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Re: Capacitor energy experiment



I will try asking Bob's question in a different way. Suppose I have two 1
F capacitors, each with potential difference 1 V. What physical difference
between the capacitors makes one of them "store" 1 J while the other
"stores" 0.5 J?

Michael Burns-Kaurin
Spelman College





David Rutherford
<drutherford@SOFT To: PHYS-L@lists.nau.edu
COM.NET> cc:
Sent by: Forum Subject: Re: Capacitor energy experiment
for Physics
Educators
<PHYS-L@lists.nau
.edu>


02/12/2003 08:20
PM
Please respond to
Forum for Physics
Educators






Bob LaMontagne wrote:

David Rutherford wrote:

Bob LaMontagne wrote:

For the capacitor, excluding mechanical deformation energy losses,
etc., the
energy is in the field - the electrons cannot store energy internally
- they
are point objects. Others on the list have calculated the thermal KE
associated with the electrons themselves and it was negligible
compared to
1/2 CV^2.

I'm not interested in what list members may have calculated;
calculations cannot be used as evidence. Obviously the energy
associated
with the heat dissipated by the resistor, during charging, is not
negligible, it is 1/2 CV^2 (I'm assuming it has actually been
measured).


David - If you reread what I said - I'm not refering to the resistor at
all. I am
simply challenging you - since your accepted an energy of 1/2 CV^2 in the
capacitor in your recent posting - to show how an identical capacitor
with the
same charge (and therefore field) could have any other value for it's
energy even
if it were charged without a resistor, i.e. from another capacitor - or
an
inductor in a switched circuit - or any of the other proposals that
people have
brought up on this list.

I _only_ agree that the energy in the capacitor is 1/2 CV^2 if the
energy dissipated during the charging of the capacitor is 1/2 CV^2.
Otherwise I _don't_ agree.

Again, I'm curious only about your claimed value for the
energy stored in the capacitor - even after it has been disconnected from
the
charging circuit and removed far from its location.

It depends on how much energy was dissipated during the charging of the
capacitor.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf