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

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

> I noticed this interesting comment at
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html
>
> regarding the energy stored on a capacitor:
>
> "From the definition of voltage as the energy per unit charge, one might
> expect that the energy stored on this ideal capacitor would be just QV.
> That is, all the work done on the charge in moving it from one plate to
> the other would appear as energy stored. But in fact, the expression
> above shows that just half of that work appears as energy stored in the
> capacitor. For a finite resistance, one can show that half of the energy
> supplied by the battery for the charging of the capacitor is dissipated
> as heat in the resistor, regardless of the size of the resistor."
>
> If this is true, doesn't it imply that, when the capacitor is
> discharged, only half of the energy stored on the capacitor will be
> dissipated as heat in the resistor, therefore, the calorimeter
> experiment would only give _half_ of the energy stored on the capacitor?

No.  This is an understandable confusion, but it is very easily
resolved if you do the energy accounting for both the charging and
discharging processes.

The charging case is usually well-modeled as an ideal (i.e. constant
voltage) battery connected in series with an (initially open) switch,
an (initially uncharged) capacitor and a resistor.  After the switch
is closed, the battery pumps the charge Q through a constant voltage
V, thereby supplying a total energy QV.  Meanwhile, the voltage
across the capacitor rises from 0 to V while the voltage across the
resistor falls from V to 0.  It is simple to perform the required
integrals that show that a total energy of QV/2 is delivered to the
resistor (which dissipates it as heat) and a total energy of QV/2 is
delivered to the capacitor (which stores it in the form of
electrostatic energy.)  Pleasingly (and completely unsurprisingly),
the energy supplied by the battery is completely accounted for by the
energy delivered to the resistor and capacitor.

The discharging case is similarly well-modeled as an (initially
charged) capacitor in series with an (initially open) switch and a
resistor.  After the switch is closed, the capacitor discharges
through the resistor.  The charge Q on the capacitor flows as the
voltage across both the capacitor and resistor falls from V to 0.  It
is simple to perform the required integrals that show that a total
energy of QV/2 is removed from the capacitor (which had stored it in
the form of electrostatic energy) and a total energy of QV/2 is
delivered to the resistor (which dissipates it as heat.)  Pleasingly
(and completely unsurprisingly), the energy supplied by the capacitor
is completely accounted for by the energy delivered to the resistor.

Please consult any introductory physics textbook for more details.

--
   A. John Mallinckrodt        http://www.csupomona.edu/~ajm
   Professor of Physics      mailto:ajm@csupomona.edu
   Physics Department         voice:909-869-4054
   Cal Poly Pomona              fax:909-869-5090
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