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Homework Help: Why does charge accumulate at the sharpest point?

  1. Nov 18, 2005 #1
    I'm sure I used to have a really simple answer to this. But I've long since forgotten.

    BTW This is not my homework.

  2. jcsd
  3. Nov 18, 2005 #2
    I believe the idea is that at least for a conductor, the charge exists evenly on the surface. So when you have a sharp point, you may have charges that are very close to each other in space, even though they may be equally spaced on the surface.
  4. Nov 19, 2005 #3
    The local charge density varies inversely as the square of the radius of curvature.
  5. Nov 19, 2005 #4
    Yes, of course. But why??
  6. Nov 19, 2005 #5


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    So that the potential may satisfy the Laplacian outside the conductor.
  7. Nov 20, 2005 #6


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    Here's a simple model.
    Consider two conducting spheres, with unequal radii A and B, connected by a long thin conducting wire... so the spheres are at the same potential. How do free charges distribute themselves on the two spheres?
  8. Nov 20, 2005 #7
    V_A & = \frac{\sigma_A 4\pi A^2}{4\pi\epsilon_0 A} \\
    & = \frac{\sigma_A A}{\epsilon_0} \\
    V_B & = \frac{\sigma_B B}{\epsilon_0} \\
    \sigma_A A & = \sigma_B B

    This suggests that the charge density is higher on the smaller sphere. Quite nice, but I think there is probably a simpler explanation.
  9. Nov 20, 2005 #8

    Physics Monkey

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    jdstokes, have you thought about what would happen at a true corner in a conductor? You know the electric field is always normal to the surface of a conductor and continuous outside the conductor, but what happens at a corner? Imagine you sit just outside the corner: if you move a little to the left you get a field parallel to one surface normal, move a little to the right you get a field parallel to the other surface normal (the surfaces meet at an edge so imagine two planes intersecting). The field seems discontinuous just outside the corner. Of course there are physical limits, but the point is that the electric field is varying really fast, so what does this tell you about the charge density?
    Last edited: Nov 20, 2005
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