On Thu, 15 Jul 2010 23:37:35 -0500, flipper wrote:

On Thu, 15 Jul 2010 17:59:57 -0700, Jim Thompson
wrote:

On Thu, 15 Jul 2010 18:45:33 -0500, flipper wrote:

On Thu, 15 Jul 2010 13:35:09 -0700, "Paul Hovnanian P.E."
wrote:

My solution for the missing energy.

I'm not sure what inspired the analysis but you don't need two
capacitors to express the 'conundrum' as you've got it in the very
first term for E in a charged capacitor.

From the definition of C, q, V, and E one might expect E, in an
'ideal' capacitor, to be qV or, by substitution, C*V^2 but, as you
point out, it's commonly known to be .5*C*V^2.

Where did the missing energy go?

The answer is the same, dissipated in the R 0, and is inherent to
the charging of a capacitor whether it is from a battery or, in your
case, another capacitor.

Trying to postulate an 'ideal' circuit with R=0 leads to the
impossibility of an instantaneous charge of infinite current and with
electron mobility limited by the speed of light the universe, as we
understand things, simply can't do it.

Flipper, _In_the_limit_ as R-0 the exact same amount of energy is
lost as with a finite R. Try it, you'll like it :-)

I understand your point and one of the endearing things about math is
you can calculate the impossible but in this case I think it is more
confounding than illuminating as most people will likely have
difficulty estimating the dissipation of infinite current through 0
ohms.

So why bother confounding the matter with a singularity that cannot
exist?

...Jim Thompson

What singularity? No singularity needed. Thimk.