h/Completing the proof of
Gauss’ theorem.the information reaches the right side. This would seem certain to lead
to a violation of Gauss’s law. How can the ideas explored in discussion
question C show the resolution to this paradox?10.6.4 Proof of Gauss’ theorem
With the computational machinery we’ve developed, it is now
simple to prove Gauss’ theorem. Based on additivity by charge, it
suffices to prove the law for a point charge. We have already proved
Gauss’ law for a point charge in the case where the point charge is
outside the region. If we can prove it for the inside case, then we’re
all done.
If the charge is inside, we reason as follows. First, we forget
about the actual Gaussian surface of interest, and instead construct
a spherical one, centered on the charge. For the case of a sphere,
we’ve already seen the proof written on a napkin by the flea named
Newton (page 641). Now wherever the actual surface sticks out
beyond the sphere, we glue appropriately shaped pieces onto the
sphere. In the example shown in figure h, we have to add two
Mickey Mouse ears. Since these added pieces do not contain the
point charge, the flux through them is zero, and additivity of flux
by region therefore tells us that the total flux is not changed when
we make this alteration. Likewise, we need to chisel out any regions
where the sphere sticks out beyond the actual surface. Again, there
is no change in flux, since the region being altered doesn’t contain
the point charge. This proves that the flux through the Gaussian
surface of interest is the same as the flux through the sphere, and
since we’ve already proved that that flux equals 4πkqin, our proof
of Gauss’ theorem is complete.
Discussion Questions
A A critical part of the proof of Gauss’ theorem was the proof that
a tiny cube has zero flux through it due to an external charge. Discuss
qualitatively why this proof would fail if Coulomb’s law was a 1/ror 1/r^3
law.
10.6.5 Gauss’ law as a fundamental law of physics
Note that the proof of Gauss’ theorem depended on the compu-
tation on the napkin discussed on page 10.6.1. The crucial point in
this computation was that the electric field of a point charge falls
off like 1/r^2 , and since the area of a sphere is proportional tor^2 ,
the result is independent ofr. The 1/r^2 variation of the field also
came into play on page 644 in the proof that the flux due to an out-
side charge is zero. In other words, if we discover some other force
of nature which is proportional to 1/r^3 orr, then Gauss’ theorem
will not apply to that force. Gauss’ theorem is not true for nuclear
forces, which fall off exponentially with distance. However, this is
theonlyassumption we had to make about the nature of the field.
Since gravity, for instance, also has fields that fall off as 1/r^2 , Gauss’
theorem is equally valid for gravity — we just have to replace mass
Section 10.6 Fields by Gauss’ law 647