Difference in potential energy of two charge configurations

[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.

 

[kuruman]

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

Are $V_1$ and $V_2$ the electrostatic potential energies associated with each distribution?

If we consider a gaussian surface inside the ring, $E=0$

How do you figure? The electric flux out of the Gaussian surface is zero, but this doesn't mean that the electric field everywhere must be zero as a result. For example, what if the configuration has N = 2?

 

[theideasmith]

Which is really difficult to solve directly.

After fixing up my algebra, I ended up with the equation

$$\begin{aligned}
U_2-U_1 &= 0 \\
&= L\frac{N-1}{2}\sum_{j=1}^{N-2}{\csc\left(\frac{j\pi}{N-1}\right)}
+ L(N-1)2
- L\frac{N}{2}\sum_{j=1}^{N-1}{\csc\left(\frac{j\pi}{N}\right)} \\
&= L\left(\frac{N-1}{2}\sum_{j=1}^{N-2}{\csc\left(\frac{j\pi}{N-1}\right)}
- \frac{N}{2}\sum_{j=1}^{N-1}{\csc\left(\frac{j\pi}{N}\right)}
+ 2(N-1)
\right)
\end{aligned}$$

and numerically solved for $N$ by plotting the series for $U_1$ and $U_2$.

I got $N=12$.

Lesson: Sometimes brute force is necessary.
Full solution: http://theideasmith.github.io/2017/02/14/Point-Charge-Configuration.html