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If we have a point charge and outside of it we have a non-conducting Gaussian sphere, then Gauss's law says that the net flux should be zero. I agree that the total field lines coming in are equal to the number of field lines coming out, but isn't it true that the magnitude of the electric field at those segments of the sphere where the field is coming in is greater than those where it is coming out? So when we integrate, wouldn't these areas, having a greater electric field, contribute more to the net flux? I don't see why they would cancel each other out.

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  • $\begingroup$ Flux depends on area. Those segments of the sphere with a high electric field should have lower area than the segments with lower electric fields. $\endgroup$ – Mark Fantini Jan 31 '15 at 2:13
  • $\begingroup$ Well, half of the sphere would have field lines coming in and these would be more intense, wouldn't they? $\endgroup$ – Simeon Jan 31 '15 at 2:15
  • $\begingroup$ Why would they be more intense? $\endgroup$ – Mark Fantini Jan 31 '15 at 2:16
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    $\begingroup$ in highly qualitative terms, if the charge is positive, the field going inward is more intense, but on a smaller surface, while the outgoing field is weaker, but on a larger surface element. $\endgroup$ – Phoenix87 Jan 31 '15 at 2:16
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    $\begingroup$ You can't have a state with parallel, non-constant electric field without a local charge density. In terms of field lines, the spacial density represents the strength, so the only way they can be parallel and getting weaker is if some fraction of the lines end which only happens on charges. $\endgroup$ – dmckee Jan 31 '15 at 3:36
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I agree that the total field lines coming in are equal to the number of field lines coming out

To be sure, do you agree that no field line entering fails to exit or any field line exiting fails to enter? That is, do you agree that no field line originates or terminates within the volume enclosed by the sphere?

but isn't it true that the magnitude of the electric field at those segments of the sphere where the field is coming in is greater than those where it is coming out?

That's essentially irrelevant. If no field lines entering terminate within the volume and no field lines leaving originate within the volume, the net flux through the surface is zero.


What would happen, however, if instead of a point charge, we had a uniform electric field (with parallel field lines), which falls of as 1/r. Then in that case, there must be more flux coming in than out?

Generally, a uniform electric field means that there is no spatial dependence; an electric field that falls of as $\frac{1}{r}$ is thus not a uniform electric field.

Let's stipulate that the electric field is of the form

$$\vec E = \frac{k}{x}\hat x$$

See that the field lines are parallel to the $x$ axis and fall off as $\frac{1}{x}$.

The divergence of this field is then

$$\nabla \cdot \vec E = -\frac{k}{x^2} = \frac{\rho(x)}{\epsilon_0}$$

So, while you're correct that the net flux through a closed surface is non-zero, it's also clear that there is a non-zero charge density enclosed since $\rho(x) \ne 0$.

Thus, by Gauss' law, there must be a non-zero flux through the closed surface.

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