# Conservation of Charges

1. Apr 6, 2009

### saltine

When you share charges between two identical capacitors (by letting half of the charges in one capacitor to spill to the other) the over energy is halved. Where does that energy go? In what way is it lost?

Example circuit:

Before t=0, capacitor C1 is fully charged. The energy that it stores can be computed by E = 0.5QV. The voltage across the capacitor is 5volt. When the circuit toggles, half of its charges would spill to C2 (which is an identical capacitor). The voltage across either capacitor will be 2.5volt. The total energy after toggling is ETotal = 2( 0.5(0.5Q)(0.5V) ) = 0.5E. Half of the energy is lost.

But in what form is the energy "lost?"

This is a similar situation using water tanks:

The potential energy in Fig6a is Ea = mgh. The potential energy in Fig6b is Eb = 2( (0.5m)g(0.5h) ) = 0.5 Ea. Half of the potential energy is lost.

When an object freefall, potential energy is converted into kinetic energy. So there is no net lost of energy. But where did the energy go in the case of the water tank? Is it true that the energy is still lost through kinetic energy, because, if the water must obviously move to get to the second tank, and while it is moving, there is additional kinetic energy (aside from its temperature), which the water gets from its potential energy?

What about the case of the capacitors? Is the energy lost simply because the charges moved (they moved because the potential energy gave them kinetic energy to move down the gradient)?

Does the energy lost have anything to do with electromagnetic radiation?

2. Apr 6, 2009

### Mapes

The energy is lost to thermal energy (via resistive heat generation in the first case and turbulent and frictional heat generation in the second case). I noticed you drew the water levels as equal in your Fig. 6b instead of having all the water rush to the right side (with no loss of energy) then all back to the left side (again, with no loss of energy); so at some level (maybe not explicitly), you've assumed that damping exists in the system. The same thing applies to the voltage in the capacitor problem; by assuming the voltages end up as equal, you're implying an irreversible system, which means that energy will be thermalized.

The lost energy doesn't involve radiation except inasmuch as hotter objects radiate more and at higher frequencies.

3. Apr 6, 2009

### Bob S

Put a resistor in series with C2, and calculate the power loss (integrate V(t)2/R dt from zero to infinity) when the switch to C2 is closed. Now slowly let R go to zero. What happens to the energy dissipated in R?

4. Apr 6, 2009

### Staff: Mentor

See this thread, especially the detailed derivation in post 9:

5. Apr 6, 2009

### saltine

I remember a demonstration of the two tank problem using a U-shape tube. The water level was clearly oscillating.

In this figure, vR = e1 = e2. From KCL I get these equations:

$$\frac{1}{R}v_R + C_1\frac{dv_{C1}}{dt} = 0$$
$$\frac{1}{R}v_R - C_2\frac{dv_{C2}}{dt} = 0$$

In my case, $$C_1 = C_2 = C$$, and $$\frac{dv_R}{dt} = \frac{dv_{C1}}{dt} - \frac{dv_{C2}}{dt}$$. So

$$\frac{2}{R}v_R + C(\frac{dv_R}{dt}) = 0$$

$$v_R = e^{-\frac{2}{RC}t}+K$$

Since $$v_R$$ is initially 5volt, K = 4. But this cannot be correct because as t goes to infinity, $$v_R$$ is supposed to be 0. What went wrong?

Re: DaleSpam

https://www.physicsforums.com/showpost.php?p=1829464&postcount=9"

How do you get (2) in your solution?

Last edited by a moderator: Apr 24, 2017
6. Apr 6, 2009

### Staff: Mentor

With Mathematica

7. Apr 6, 2009

### Staff: Mentor

Why don't you just try the general solution I linked to, just to see if it works.

8. Apr 6, 2009

### saltine

(That is because I already know that it works!)

Ok I found what was wrong. That constant from the integral was there before the natural log, so it is a multiplier to the exp.

$$V_R = 5 e^{-\frac{2}{RC}t}$$

Power is VR^2/R = $$\frac{25}{R}e^{\frac{4}{RC}t}$$

Total Energy dissipated by resistor is:

$$\int_0^{\infty} E dt = \frac{25C}{4}$$

The initial energy was $$0.5 CV^2 = \frac{25C}{2}$$. So half of the energy is lost eventuall regardless what R is.

But in this derivation, wouldn't it mean that the charge will not oscillate between the two capacitors? Because VR is just an exponential decay. But in the water tube demonstration it would oscillate.

9. Apr 6, 2009

### Staff: Mentor

That is correct, it will not oscillate with two capacitors. If you want some oscillation then you need to include an inductor. I think I may have worked out that situation later in the same thread, but I don't remember for sure.

10. Apr 6, 2009

### saltine

Yes, you did. Can I say that adding inductors would not cause more energy loss because in an RLC circuit, the resistor is the only way to lose energy.

11. Apr 7, 2009

### Staff: Mentor

I would say that the energy is dissipated via both resistive heating (as described) as well as some amount of electromagnetic radiation due to the shifting charges.

12. Apr 7, 2009

### Staff: Mentor

Correct.

13. Apr 7, 2009

### Mapes

Whoops, I missed the issue of radiation from moving charges in my answer above. Obviously I was too absorbed in standard circuit elements and water flow. In any case, radiated EM waves aren't solely sufficient in explaining the energy loss, as Doc Al says.

14. Apr 7, 2009

### saltine

How is the loss due to EM radiation account in the total energy? What would you need to change to make it the dominant form of energy loss?

When you wiggle a charge in space, its electrical field lines wiggle. In the direction perpendicular to the motion, there is electromagnetic wave.

In the case of the capacitors, the charges only flow from C1 to C2 once. So there won't be an actual wave. It would be more like a shift?

15. Apr 7, 2009

### Staff: Mentor

I should emphasize that my analysis also was strictly based on circuit theory and explicitly does not consider the radiation of any energy.

Saltine, if you want radiation to be the dominant source of energy loss than you should get a good book on antenna design. Unfortunately I will not be of much help for that nor will those equations I derived previously.

16. Apr 8, 2009

### raknath

Hi just a quick question on this, when we talk of induction playing a part in the transfer of charges from Capacitor A to B, dont we take air resistance and medium permittivity into account