Re: discharging a capacitor

["John S. Denker" <jsd@MONMOUTH.COM>]

At 02:31 PM 3/6/99 -0500, Donald E. Simanek wrote:
> a more fundamental (and important) question:
>
> In any capacitor (forget the dielectric) why does the mere act of
> connecting wires to its plates and bringing them together cause *any*
> electrons to take the long route around the wires to discharge the
> capacitor? The field in the vicinity of the electrons must change, and
> throughout the wire, just from the act of connecting the wire. Why? What
> are the *details*? What's going on within the wires?

Short answer:  Capacitors have fringing fields, and wires have finite
conductivity.



Longer answer follows:

Consider the following step-by-step analysis:

Step 1:  We start out with two capacitors, as shown in figure 1.  Each is
charged so as to have a one-volt drop across each.  We like to think, based
on Kirchhoff's laws, that all E fields are inside capacitors, but that is
not strictly true.  Each capacitor will have a fringing field, which will
be important to our story.  By symmetry you can convince yourself that at
point x, the fringing field has no component in the up/down direction, so a
test charge at that point would feel no up/down force.


                     a   b
                     a   b
                     a   b
       (+++) aaaaaaaaa   bbbbbbbbbb (---)
                     a   b
                     a   b
                     a   b

             x

                     c   d
                     c   d
                     c   d
       (+++) ccccccccc   dddddddddd (---)
                     c   d
                     c   d
                     c   d


                    Figure 1
======================================================

Step 2: Now consider the situation after we have decreased the gap in the
top capacitor, as shown in figure 2.  This was done under conditions of
constant charge, so the E field in the gap remains constant, so the voltage
varies in proportion to the distance.  The fringing field for the top
capacitor is now less intense than for the bottom capacitor.  A positive
test charge at point x will feel a force with an upward component.


                      a b
                      a b
                      a b
        (+)   aaaaaaaaa bbbbbbbbbb   (-)
                      a b
                      a b
                      a b

             x

                     c   d
                     c   d
                     c   d
      (+++)  ccccccccc   dddddddddd   (---)
                     c   d
                     c   d
                     c   d


                    Figure 2
======================================================

Step 3: Connect the capacitors using pieces of high-resistance wire.  At
first there will be no change, but over a times on the order of the RC
time, charges will flow (because of the fringing fields) to establish the
equipotentials (Va = Vc) and (Vb = Vd).  Note that if you use nonresistive
wire, the whole question doesn't make sense -- you can't discuss the forces
on electrons in a wire if the wire is a perfect equipotential.


                      a b
                      a b
                      a b
              aaaaaaaaa bbbbbbbbbb
              r       a b        r
              r       a b        r
              r       a b        r
              r                  r
            x r                  r
              r                  r
              r      c   d        r
             r       c   d        r
             r       c   d        r
             ccccccccc   dddddddddd
                     c   d
                     c   d
                     c   d


                    Figure 3
======================================================


Step 4: Relabel the diagrams so that the bottom capacitor remains a real
capacitor, but the top "capacitor" is a model for the gap between the
contacts of a switch that is being closed.  Make the plate area of the top
capacitor much smaller, so it doesn't hog all the charge.

You can figure out the rest............


> Students actually ask, and are troubled by this question. Few books even
> address it. Most physics teachers (and I have tested this) are speechless
> when asked this question "cold". Or they mumble some totally insufficient
> "answer".

That's understandable.  The analysis seems to require rescinding two
beloved simplifying assumptions.  That's hard for people to do.

Cheers --- jsd