John E. Davis wrote:
On Tue, 13 Mar 2007 13:57:18 -0800, Richard Clark
wrote:
Was this thanks for his misreading Gauss where it should have been
Maxwell?

I do not understand your comment. If you go back and look at my first
post on this subject (Message-ID ),
you will see that I equated Gauss's law with the first Maxwell equation.

Gauss's law is commonly stated as:

The electric flux through a closed surface is proportional to the
amount of charge enclosed by the surface.

As I wrote before, this also happens to be the integral form of the
first Maxwell equation:

div E(x,t) = 4\pi\rho(x,t)

While Gauss may have stated this law in terms of static charges, and
it finds most applications in the static case, the law also holds for
the dynamic case. This is why physicists equate Gauss's law with the
integral form of the first Maxwell equation. And as evidence of this
association, you indirectly pointed out in Message-ID
that Feynman equated the
two in the table 15-1 of volume II of his lectures.

--John

Using the MKSA system, Gauss' law is expressed as div D = rho. Art can
take the time derivative of both sides, if he wants to, in which case he
gets div d(D)/dt = d(rho)/dt. This doesn't mean much except that it's
what you end up with when you take the divergence of both sides of
the Maxwell equation curl H = j + d(D)/dt, and then apply the equation
of continuity where it fits. (You have to pretend the 'd's'
in each equation are the funny little Greek letters that signify
partial differentiation.) Feynman didn't like to use the magnetic field
intensity vector H or the electric flux density vector D
so he used their B and E equivalents in his presentation of Maxwell's
equations in his _Lectures on Physics_. I guess you could start an
argument over whether or not H and D have physical significance, but
don't ask me to join in.
John, I think you might want to re-think your equation div
E(x,t)=4\pi\rho(x,t).
73,
Tom Donaly, KA6RUH