Small conservatin of energy issue

[idobox]

Hi everyone,

I am having a small conceptual problem with conservation of energy and linear accelerators.

The problem is simple: If I use a parallel plate capacitor to accelerate a particle, the particle gains kinetic energy, so something should loose energy, but what?
The obvious answer is it's converting potential energy into kinetic energy, but that still doesn't satisfy me.

If we start with an uncharged capacitor, the potential is flat. Then when we invest energy to charge it, accelerate the particle, and discharge the capacitor, the potential is flat again. I expect the energy retrieved to be lower than the energy invested (the difference being diverse losses plus the kinetic energy of the particle) but I don't really see what EM process accounts for the charge difference.

[eSOANEM]

The energy does come from the potential energy. That may not satisfy you, but that's how it is.

In your second discussion, we initially have two neutral plates. We then invest a small amount of energy (lets call it E) to charge it to one plate with a +1 charge (it has 1 hole) and one plate with a -1 charge (it has 1 excess electron).

If we then say that the cap is going to discharge, the electron will gain some kinetic energy as it falls into the hole.

The gradient theorem implies that this energy is the same as the energy required to separate them (which was E).

We now have the same situation as we did initially except that, at the beginning, we added E to its total energy and this energy is still here and will probably be lost as heat.

Interestingly, if we're still talking about a situation similar to the first where the electron moves between the two plates, this capacitor will probably function more like a resistor. The charge on the plates is linear with the voltage across it (because it's a cap) but, assuming each electron has an equal chance of independently jumping across the gap between the plates, there is a current through the cap proportional to this voltage. This gives us that V=k*I which looks suspiciously like Ohm's law.

I don't really see what EM process accounts for the charge difference.

I'm not sure what you mean by "charge difference". Were you thinking of the charged cap as having a -ve plate and the other plate being neutral? If so, the answer is that there isn't one, because this doesn't happen. Kirchoff's current law can be rephrased to say that the total charge contained within any component is constant. As such, when one plate of the cap gains a -ve charge, the other gains a +ve charge.

[idobox]

I accept that kinetic energy comes from potential energy, but I have trouble understanding where potential energy comes from.

If I charge a capacitor, I store energy E= 1/2.C.U^2, so charge (of the capacitor) and energy are linked.

Let's do the scenario with more details.
I start with a large parallel plate capacitor made of metallic grids, and a charged particle very far away moving toward the capacitor.
I invest energy to charge the capacitor, and wait for the particle to pass through and be accelerated.
Then I drain the energy from the capacitor, let's say into a resistor.

In this scenario, the particle has gained (or lost) some kinetic energy. So I expect the energy I drain from the capacitor in the end to be different from what I put in in the beginning. Since the value of C hasn't changed, it must mean the charge/voltage across the capacitor must be different. My question is, how do EM explain that a moving particle makes charges go from one plate to the other?

[eSOANEM]

Ah, sorry, I completely misunderstood the question initially.

I think I see the problem now although I don't think I'm able to provide a proper solution.

I suspect that the solution will be to do with the electric field, and magnetic field from the cap charging interfere with the magnetic field produced by the electron's motion but I'm not sure and don't have enough experience with proper EM to offer a particularly good solution.

[Qaanol]

Initially, there were two plate with no net charge, and a distant charged particle. The electric potential from the plates is 0 throughout all space.

You first pumped charge from one plate to the other. In order to do so, you had to do work to overcome the electric forces on the charges you moved. Those electric forces came from all the other charges on the plates, as well as all the other charges in the universe, which in this case is just that one distant particle.

After separating the charges on the capacitor plates, all of space to the left of the capacitor is at positive voltage due to the capacitor, say 5V, because it is closer to the positive plate than the negative. Similarly, all of space to the right of the capacitor is at negative voltage, in this case -5V. In between the plates there is a linear voltage gradient.

Since electric field is the gradient of voltage, and voltage is constant outside the capacitor, there is no emf due to the capacitor except in between the plates. This is as we expect.

However, in order to reach this state, you had to “lift up” the distant particle, from 0V to 5V, and that took energy. Indeed, you had to “lift up” all the particles with the same charge as the side of the capacitor they are closer to. Why didn’t it “seem” like it took extra work to charge the capacitor? Because in real life there is an approximately equal distribution of positive and negative charges on either side of the capacitor, so the same amount of work done by the capacitor, is also done on the capacitor, as it charges.

