I searched the archives to see if this has been discussed before because
it seems it should have been. I didn't see it. My apologies if it has
already been discussed.
When analizing a circuit containing a charged capacitor, a resistor, and
an initially-open switch, the analysis in many textbooks goes like
this...
(1) close the switch
(2) apply the loop theorem and get q/C + IR = 0
In some books this is written -q/C -IR = 0. Normally I would say this
is because they went around the loop in the other direction, but as I
explain below, it seems neither is correct.
(3) Replace I with dq/dt
(4) Rearrange to get dq/q = -dt/RC
(5) Integrate to get q(t) = Qexp(-t/RC) where Q is the initial charge.
Of course (5) is the correct answer, but (2) is not what I (and
students) get from the loop theorem. Proper application of the loop
theorem yields opposite signs for the two terms rather than the same
sign. Then, following through steps 3 to 5 yields q(t) = Qexp(+t/RC)
which is the incorrect answer.
A couple sources I have do the loop theorem in step (2) as q/C - IR = 0
or -q/C + IR = 0 depending on whether you do the loop with the current
or opposite the current. Then they get the signs both the same by
applying some handwaving in step (3). The handwaving goes like this...
I = dq/dt, but since q is decreasing we have to write I = -dq/dt.
It seems to me that the textbooks that get the same sign for the two
terms from the application of the loop theorem in step (2) are just
plain wrong. No matter which way you navigate the loop, one delta-V
will be positive and the other negative. Indeed, when analyzing the
charging of the circuit (with battery in the loop) most books do the
loop theorem properly... E - q/C - IV = 0 going in the current
direction. Serway and Jewitt actually say (for charging), "For the
capacitor, notice that we are traveling in the direction from the
positive plate to the negative plate; this represents a decrease in
potential." (page 874) But then on page 876 where they analyze the
discharge they use the same equation and just drop the E because the
battery has been removed. The problem is, when the battery is removed
(and replaced with a wire) the current flows the opposite direction. In
this case (current direction) we still have a potential drop across the
resistor, but we now have a potential increase across the capacitor (if
we apply the same reasoning as stated in the quoted portion from page
874). Serway and Jewitt just seem to ignore this, i.e. they make no
comment about which way they went around the loop and how -q/C -IR could
possibly be zero if q,C,I,R are all unsigned numbers.
One the other hand, even though I prefer saying I = -dq/dt in step (3),
the problem with setting I = -dq/dt is that it certainly seems to
students that some handwaving is being done just to get the correct
answer. Indeed, when current is typically defined as dq/dt we usually
explain that it doesn't really mean charge is changing, but rather a
certain amount of charge dq passes a point in time dt. From that
definition of I = dq/dt it seems difficult to justify setting it to
minus dq/dt in the capacitor problem.
Are any of you bothered by this? Do any of you have a clean way out of
this?
Michael D. Edmiston, Ph.D.
Professor of Physics and Chemistry
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu