of ##U_2## determines the potential for the lone particle in the center.

Now we solve for the ##N## at which ##U_1 = U_2##

\begin{aligned} U_1 - U_2& = \Delta U \\ &=N\frac{q^2}{32\pi\epsilon_0R} \sum_{j=1}^{n-1}{ \csc\left(j\frac{\pi}{N}\right) }\\ &+(1-N)\frac{q^2}{32\pi\epsilon_0R} \sum_{j=1}^{n-2}{ \csc\left(j\frac{\pi}{N-1}\right) }\\ & +(1-N)L \end{aligned}

Which is really difficult to solve directly.