A parallel-plate capacitor is filled with air. Each plate has area 10 cm

^{2}. The separation between plates is 2.0 cm. The top place stores +200 nC of charge, while the bottom plate stores -200 nC of charge.
Above is just the stem of the question I'm writing for the upcoming edition of

*5 Steps to a 5: AP Physics C*. Consider it a "goal-less" problem for a moment. What could be figured out?
The first and most obvious answer is to calculate the capacitance of the capacitor using

*C = εA/d*. And yes, that is a calculation fraught with pitfalls, as the plate area and separation have to be converted to SI units.* But carrying out this sort of calculation is not the point of introductory physics. If you really want a correct calculation, don't ask a first-year high school student with a calculator - use Wolfram Alpha. More importantly, I don't expect even strong students to be able to calculate the right answer in about a minute.
*

*Or I suppose you could convert epsilon-naught to F/cm. Whatever floats your boat.*

Better, more authentic questions might involve semi-quantitative reasoning. What happens to the capacitance when I double the plate area? Plate separation? Charge storage? Using

*C = εA/d,*you can see quickly that the capacitance doubles, halves, or remains the same because the relevant equation is in the numerator, denominator, and not involved, respectively.
A further step could involve graphs - what would an experimental graph of capacitance vs. plate area, plate separation, or charge storage look like? Linear, inverse, and horizontal, respectively.

Or, we could go one step beyond single-equation reasoning to ask how the electric potential across the capacitor changes depending on plate area, plate separation, or charge storage. This question involves combining two fundamental relationships:

*C = εA/d*for the capacitance, and*Q = CV*for the voltage. In this case doubling plate area doubles capacitance, but halves voltage; doubling plate separation halves capacitance and doubles voltage; doubling charge storage doesn't change capacitance but doubles voltage.
(And you could even ask about the electric field inside the capacitor, which is uniform and equal to

*V/d*. Or the field or potential at several positions inside the capacitor. Or...)
A fun exercise might be to present your class with the stem above and some whiteboards with, say, 10 minutes remaining in class. Then tomorrow's daily quiz would include two multiple choice questions based on this stem, with only two minutes to answer. I'll bet you get (a) good conversation amongst the students, (b) mostly right answers on the quiz, and (c) everyone finishing the quiz with lots of time to spare.

For the "beyond single-step equation reasoning", do you have to specify whether the capacitor is connected to a constant voltage source or not?

ReplyDeleteDepends on what you're asking. If it's an isolated capacitor, the charge on the plates is constant; that's what I'm assuming above. But nothing wrong with connecting to a constant voltage source, in which case the charge can change. See... billions and billions of ideas!

ReplyDelete