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Flux Through a Sphere Containing a Dipole

  1. Sep 9, 2013 #1
    So I am probably understanding something wrong but would gauss's law [itex]\frac{Qinside}{\epsilon0}[/itex] for a sphere surrounding an electric dipole, with a point charge just outside of the sphere.

    If you imagine this scenario without the outside charge, the field lines through the surface all come back in and some don't leave the sphere making the flux = 0. Which is consistent with [itex]\frac{total charge inside}{epsilon0}[/itex] which is = 0. BUT If you introduced a negative charge just outside the sphere some of the electric field lines would terminate there, making the flux a positive number (not all the field lines that leave through the surface return). While the equation [itex]\frac{total charge inside}{epsilon0}[/itex] would still be 0.

    What is wrong with my mental picture? Why does this scenario seem to violate Gauss's Law?
  2. jcsd
  3. Sep 9, 2013 #2
    The flaw in your logic is that the field lines from the outside charge will not cause any net flux--any field lines going into the sphere due to the nearby charge will be compensated by field lines going out of the sphere.

    One way to see that this is true is to invoke the principle of superposition. Go ahead and calculate what the net flux into an empty sphere is due to a charge sitting outside it. You will indeed find it is zero--and that is because the field lines going into the sphere must leave the sphere, so there is cancellation. You already argued why the dipole's field has zero net flux through the sphere. We can use superposition to say that the field with a dipole in the sphere and a charge outside the sphere is exactly the sum of the dipole's field and the outside charge's field. Since both of those two contributions to the field both have zero net flux into the sphere, they sum to give zero net flux into the sphere.

    If you are worried that the field lines from the outside charge will hit the dipole and "terminate" at the dipole, then what will happen is that the dipole will also emit another field line to exactly compensate. If any field lines were to enter the dipole without some corresponding exiting field lines, that would be a contradiction, since it would imply the dipole has a net charge.
    Last edited: Sep 9, 2013
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