Understanding Gauss's Law: When to Apply and When it Fails

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Gauss's Law is applicable in scenarios where the symmetry of the charge distribution allows for simplification, but it can fail with discontinuous charge distributions like dipoles. When applying Gauss's Law, the total electric flux through a closed surface is determined by the enclosed charge, which can lead to confusion if the charge is zero, as in the case of a dipole. It is crucial to understand that a zero net charge does not imply a zero electric field at every point on the surface. Instead, the law indicates that the electric flux entering and exiting the surface is balanced. Proper application requires recognizing the nature of the electric field and charge distribution involved.
dipole
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This is not a homework question.

I'm just having trouble understanding when it's appropriate to apply Gauss's law or not, and under what conditions does it fail.

For example, suppose I have two point charges forming a dipole. If I draw a cylinder around them, then the total charge enclosed is zero, implying that the field is zero.

However, this is obviously not true for a dipole.

So would I be using Gauss's law incorrectly in this case, or does it not applying to discontinuous charge distributions, or what?
 
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dipole said:
This is not a homework question.

I'm just having trouble understanding when it's appropriate to apply Gauss's law or not, and under what conditions does it fail.

For example, suppose I have two point charges forming a dipole. If I draw a cylinder around them, then the total charge enclosed is zero, implying that the field is zero.

No, it implies that the closed surface integral of a certain projection of the electric field is zero. Not that the field is zero at each point on the enclosing surface.
 
dipole said:
For example, suppose I have two point charges forming a dipole. If I draw a cylinder around them, then the total charge enclosed is zero, implying that the field is zero.
Gauss's Law would say ##\oint \vec{E}\cdot d\vec{S} = 0##. This means the electric flux into a volume is balanced by the electric flux out of the volume. It doesn't mean ##\vec{E}=0##, which would mean there is no electric flux at all into or out of the volume.
 
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