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theideasmith
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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.
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.