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Re: New capacitor problem

This is a common problem when students forget the "true" meaning of the
elements of a formula. The defining equation for the linear capacitor is
Q = CV
This can be read (and should be read) as the magnitude of the charge, Q,
(the EFFECT) that resides on either plate of a capacitive device is
directly proportional to the potential difference, V*, (the CAUSE) between
the plates and the proportionality CONSTANT is the capacitance, C, of the
device.

Simple mathematical manipulation yields a equation from which the
capacitance may be (and often is) calculated (assuming that Q and V are
known):
C = Q/V
For too many students, they have learned (incorrectly) that this equation
should be read as "the Capacitance is directly proportional to the charge
on either plate and inversely proportional to the potential difference
between the plates." The "directly proportional ..." part being why
students will come up with zero for the capacitance for an uncharged
capacitor.

What they have forgotten is that the capacitance is really a
proportionality constant for a linear device and, as such, can neither be
proportional nor inversely proportional to anything. In another context,
the same student might look at the equation
R = V/I
and conclude (obviously incorrectly) that the resistance is directly
proportional to the voltage and inversely proportional to the current. Or
equally absurdly, after solving a form of Newton's 2nd Law for mass
m = F/a
conclude that the mass is directly proportional to the force and inversely
proportional to the acceleration.

I've seen students with this problem for years and I have no real cure for
the incorrect assertions. The only thing that I have found helpful is to
use the m,F,a analogy to show how dangerous this kind of logic can be and
then talk about the structural problems a capacitor would have to have for
C to be a function of Q &/or V. I'd be lying if I were to tell you that
this solves the problem. Even after a fairly extensive discussion of the
problem, I will still find fully 25% that will miss such a question on an
exam.

BTW: I'm told that my students are the cream of the crop!


On Sat, 29 Mar
1997, Donald E. Simanek wrote:


I'm trying an experiment of mailing to two groups of which I am a member.
Some of you who are members of both may get two copies. If you are not
members of both and reply without deleting the other one in the "To:"
field, it should bounce. Am I volating any rule of netiquette here? I'd
hate to bring both groups crashing down. :-)

This is not a problem for you folk, but one to try on your *students* to
see whether they grasp the capacitor concept.

1. A capacitor is charged with 6 microcoulomb, and the potental of the
capacitor is measured and found to be 3 volt. What is its capacitance?

2. This same capacitor is now carefully discharged completely. What is its
capacitance now?

You'd be surprised how many students will answer "zero" to part 2. What is
the flaw in their thinking? What is incomplete or incorrect about their
conceptual understanding of capacitance.

-- Donald

.....................................................................
Dr. Donald E. Simanek Office: 717-893-2079
Prof. of Physics Internet: dsimanek@eagle.lhup.edu
Lock Haven University, Lock Haven, PA. 17745 CIS: 73147,2166
Home page: http://www.lhup.edu/~dsimanek FAX: 717-893-2047
.....................................................................




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