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Difference in potential energy of two charge configurations

  1. Feb 14, 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
  3. Feb 14, 2017 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Are ##V_1## and ##V_2## the electrostatic potential energies associated with each distribution?
    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?
    2charges.png
     
  4. Feb 14, 2017 #3
    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##.
    MDdbAwt.png
    I got ##N=12##.

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