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When can a charged sphere be treated as a point charge?

  1. Oct 27, 2011 #1
    Not much more to add to the title. I was just wondering, under what conditions can a charged sphere be treated as a point charge?
     
  2. jcsd
  3. Oct 27, 2011 #2

    BruceW

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    The electric field outside a (uniformly) charged sphere is equal to the electric field that would have been created by a point charge with the same total charge.
     
  4. Oct 27, 2011 #3

    Doc Al

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    When the charge distribution is spherically symmetric, the field outside the sphere is exactly that of a point charge at the center of the sphere. (Where the charge of the point charge equals the total charge of the sphere.)
     
  5. Oct 27, 2011 #4
    So in cases when the charge is uniformly distributed over its volume or its surface, the sphere can be treated as a point charge?
     
  6. Oct 27, 2011 #5

    BruceW

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    yep. Doc Al's response is slightly more accurate than mine - as long as the charge distribution is symmetric, the electric field outside will be same as that created by a point charge.

    So any symmetric distribution you can think of will have this remarkable property.

    Edit: For clarity, symmetric just means that the charge distribution will only depend on the radius. (Well, that's probably not the technical meaning of symmetric, but that's what it means in this example).
     
  7. Oct 27, 2011 #6
    Apply the Gauss law in Maxwell's equations. Integral form, ∫E·dA=Q/ε0, where Q is the charge bounded inside the close surface. For any spherical Gaussian surface, E4πr2=Q/ε0, and is exactly the same as coulomb's law for point charge.

    This means you can apply to whatever the charge distribution, it can be modelled by a equivalent point charge.
     
  8. Oct 27, 2011 #7
    "When can a charged sphere be treated as a point charge? "

    This is only true if the sphere is isolated from everything else. If another charge, or another object held at some potential, is brought near the sphere, it will induce surface charges in the sphere that are not spherically symmetric, and therefore you can no longer treat it as a point charge. But if the radius of the sphere is very small compared to characteristic lengths of the system, then you can treat it as a point charge even if it is not isolated because the induced charges will be spherically symmetric as an approximation if the sphere is very small.
     
  9. Oct 27, 2011 #8
    But it can still be viewed as an point charge by taking a new gaussian surface where the centre shifts to the new equivalent point charge and take a larger sphere. Everywhere outside the surface the model is still valid.

    But when there is an external charge the field is again distorted. Actually the net field should be modelled by another equivalent charge and a even larger sphere as gaussian surface.
     
  10. Oct 27, 2011 #9
    I think you are confused. The Guassian-surface trick only works if you have a constant field over the Gaussian surface which will only happens if the charge density if there is symmetry (a spherical Gaussian surface requires a spherically symmetric charge density, a cylindrical Guassian surface requires...,etc). As soon as the symmetry is lost, the Guassian-surface trick no longer works and you have to go to more advanced methods (method of images, Green functions, etc).
     
  11. Oct 27, 2011 #10

    BruceW

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    No, no, no. Even if you shift the centre and use a larger gaussian sphere, the electric field at each point on that gaussian sphere would not be equal. (Because the distribution is no longer symmetric).
     
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