Re: Gauss' law and displacement current

["Carl E. Mungan" <mungan@USNA.EDU>]

> I think the charges that are causing the "electric field in the
> wire" are located outside of the Gaussian surface you have described.
> (In particular, the battery is located outside the surface.) So the
> electric field lines may "enter" the Gaussian surface through the
> wire, but they continue beyond the charges on the capacitor face and
> "exit" the Gaussian surface somewhere on its other side, leading to no
> net change in the integrated flux.

This is the part I don't understand. When you close the switch and
turn on the current, why does the electric field inside the capacitor
change (in order to give you extra field lines exiting there as you
state) even though the charge on the capacitor hasn't changed?

This is why I think I haven't accounted for the charges properly. You
can't change the fluxes without changing the charges. I'm still
stumped. More help anyone?

In case anyone missed my question the first time around, here it is
again. Consider two identical RC circuits, with both capacitors half
charged up. In circuit A, there is no current running into the
capacitor. In circuit B, there is. For both circuits, surround the
positive plate with a Gaussian surface cutting across the wire
connected to the plate. The right-hand side of Gauss' law seems to be
the same for both circuits because both capacitors carry the same
charge; the left-hand side is different for both circuits because the
electric field in the lead wire decreases the net flux for circuit B.
(Since both capacitors carry the same charge, I assume the electric
field inside both capacitors is the same.) What's wrong with my
analysis?
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/