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I Delta otential energy between two charge configurations

  1. Feb 13, 2017 #1
    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.
     
  2. jcsd
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