Given good quality real world components I could charge up one capacitor to 100volts, connect through a series of wires of different lengths and diameters, some wound as formal inductors, some straight to a second similar capacitor and then disconnect.

In each and every case I would walk away with two capacitors charged to pretty close to 50 volts.

Are you suggesting that somehow the energy is not 5000C before the second capacitor is connected and 2500C after ?

No, this is precisely what happens.

With real world components then the missing energy has almost all been dissipated in losses in the conductors and dielectrics. It's possible that with long wires you may have launched some into the aether.

Something similar to this happens billions of times a second in your computer, the dominant loss is CMOS devices is in driving load capacitances. It doesn't matter how small they make the resistance of the FETs, the power lost in taking the load capacitor from 0V to 3V (or whatever) and back again is constant.

 

the missing energy has almost all been dissipated in losses in the conductors and dielectrics.

If this were the case then I could compensate for this loss with a top up. This was why I specified good quality compoents. If I arrange the wiring to be less than 1 milliohm and the capacitor resistive losses to be < .1% I can pre-charge the first capacitor to 101 volts.

To a first approximation the resistive and dielectric losses should be independant of the precharge, so according to your theory I should end up with 100 volts remaining on the capacitors!

Of course I don't. So I must conclude that there is another mechanism in play.

 

If this were the case then I could compensate for this loss with a top up. This was why I specified good quality compoents. If I arrange the wiring to be less than 1 milliohm and the capacitor resistive losses to be < .1% I can pre-charge the first capacitor to 101 volts.

To a first approximation the resistive and dielectric losses should be independant of the precharge, so according to your theory I should end up with 100 volts remaining on the capacitors!

Of course I don't. So I must conclude that there is another mechanism in play.

No this is wrong. Do the analysis of the circuit with the two capacitors and a resistor R between them. You will find that no matter the value of R (as long as it is not zero), all the missing energy goes into R, and the smaller you make R the quicker it gets there. So it doesn't matter how good your components are, unless they are perfectly conducting and perfect dielectrics, all the missing energy goes into the losses.

 

You will find that no matter the value of R (as long as it is not zero), all the missing energy goes into R, and the smaller you make R the quicker it gets there.

I'd like to make a clarification about your comment. I'm sure you are aware of this, but others might misinterpret your point here.

Your comment is true. However, the circuit theory calculation can also be misleading sometimes. The balancing of energy in a calculation might lead one to think that circuit theory is giving the right answer, when it might not be in some cases.

As you make R smaller, the discharge rate gets smaller. Eventually, the current pulse time could get small enough to induce radiation loss. Hence, circuit theory would correctly calculate the missing energy, but still would not be giving you the most accurate answer. If one were to measure the power loss in the resistor, there could be a discrepancy. Then, we still would have to ask: "where is the missing energy?". A more detailed calculation, with Maxwell at your side, will get it right.

 

I'd like to make a clarification about your comment. I'm sure you are aware of this, but others might misinterpret your point here.

Your comment is true. However, the circuit theory calculation can also be misleading sometimes. The balancing of energy in a calculation might lead one to think that circuit theory is giving the right answer, when it might not be in some cases.

As you make R smaller, the discharge rate gets smaller. Eventually, the current pulse time could get small enough to induce radiation loss. Hence, circuit theory would correctly calculate the missing energy, but still would not be giving you the most accurate answer. If one were to measure the power loss in the resistor, there could be a discrepancy. Then, we still would have to ask: "where is the missing energy?". A more detailed calculation, with Maxwell at your side, will get it right.

Radiation loss will not necessarily occur. Enclose the experiment in a conducting shield. Heck, you could even construct the capacitors so that they inherently shield the experiment - make two concentric spherical capacitors, one from radius of R1 to R2 and the second from R2 to R3. You can arrange the radius's to get any capacitance you want. Charge the inner one and drill a small hole through the R2 surface where you connect a conductor from the R1 shell to the R3 shell. Place a small resistor in this connection with your 'switch', then you find that all the energy loss is dissipated in this resistor no matter how small it is as long as it is > 0. I can't see any radiation loss.

 

You will find that no matter the value of R (as long as it is not zero), all the missing energy goes into R, and the smaller you make R the quicker it gets there.

Really?

I would be interested to see your energy calculation supporting this view.

 

Radiation loss will not necessarily occur.

