# I Delta otential energy between two charge configurations

1. Feb 13, 2017

### theideasmith

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.