Magic capacitors!

radiation. See:

The presumption is radiation is able to leave. Expressing radiation as = a lumped constant equivalent series resistance, assume zero ESR. In = other words, inside a superconducting box, so there is no way to measure = the resonance (it's a theoretical problem, that's okay) and it resonates = forever.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

I don't see what this has to do with the problem though... I'm sure you do... of course this assumes you are right.

You have no clue. The "lost" energy must go somewhere. It goes into moving the charge from one plate to another. If there is no resistive force then no energy is lost in the resistance. If there were a resistance then it would reduce the overall voltage and that loss would be converted into heat in the resistance.

When the switch is closed the negative charge must move from one plate to another. Hence a force exists and because of the finite length between the plates work is done.

Regardless of how the caps are connected the there will be work done in moving the charge. The link provided by phantom shows that in the ideal case the energy is radiated away. This should make absolute sense as the medium surrounding the wire must absorb the energy(hence it is radiated).

Your "proof" is just pure nonsense. The idea is that you must put in half the energy to move half the charge(which is 1/4 the original energy) and regardless of how the energy is dissipated it must be done in some way. For an ideal wire it obviously can't be done through heat.

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

I'm glad my post got what it was suppose to get out. I kinda feel like Breitbart.

That's the presumption I made. Until you just now said so, nobody said we were inside a superconducting box.

zero ESR. In other words, inside a superconducting box, so there is no way to measure the resonance (it's a theoretical problem, that's okay) and it resonates forever.

Initial condition:

C1 2F, 1 volt, 1 joule, 2 coulombs

C2 1F, 0 volts, 0 joules, 0 coulombs

Now remove all the energy from C1 and deliver it to C2. An inductor will move the energy nicely.

Now

C1 has 0 volts, 0 joules, 0 coulombs

C2 has 1.414 volts, 1 joule, 1.414 coulombs.

John

multivariate calculus problem, and therefore it matters which variables we take to zero first, or in what proportion.

exist, so we shall take simple directions (R and L axes, respectively).

case, the energy remains (E = Eo).

undefined. QED.

Which is why I didn't say "model R => 0".

Nope. Tautology is not a mathematical model.

Model infinite current and zero time. The equations blow up (Larkin's "singularity").

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

So, is this like "Wait 30 days and get a different answer?" , I play this all the time with those management types.

Cheers

Where did the 0.707 coulombs "go", John? ...Jim Thompson

The simple answer is that for zero resistance and zero inductance the voltage and current would have to be in phase, so as you switch there's an instantaneous drop in voltage across a zero ohm resitor and hence infinite current for an infinitesimily small time. There is no solution in this case.

Mark.

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A 'hypothetical mathematical model' doesn't have to conserve energy or charge or correspond to reality. That makes it ... uninteresting. Any transfer of electrical energy is accompanied by a Poynting vector which means the B fields are NOT zero, and that means negligible inductance is never physical. Negligible resistance is sometimes possible, though (superconducting resonators have Q values in the tens of thousands).

Mathematicians have a sensible way to deal with singularities; they just say "that's undefined."

Sure, create your model. Compute 1/0. Enjoy.

John

And do we also assume it does not oscillate nor radiate?

If you have a resistor between the capacitors,half the energy is dissipated. The math does NOT depend on the resistance value.

In the case of zero resistance, you get infinite current and a huge spark that radiates the equivalent energy.

However, you can charge the second capacitor more efficiently with an inductor in a resonant circuit.

Sorry Don. that does not stand up to the physics or the arithmetic.

Energy may be radiated, but not by the switch. Nor is there a requirement for a spark.

Not part of the question.

Sorry, JosephKK, But it does. Shame :-(

Do the math JosephKK. You are a disappointment :-(

Newbies. If you think differently, go into sales ;-) ...Jim Thompson

=A0 =A0 =A0 =A0...Jim Thompson

You are an idiot. If the two contacts are to be closed, and the resistance is zero, there WILL be a spark BEFORE closure occurs. And in ANY case of ANY spark, there WILL be radiation.

As if you had any grasp of even that.

A 'spark' needs some medium to form, in air this spark is caused by electron avalanche. Since it involves electrons moving and being accelerated it will take time. The question didn't say how fast the switch was closed, it could be much faster than than the time for the avalanche to happen. Also there's no requirement for any air at all, a vacuum is a pretty good insulator. With relatively low voltages and an absense of any free electrons (like no sharp points like dust) there wouldn't be a spark in a vacuum:

Mark.