# Capacitor discharging on another - energy considerations

1. Oct 20, 2011

### DaTario

Hi All,

Going directly to a concrete numerical problem: Consider two capacitors in series in a one-loop circuit with a switch S. S is initially open. At t=0 S is closed.
Let $C_1 = C_2 = 2F$, the capacitances.
Let $Q_1(t=0) = 4 C$ and $Q_2(t=0) = 0 C$ (second capacitor initially discharged)

At equilibrium, after S is closed we have
$\frac{Q_1}{C_1 }= \frac{Q_2}{C_2}$ which by simmetry implies that
$Q_{1f} = Q_{2f} = 2C$

Now comes the problem. Initially the enegy content is located at capacitor 1.

$E_i = \frac{{Q_1}^2}{2\, C_1 } = 4 J$

At the end of the process (flow of charges until the equilibrium) there is energy in both capacitors:
$E_f = \frac{{Q_{1f}}^2}{2\, C_1} + \frac{{Q_{2f}}^2}{2\, C_2 } = 1+1 = 2 J$

Is the missing 2 Joules distributed in space in the form of a propagating EM wave since there were accelerated charges in the process?

Does this system provides us a way to infer (derive, calculate) the law of emission of radiation of an accelerated charge?

If we add a resistor we will have an extra energy cost with the term:

$\Delta E = \int R (i(t))^2 \; dt$ due to Joule effect. Is it correct?

Best wishes

DaTario

2. Oct 21, 2011

### Delta2

In real world there will always be some small ohmic resistance R (in the capacitors and in the connecting wires) therefore there will be this substantial extra energy cost term that you mention and most of the energy will be lost as heat in the resistor (and some small portion of energy as EM radiation).

In theory with R=0 it seems that we have to accept that all the energy will be lost as EM radiation. Probably we can assume a radiation resistance R and model the circuit as this radiation resistance was ohmic (i.e $$IR+\frac{Q1}{C1}+\frac{Q2}{C2}=0$$).

However to calculate this radiation resistance R, extra assumptions should be made and the geometry of the circuit and the capacitors should be taken into consideration. Due to these reasons i doubt if this system could be taken as a way to calculate the emission of radiation from an accelerated charge.

3. Oct 21, 2011

### DaTario

Yes, I agree with you. I now see that the acceleration of charges in this circuit without resistance would be infinite so that the amount of energy lost due to emission of radiation seems to be also infinite. But this so well defined amount of 2 joules lost still impresses me.

Note that the equilibrium situation of charges in the capacitors does not depend on the existence of a restistor in this circuit. So the term due to Joule effect seems to be really an extra term in this discussion. No conservation of energy has its place, in this problem, whatever may be the value of the resistance. The effect of this resistance is not to justify this non conservation of energy, but just to increase it in magnitude.

Perhaps the emission power is infinite but as the time it take this system to reach the equilibrium tends to zero (without resistance), the energy lost in this kind of Bremstrahlung phenomena is finite.

The question(s) seems to be still on its(their) foot: why conservation of energy does not hold in this system? Where did this part of the total energy go to?

Thank you for you attention and response,

Best wishes

DaTario

Last edited: Oct 21, 2011
4. Oct 21, 2011

### DaTario

I think I must add this (have just think of it)

Maybe the abscence of a resistor is not so important as the abscence of an inductance. As a circuit, whatever may be is shape, the topology forces the existence of some self-inductance. And this will provide some inertia to the current (in abscense of resistance, for example) making it to oscilate from one capacitor to the other idefinitely, creating detectable kinetic energy in this system which, in its turn, will make the conservation law valid. Does it seems to be the correct way to handle this problem?

Bast wishes

DaTario

5. Oct 22, 2011

### Per Oni

Perhaps this helps:

Imagine your 2 capacitors consisting out of one common bottom plate and 2 equal but separate top plates (each with area A) placed at the same height L from the bottom plate. Say one of the top plates is charged with an opposite charge on the bottom plate.

Consider first the energy content: W = 1/2 q^2 L/A εo. If you now shove the 2 top plates together the total area becomes twice as big, the charges can spread out, so the energy goes down by the same factor.

To explain what happens, the distance between like charges are increasing. Perhaps it’s comparable to the energy loss of an expanding gas. Anyway you end up with the field line density of ½, but the same q and L.

This shows that electro static energy of a many charges system is not only separation of opposite charges alone but is also affected by the distance between equal charges.

The time involved in the shove plays no role and (normally) the lost 2 Joules warm up the plates.

6. Oct 22, 2011

### Delta2

Yes this is a much better way to handle the problem. I myself forgot about the self inductance of any closed loop like the one in this problem with the 2 capacitors. But even if we include the inductance L of the circuit we still miss one thing: That is that the capacitor and the inductor are not perfect i.e they cannot perfectly hold inside them all the energy as magnetic field energy (in case of the inductor) or electric field energy (inside each of the capacitors) There will always be some small portion of electromagnetic field energy in the space surrounding the circuit (and outside the closed loop) and this energy is the EM radiation.

But we assume in circuit theory that capacitors and inductors are perfect i.e that no electric field exists outside a charged capacitor and that no magnetic field exists outside the spiral of an inductor , that is we assume that there are no radiation losses in any circuit. If we treat this circuit as an LC circuit without radiation losses then indeed the energy will oscilate between electric field energy in one capacitor and magnetic field energy (inside the closed loop) then again electric field energy in the other capacitor. Current will go at a high value but will not be infinite (the current will be high enough as such that the magnetic flux through the circuit loop is high enough to give magnetic field energy equivalent to the electric field energy initially stored in capacitor). In general it will behave as a series RLC circuit with very low value for L and for R=0. Because R=0 the damping factor a=0 will be zero and the circuit will be underdamped. Check wiki pedia analysis for RLC circuits. http://en.wikipedia.org/wiki/Rlc_circuit#Underdamped_Response

Last edited: Oct 22, 2011
7. Oct 22, 2011

### DaTario

I agree with you that the circuit can be treated as an RLC with zero R and small L. I also agree that the very dynamics of this RLC implies the emission of radiation and so, the oscillations are not to last forever. One thing that is important to note is that if we insert a resistance the Joule effect losses will serve as an extra channel for the energy to leave trhe system. In a certain sense, Joule effect is also radiation due to charges which are accelerated, but in the context of a sort of solid state physics.

Thank you all for the contributions,

Best Regards,

Dario

8. Oct 22, 2011

### fizzle

You could also analyze this problem from a transmission line viewpoint: you have two identical parallel plate lines with a switch in between. Lets call the charged line C and the uncharged line U. The lines are the same length, L. You can consider the initially charged line C to contain oppositely moving em waves where the electric fields add and magnetic fields cancel. If the total electric field is E, when the switch is engaged a pulse of strength E/2 and length 2L travels down U.

At the instant the leading edge of the pulse reaches the open circuit at the end of U, the entire length of both lines contain a wave of E/2 and B/2 (the total energy is split between electric and magnetic). When the wave reflects positively off the open circuit, the E fields add and magnetic fields subtract in U. By 2L/c, the overlap is complete in U. It now contains all the charge and C is uncharged. This process repeats over and over again.

Of course, this is the ideal case that ignores all the messy details of real lines: various losses, etc. The nice thing is that you can actually do experiments like this with a scope and some lines.