Re: There's work, and then there's work

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

David Rutherford wrote:

> Bob LaMontagne wrote:
> > David Rutherford wrote:
> >
> > > Could someone please refer me to experimental evidence that the total
> > > energy stored by a capacitor is 1/2 CV^2. Thanks.
> >
> > That's an easy one. In General Physics II the students charge a capacitor to
> > an arbitrary voltage, V. They then discharege it through a large resistor,
> > measuring the current as a function of time. They then graphically integrate
> > i*dt to get the total charge,Q. They calculate the value of 1/2 Q*V, which is
> > found to be identical to 1/2 C*V^2. Not as direct as a calorimeter experiment,
> > but the point is that the value 1/2 C*V^2 is not only well know, it has been
> > verified often - even by introductory students.
>
> It seems to me that you could have, just as easily, started with QV and
> ended up with CV^2, since according to the conventional definition, Q =
> CV, you can say that
>
> QV = (CV)V = CV^2
>
> or you could have said that
>
> 1/4 QV = 1/4 CV^2
>
> How does the way you did it prove that the correct value for the energy
> stored in a capacitor is 1/2 CV^2?

Yes, if you take it that literally, you're correct. Please excuse my shorthand - it
was an abbreviation for the quick response to students when they ask why the energy
is 1/2QV instead of QV - one responds that the potential varies from 0 to V and the
average is V/2.

More explicitly, what the students do in the lab is the following: They measure the
slow discharge of the capacitor through a resistor, R. They record the current at
regular time intervals, dt, say 5 seconds. They then graph the current as a function
of time and obtain the total charge. But, more importantly, they do a semilog plot
of i vs t. This gives them an explicit equation for the current as a function of
time: i = V/R exp(-t/RC). This is a direct fit to their data and makes no
assumptions. They can then find the energy dissipated in the resistor by forming the
integral of i^2 * R * dt. They substitute their experimentally determined function
for i and perform the integration, which results in 1/2 C V^2.

Again, this is not quite as satisfying as Ludwik's suggestion of just measuring the
temperature rise using  a calorimeter, but it does suggest the answer you will get
if you do it.

Sorry for not being as complete in my previous response as I could have.

Bob at PC