In any case, the act of charging the capacity took extra energy due to the distant particle, and that is exactly the amount of kinetic energy the particle gains when it falls through the voltage gradient. But wait! The particle started at 0V, was raised to 5V, and now it drops all the way to -5V, so it ends up with twice as much kinetic energy as it was imbued with, right?

Well not quite. As you were charging the capacitor, your had to lift the charges in the capacitor up a higher and higher voltage difference across the gap, until at the very end you were lifting them the full 10V. The work done against the electric field of the distant particle, by your act of moving charges across the capacitor, ends up occurring through each voltage difference from 0V when you start charging the capacitor, to 10V when you finish.

And when the distant particle later reaches the capacitor and falls through the voltage gradient, it picks up kinetic energy through that same drop of 10V.

Another way to think of it is, when the distant particle was on one side, its charge was “helping” make the capacitor have a larger gradient, because it was on the side with the plate matching its charge. After the particle passed through the capacitor, its charge was “hindering” the capacitor, so there was less of a voltage drop.

Thus, as the particle picked up kinetic energy between the plates of the capacitor, it was in equal proportion decreasing the voltage drop across the capacitor, simply by changing its own location because it has a charge.

[eSOANEM]

Why didn’t it “seem” like it took extra work to charge the capacitor? Because in real life there is an approximately equal distribution of positive and negative charges on either side of the capacitor, so the same amount of work done by the capacitor, is also done on the capacitor, as it charges.

I don't see how that changes things. As I see it, as the voltage across the cap rises, the voltage outside the cap will rise too but that the rise in the voltage will only travel at c. Because of this, as the fact that the rate at which the voltage across the cap increases is dV/dt=I/C, the emf outside the cap dV/dr=(I/C)*dt/dr=I/(C*c).

If there is an equal distribution of +ve and -ve charge around the plates, this means that the cap will feel no resultant force from Newton's third law, but does not affect the work done which will depend on the motion of the charges which, assuming a standard normal distribution in each direction, will result, in a positive amount of work being done on those charges around the capacitor.

Well not quite. As you were charging the capacitor, your had to lift the charges in the capacitor up a higher and higher voltage difference across the gap, until at the very end you were lifting them the full 10V. The work done against the electric field of the distant particle, by your act of moving charges across the capacitor, ends up occurring through each voltage difference from 0V when you start charging the capacitor, to 10V when you finish.

What? The distant particle never went up to 10V no matter what the particles on the cap went through. It's quite right that it goes up 5V then down 10V (leaving it -5V from its starting point) because the cap is still charged. It's only once the cap is discharged that it will be raised 5V and return to the same potential as before.

...

Thinking about it some more, it seems to me that, whilst 1/2CV^2 is the energy stored in the electric field (and therefore recoverable), when charging and discharging the capacitor, additional energy will be spent doing work on the surrounding particles. I think it is this work which accounts for the missing energy.

[idobox]

So basically, the potential is never really flat because of the particle, and the fact it goes from one side to the other, and change the potential, resulting in electrons moving through a different one when charging and discharging the capacitor?

[Qaanol]

Thinking about it some more, it seems to me that, whilst 1/2CV^2 is the energy stored in the electric field (and therefore recoverable), when charging and discharging the capacitor, additional energy will be spent doing work on the surrounding particles. I think it is this work which accounts for the missing energy.

Yeah, I flubbed up my explanation, but this is essentially the idea.

Really it’s more like, you have to do extra work on the particles within the capacitor as you move them, on account of the external particles. When the external particles are on average neutral, the amount of extra work you have to do is increased and decreased by the same amount, so it comes out with the textbook formula. But for any specific charged external particle, you still had to do extra work due to that particle’s existence, and the energy from that work you did gets stored in the electric field. For “bookkeeping” purposes, the energy is tallied as potential energy of that specific particle on account of it being raised from 0V to 5V (or whatever).

When that specific particle falls across the capacitor, it converts that stored potential energy to kinetic energy, and as you rightly mention, converts even more potential to kinetic when it falls all the way to -5V. The voltage drop across the capacitor decreases a minuscule amount from that particle switching from one side to the other, which accounts for the slight drop in stored potential energy in the electric field.