I didn't mean that radiation loss will necessary occur. I was clarifying that it can occur in some situatons, (generally, I mean). In those cases, circuit theory would not show an energy discrepency, but would give you the wrong anwer about all energy dissipating in the resistance. This shows that when you employ circuit theory, you need to consider frequency and circuit dimensions to make sure you are in the "circuit theory" realm.

I'm just clarifying since someone else could take your statement (which I quoted in my previous post) out of context. There are so many sub-cases we are talking about now (shielding, no shielding, inductance included, etc), that it gets confusing to know what statements are general, and which are specific to an example.

 

But what's the point of this whole lot of arguments and counter arguments if one fails to reach a common ground of reasoning or consensus for that matter.

When one considers an 'ideal' world, let's consider it 'ideal' in puritical terms and then continue with the discussions...or else this discussion will go on and on upto 'infinity'!

 

But what's the point of this whole lot of arguments and counter arguments if one fails to reach a common ground of reasoning or consensus for that matter.

When one considers an 'ideal' world, let's consider it 'ideal' in puritical terms and then continue with the discussions...or else this discussion will go on and on upto 'infinity'!

This is exactly why I cringe whenever someone brings up the notion of "ideal."

 

This is exactly why I cringe whenever someone brings up the notion of "ideal."

But one must admit, it's much more "fun" this way!

 

Really?

I would be interested to see your energy calculation supporting this view.

Two capacitors C, one charged to V the other uncharged, series resistor R. The time constant T = RC/2. The current is a decaying exponential. The initial current is V/R. So I = V/R exp(t/T). You can work out that the energy dissipated in R is V^2C/4.

Without doing the integral you can see that this is sensible

Energy approx = I^2R x time = (V/R)^2 x R X RC/2 = V^2C/2 independent of R.

The constant is wrong as the calc assumes that the current is constant at the original value for T.

 

Energy approx = I^2R x time = (V/R)^2 x R X RC/2 = V^2C/2 independent of R.

This formula Energy = V^2C/2 is the entire energy in the charged capacitor.

But this capacitor is discharged to V/2 volts, not zero.

Your calculation implies that the entire energy of the charged capacitor passes through the resistor, which we know is not the case.

You should remember that as the voltage at one end of the resistor is falling exponentially it is rising equally at the other until there is zero PD across the resistor.

If it's any consolation I did this calculation this morning and got the same result as you, but I haven't yet figured the error.
I am posting it anyway in case someone else can see it.

 

This formula Energy = V^2C/2 is the entire energy in the charged capacitor.

But this capacitor is discharged to V/2 volts, not zero.

Your calculation implies that the entire energy of the charged capacitor passes through the resistor, which we know is not the case.

You should remember that as the voltage at one end of the resistor is falling exponentially it is rising equally at the other until there is zero PD across the resistor.

If it's any consolation I did this calculation this morning and got the same result as you, but I haven't yet figured the error.
I am posting it anyway in case someone else can see it.

There's no error. The final voltage across the resistor is zero, not the final voltage across the capacitor.

edit - I take that back - just checked your calcs, there is at least one error, the timeconstant is RC/2 as you have two capacitors in series.

 

But you claim that the entire energy of the charged capacitor passes through the resistor, yet there is some left in the capacitor?

 

But you claim that the entire energy of the charged capacitor passes through the resistor, yet there is some left in the capacitor?

You are concerned about the line

Energy approx = I^2R x time = (V/R)^2 x R X RC/2 = V^2C/2 independent of R

this is just an order of magnitude approximation to show that R drops out. It estimates the energy as the power at t=0 times RC/2. It ignores the shape of the discharge curve so there is a missing constant, but that is independent of R.

If this confuses you, just go to the current waveform

I = V/R exp(-t/T) where T = RC/2 and do the integral.

It's easy to see that this is the waveform as it must be a decaying exponential, the starting current is determined from the voltage across the resistor and the time constant is RC/2

 

Changing the time constant will correct my figuring.

Thank you for the discussion

 

Well, it's been a while since anyone tossed in any comments on this. Here are mine. The OP said there was no resistance. Nothing was said about the switch, so assume it is a perfect switch with no losses. Then this becomes an unphysical problem, as an infinite current will flow when the switch is closed. An infinite magnetic field would result and would change over time. Clearly, energy will be propagated away as an electromagnetic wave, but we don't have the tools to calculate what will happen. Since it's a physically unrealizable situation, it can't be duplicated in the laboratory. But I believe this qualitatively shows that the electromagnetic characteristics would have to be taken into consideration. The details, of course, would then depend on exactly how the switch behaves; some energy could go into vaporizing switch material and some into the magnetic field.