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Point charge inside solid sphere

  1. Aug 15, 2010 #1
    I'm curious. Say you have a solid, positively charged sphere with a uniform charge distribution. Now suppose you place a single, negatively charged electron inside the solid sphere. How will the electron behave? I assume the electrostatic forces exerted on the electron will ultimately tend it towards the center, where it will reach a sort of equilibrium and remain there motionless?
  2. jcsd
  3. Aug 15, 2010 #2
    Sort of.

    In fact, you couldn't place it so carefully that it wouldn't have some residual momentum.
    It would move around, and the other charges involved would move in response. It's a pretty chaotic situation with no real stable equilibrium, like balancing a pencil on it's point.

    Inevitably it would come into close proximity with one of the other charges and they would neutralise one another.
  4. Aug 15, 2010 #3
    the electric field inside a uniformly charged sphere is exactly zero. this was first proven by Newton for gravitational forces.
  5. Aug 15, 2010 #4
    As long as you don't disturb the status quo by trying to stick a charge in the middle.

    Of course, it returns to zero after the intruder has been dealt with.
  6. Aug 15, 2010 #5
    He didn't say the sphere was conductive, so I assumed that the charges are rigidly stuck on the sphere.
  7. Aug 15, 2010 #6
    Hmmm... theoretically yes, but is that physically possible do you think?
  8. Aug 15, 2010 #7
    This problem is theoretical. Also, if it was a conductor, then there is the Thompson's theorem that a point charge can never be found in a state of stable equilibrium in an electrostatic field.
  9. Aug 15, 2010 #8
    You didn't mention the approximations you want to assume. In classical electromagnetism, neglecting the electron's field, inside the sphere you have field directed radially with amplitude proportional to the distance from the center. So it's like the electron is linked to the center with an ideal spring. The resulting motion is an ellipse. In time the electron will lose energy and spiral down the center.
  10. Aug 16, 2010 #9
    It's true only if the sphere is a conductor. But if the sphere is an insulator, then you can easily prove (by Gauss law) that the electric field inside the sphere is proportional to the r, the distance from the center (r<R radius of the sphere). Hence, exactly in the center of the sphere, the electric field is zero and because the electron is point charge we can assume that it will remain in the center in equilibrium.
  11. Aug 16, 2010 #10
    Ok, ok, I thought "sphere" means a spherical shell, not a uniformly charged ball.

    If it is a uniformly charged spherical shell, then the electric field inside is zero.

    If it is a uniformly charged ball, then the electric field inside is radial and the magnitude is proportional to the distance from the center, reaching its maximum value:

    E = k_{0} \, \frac{Q}{R^{3}}

    on the surface of the ball [itex]r = R[/itex].
  12. Oct 19, 2010 #11
    This problem is quite similar to one that I've published on. However, I would never assume that any three-dimensional system consisting of more than 4 point charges is "uniform". (What is meant by "uniform"?!)

    In this case, perhaps we are talking about the original Thomson problem (or, perhaps the jellium model that many today love and adore) in which Thomson assumes that the atom consists of negatively charged electrons (corpuscles) in a "uniform" positively charged volume. But, with our knowledge of atomic structure today, having particulate structure, what does "uniform" mean? With five point charges, one can never place all of them in a three-dimensional volume such that they are all equidistant from each other. At best, you could find a configuration in which they may all have the same electrostatic energy.

    ..basically, my argument is that this question is physically impossible, not so much because of so-called 'quantum-fluctuations' that prevent a negative point charge charge from actually finding, and residing at, the origin of the sphere, but because there is no such as a perfectly "uniform" volume of charge. ...those charges are particles.

    Now, if we are discussing "an approximation to uniformity", then the greater the number of charges there is in the positive sphere, the better that approximation. Then, we might as well consider a metallic system because all the spatial symmetry properties of the "uniformity" must be "washed out".

    While this may seem like a trivial point to make, I assure you that it makes a world of difference in the final analysis. Of course, if you're looking for a statistical-natured argument (e.g. quantum or density functional), then by all means, this is a trivial point. But if you're looking for a true electrostatic solution, then this point is of utmost importance.

    Just some thoughts.

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