If you then discharge the capacitor, you’ll get back the energy you spent charging it, minus the energy required to raise that external particle up from -5V back to 0V.

[eSOANEM]

Really it’s more like, you have to do extra work on the particles within the capacitor as you move them, on account of the external particles. When the external particles are on average neutral, the amount of extra work you have to do is increased and decreased by the same amount, so it comes out with the textbook formula.

Except you don't.

The work done is the integral of F dot dx. As such, all that matters for the sign of the work done is whether, on average across all the charged particles and the whole time the force is applied, the particles move in the same or the opposite direction as the force.

If we assume an initially symmetric distribution of velocities, an equal amount are moving in same and opposite direction as the force. At the end of the period the force is applied, there has been acceleration along the force and so those moving in the same direction as the force are moving faster (and so moved further) than those moving away and there are now more of them as the slower moving particles which were moving away have now started moving in the same direction as the force.

If F.dv was 0 on average and is now non-zero positive (and has increased monotonically as it will in this case), it must be positive when averaged over the whole period. As such, the work done is positive.

This argument is, note, independent of the charge of the particles in question because the force being opposite in direction is countered by the dx changing direction (on average).

Assuming this extra energy put into the surroundings has time to thermalise before the cap is discharged, the same argument will apply on discharge and more work will have to be done on the exterior particles.

The reason for this extra work is that (for neutral surroundings) you're forcing the atoms to form dipoles and so, in effect, having to charge lots of other tiny caps. The reason the textbook formula works is that the charge on each of these caps is tiny but their capacitance would be huge (because of how close their "plates" are together) meaning that the energy stored in them is vanishingly small compared to the tolerances on the value of the main cap.

[Qaanol]

I have two metal plates of arbitrarily large area, with voltage difference V. I take one single electron from plate A and move it to plate B, ending with the same velocity it started with.

The amount of work I had to do is independent of the path I brought the electron along, and independent of any external particles, because the Coulomb force is conservative. The only thing that matters is the how much voltage change the electron moved through.

In the limit of infinite-area plates, a charged particle external to the capacitor, say a proton, acts (in total over the whole area of the capacitor) to contribute a voltage drop across the capacitor equal to what that same particle would contribute if it were part of one of the capacitor plates.

Therefore, if the external particles are collectively neutral, then they will contribute zero change to the voltage across the capacitor. When I move my one electron from plate to plate, the work I have to do is the same as if there were no external particles, because voltage change depends only on the charges in the capacitor plates.

For a given external atom, say hydrogen, I had to do extra work because of the hydrogen electron repelling my electron, but I also had extra work done for me because of the hydrogen proton attracting my electron. Over the course of moving trillions and trillions of electrons to charge the capacitor, each of them had the same amount of extra work done on it by attractive particles as done against it by repulsive particles.

Each of those external particles was raised or lowered by the same voltage, which changes its potential energy. But the ones that gained potential are exactly balanced by the ones that lost potential, so the net energy put into the external particles is zero.

Now you take one single external particle, which had its potential raised. It took energy—in the form of work done to move electrons—to raise that potential. But it didn’t “feel like” extra work was required, because another particle nearby this external one had the opposite charge, so its potential energy was lowered by the same amount, and it helped to move the electrons in the capacitor.

Anyway, this one specific external particle has a charge, and is contributing to the voltage drop across the capacitor exactly as if it were part of the nearer plate. You let it fall across the capacitor, it converts potential energy to kinetic energy, and is now moving fast. That potential energy had been stored in the electric field between the capacitor plates due to the voltage gradient.

Now this same external particle is on the other side, so it contributes to the voltage drop across the capacitor as if it were on the plate that is now nearer, which is the other plate from before. So the voltage across the capacitor is exactly as if this one charged particle moved from one plate to the other. That means the capacitor voltage is now slightly less, due to this tiny shift in charge from one side to the other.

Sure enough, the potential energy that was converted to kinetic, was actually, in fact, really, truly extracted from the electric field within the capacitor. That electric field is now weaker by the exact same amount, because the voltage drop across the capacitor is appropriately less.

[eSOANEM]

I see where I went wrong. The argument I provided only held for a medium without significant attractive forces within itself. This is clearly not the case for realistic media.

[idobox]

thank you all for your help.