Re: Gauss' law and displacement current

[Michael Bowen <fizzbowen@MINDSPRING.COM>]

"Carl E. Mungan" 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'll take a stab at it, using language that students should be able to
understand (I hope I don't oversimplify; although if I do we'll get a
response from others on the NG, which means I'll learn something too
:-) ). 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.

> 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.

I don't think this need be so; H & R, for example, state no explicit
assumption either way, and make it clear in earlier chapters that the
integral must always be taken over a closed surface. Close your
surface by including the wire, but then take into account the
additional flux that leaves other portions of the Gaussian surface,
probably mostly in the region between the cap plates.

> 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.

I think we've solved the problem above. This a different (although
possibly valid) issue. Displacement current is related to the time
derivative of the flux, not to the flux itself.

> 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

You accounted beautifully for the charge, but you missed some flux.

--MB