Re: Gauss' law and displacement current

[Eugene Mosca <emosca@PTD.NET>]

You seem to be neglecting the field both outside the conductors and outside
the region between the capacitor plates.  For example, the field inside the
current carrying wire is sourced by surface charges on the wire. These
surface charges also source a field outside the wire.  In addition, the
configuration of these charges changes when the switch is opened and the
current ceases.

Also, if the switch is closed and there is current their must be a radial
electric field inside each capacitor plate driving the current.  I presume
this field is sources by surface charges on both sides of each plate, and
that these surface charges also source a field outside the plates.  Again,
the configuration of these surface charges changes when the switch is opened
and the current ceases.

Gene

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On 4/18/01 2:30 PM, "Carl E. Mungan" <mungan@USNA.EDU> wrote:

> Consider the usual RC series circuit consisting of a battery V,
> switch, resistor R, and parallel-plate capacitor C. Close the switch
> and allow the capacitor to charge up to say Q = CV/2. Now open the
> switch. Draw a Gaussian surface which surrounds the positively
> charged plate and cuts through one of the lead wires. The flux
> through this surface is Q/e0 (where e0 = 8.85 pF/m).
>
> Now close the switch and evaluate Gauss' law an instant later. The
> charge on the plate hasn't changed. There's no net charge in the lead
> wire. So the flux should still be Q/e0. Look at the other side of
> Gauss' law. The electric field due to the capacitor hasn't changed.
> BUT!!!! There's now a current in the lead wire. This means there must
> be an electric field in the wire. So the flux can't be the same.
>
> Where's the flaw in the above reasoning?
>
> I hope you see where I'm getting this from: the standard presentation
> of displacement current in textbooks. Take at look in any text.
> Somewhere in the argument the authors will invoke Gauss' law.
> Somewhere else in the presentation, that flux over a closed surface
> becomes a flux over an open surface which excludes the wire. Most
> students don't notice this swindle, until it comes time to explaining
> why in one of Maxwell's equations the surface integral of the
> electric field is over a CLOSED surface and in the other it's over an
> OPEN surface.
>
> An easy fix for most textbooks would be to make the lead wire
> superconducting so that it can carry current with no electric field.
> After this simple case is understood, a textbook can if it wishes
> introduce a wire with finite resistivity and show that the wire now
> carries both a conduction current AND a displacement current, rather
> than giving students the impression that displacement current is a
> special property of capacitors.
>
> Meanwhile, back to my above question. I assume there's some extra
> charge I've missed in my accounting but it's not obvious to me where
> it's lurking. Carl
> --
> 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/