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Contradiction in Gauss's Law

  1. Apr 11, 2007 #1
    Consider a gaussian surface of arbitrary size and a point charge located outside of the gaussian surface at an arbitrary distance.

    Gauss's law states that the flux through the gaussian surface is zero, since there is no charge enclosed by that surface. From this we can deduce that the electric field must be zero everywhere on the surface, since the flux is equal to the integral of the dot product of the electric field and dA. But from Coulomb's Law we know that the E = kq/r^2 at any point on the surface, where r is the distance from the point charge.

    It seems to matter what gaussian surface you choose and whether or not it encloses the point charge. If you choose a spherical gaussian surface centered around the point charge, you can easily derive E = kq/r^2. So why does this contradiction occur if you choose a gaussian surface that does not enclose the point charge? Are you simply not supposed to or allowed to do that?
     
  2. jcsd
  3. Apr 11, 2007 #2

    marcusl

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    The electric field is not zero on the surface! Only the total flux integrated over the surface is zero, indicating there is no source inside. It might help to think of water flowing through the volume. Gauss's law says as much water leaves as enters, since there's no source. That's different from the flow itself, which is nonzero on the surface.
     
  4. Apr 11, 2007 #3

    jtbell

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    Imagine that the Gaussian surface is the surface of the moon, and the (positive) charge is located on the earth. On the side of the moon that faces the earth, the electric flux "goes into" the moon, and is negative; whereas on the far side of the moon, the flux "comes out" of the moon (having "passed through" it), and is positive. The net flux over the entire surface is zero.
     
    Last edited: Apr 11, 2007
  5. Apr 12, 2007 #4

    Meir Achuz

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    "From this we can deduce that the electric field must be zero everywhere on the surface, since the flux is equal to the integral of the dot product of the electric field and dA."

    A zero integral does not imply a zero integrand.
     
  6. Apr 12, 2007 #5
    That comment is misleading. A zero indefinite integral implies a zero integrand. However, a zero definite integral does not imply a zero integrand.
     
  7. Apr 12, 2007 #6

    reilly

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    Gauss's Law can be based on the volume integral of the divergene of the electric field. If there's no charge inside a closed surface, then the divergence of E is zero, and in a couple of simple steps you are done.

    Regards, Reilly Atkinson
     
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