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Re: Capacitor Misconceptions

I don't find it AT ALL surprising that students will treat C = Q/V as
described but deal _more_ correctly with m = F/a. Students have an instinct
for mass (even if they, and we can't define what it really is) and they can
easily buy into the 'mass doesn't change' maxim (excluding relativistic
effects). They don't have the slightest inkling what capacitance is, and
without considerable effort in the instructional process won't internalize
it as a property of the particular device. I know I don't do enough to
solidify this, but I start the capacitor section by having the class itself
derive a formula for capacitance by using the simple parallel plate model
and saying that we want the device to be able to _store_ a separated charge
(positive on one plate, and equal amount of negative on the other). What
physical factors are going to matter and in what way? The class can easily
come up with the Area/separation part and you can usually get someone to
suggest that what's between the plates may matter. This is a start, but
unless you spend considerable time having them calculate the capacitance of
different physical arrangements and really stress that capacitance is
fundamentally a function of the geometry of the individual device, then "C
is proportional to the charge and inversely proportional to the voltage"
should be an _expected_ reply.

rick


-----Original Message-----
From: Prof. John P. Ertel (wizard) <jpe@nadn.navy.mil>


Here is a little problem for all physics teachers when they get to the
section of E&M dealing with CAPACITANCE.

The standard defining equation for the "linear device" we call a
capacitor is
Q = C V or V = Q/C

depending on which, Q or V, you consider to be the "independent variable."

Now, do not ask the following question, unless you want something that
will keep you awake when driving long distances. If you just rewrite the
equation in the form normally mentioned in most texts for the calculation
of capacitance
C = Q/V

ask your students to describe in words what this equation means. Most, if
not all, will give something like the following:

"The capacitance is directly proportional to the charge {on board either
plate}, and inversely proportional to the voltage {difference} between the
plates."

Note: Braces are used here to show words that are sometimes omitted in
their explanation.

I'M NOT KIDDING! Most will really give this explanation. And, when you
tell them its wrong, that since capacitance is a constant, it cannot be
proportional to anything --- it is merely a ratio --- they will certainly
look at you as if you had stripped all of your gears.

What is most paradoxical about this to me is that they will relatively
rarely (although I will NOT use the NEVER word) make the analogous mistake
with the scalar, constant mass form of Newton's second law. That is, if
you write the linear form

F = m a or a = F/m

and solve for the mass

m = F/a

they will rarely say that "the mass is directly proportional to the force
applied, and inversely proportional to the acceleration produced."

And after a detailed explanation of the particulars, again, that since
capacitance is a constant, it cannot be proportional to anything --- it is
merely a ratio --- they will persist in their misconceptions. If on a
later test, you give them the equation (since most will have forgotten it)

Q = C V

and again ask them to tell you what this equation implies about
capacitance, they will solve for the capacitance and offer the same
gibberish as before. It is not only to be a misconception, it seems to be
an "invincible misconception."


ERTEL SENDS.
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