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Force on a point charge due to a sphere

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    An insulated conducting sphere of radius ##R##, carrying a total charge of ##Q##, is in the field of a point charge ##q## of the same sign. Assume ##q\ll Q##. Calculate and plot the force exerted by the sphere on ##q## as a function of distance from the center. In particular, for ##q/Q = 0.1##, calculate the point at which the direction of force reverses.

    2. Relevant equations
    The electric field due a point charge $$E_\mathrm{pt} = \frac{q}{4\pi\epsilon_0 r^2}$$
    $$\Phi(r,\theta) = \sum\limits_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l\left(\cos{\theta}\right)$$
    $$\vec{E} = -\vec{\nabla}\Phi$$

    3. The attempt at a solution
    So since the solution is wanted as a function of distance from the center of the sphere, I am using the legendre solutions to laplace's equation. The boundary conditions are ##\Phi(R,\theta) = 0## and ##E_\mathrm{in} -E_\mathrm{out} = -\sigma/\epsilon_0 \implies E_\mathrm{point} - \nabla_r \Phi =-\frac{4\pi R^2}{\epsilon_0}##
    Then once the boundary conditions are satisfied then the force is just ##F = qE_\mathrm{sphere}##.

    So for the potential of the sphere:
    $$\Phi(r) = A_0 + \frac{B_0}{r}$$
    $$\Phi(R) = 0 = A_0 + \frac{B_0}{R}\implies B_0 = -A_0 R\implies \Phi(r) = A_0\left(1-\frac{R}{r}\right)$$
    $$E_\mathrm{sphere} = -\frac{2 A_0 R}{r^2}$$
    Then, applying the boundary conditions at the surface of the sphere:
    $$\frac{q}{4\pi\epsilon_0 R^2} + \frac{2 A_0}{R} = -\frac{4\pi R^2 Q}{\epsilon_0}\implies A_0 = \frac{-q + 16\pi^2 R^4 Q}{8\pi\epsilon_0 R}$$

    Then,
    $$E_\mathrm{sphere} = \frac{1}{r^2}\left(\frac{q + 16\pi^2 Q R^4}{4\pi\epsilon_0}\right)$$
    Then from here I would just do force by multiplying by ##q## of the point charge and then find the point at which the force reverses.

    Alot to take in, but does this seem correct thus far?
     
  2. jcsd
  3. Mar 8, 2015 #2

    mfb

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    Where does that come from? I think the idea of the conducting sphere is the non-uniform charge distribution. The electric field and therefore the potential should depend on θ.
     
  4. Mar 8, 2015 #3
    Right, then I don't see how I can get the constants ##A_1, B_1## with only two boundary conditions.
    Thats why I only chose to keep ##l = 0## terms.
     
  5. Mar 8, 2015 #4

    mfb

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    There are formulas how a charge q at some distance L will influence charges on a conducting sphere. It's possible to derive them but I guess that is not necessary here.
    Q does not matter for this induced charge.
     
  6. Mar 8, 2015 #5
    Is this method correct that I am using? Or should I use an image method here?
     
  7. Mar 8, 2015 #6

    TSny

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    Your method is not incorrect, but it involves an infinite series. I would try the image method first.
     
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