# Work done by electric field

1. Oct 20, 2012

### AlchemistK

Suppose there are two parallel plates with charges of opposite polarity on the faces towards each other (a charged capacitor, disconnected) and we send a charged particle through the field between them from one end. When the particle comes out the other end, it has more kinetic energy, because electric field did work on it.

Now, does this cause rearrangement of charges on the plates in order to conserve energy by reducing electric field magnitude or is something else is happening?

TL;DR - Particle sent through electric field, gains KE, where does this energy come from?

2. Oct 20, 2012

### Q-reeus

You haven't specified direction of charged particle wrt plates geometry, but I will guess that initial velocity vi is parallel to plates at entry point say at left end of plates and midway between plates, and at almost symmetrically located exit point at right end of plates, has acquired an additional transverse component of velocity vt, with vi component unaltered assuming relativistic effects are negligible. Particle is now obviously closer to one plate than the other given it has been 'falling' in the direction of plate of opposite charge to itself during it's traversal between plates. Over this length of run, there is indeed a net gain in particle KE, and it can be explained entirely in terms of the net change in field energy of combined system - charged capacitor + charged particle - with no need to invoke any rearrangement of charge on the cap plates (which in reality wil occur to some slight but probably negligible degree). However, if that charged particle is allowed to continue on for a great distance, you will find the apparently insignificant fringing fields emanating from capacitor will act to slow that particle back down to it's initial KE - but not in general restoring the direction of initial motion, since unlike net energy, the momentum exchange will not have zeroed out.
And btw just what does "TL;DR" mean?

3. Oct 20, 2012

### Philip Wood

What a nice paradox! I think the answer lies in edge-effects. Although the particle (which, to save me getting confused, I've taken as positive) gains 'vertical' velocity as it passes through the gap, it loses 'horizontal' velocity, as it enters and leaves the field, owing to the action of the horizontal field components, EH1 and EH2at the edges. Thus there is no net gain in speed.

This certainly seems to be required by energy conservation, because, if the capacitor plates don't lose or gain charge, the capacitor can't lose or gain energy, if we compare its states before the moving charge is anywhere near it, and when the charged particle is well clear of it.

[It does have to be said that my diagram gives the 'best case scenario', in terms of where the particle enters and leaves. But even if the particle enters below halfway between the plates, making EH1 to the right, so the particle gains horizontal velocity to the right, I believe that the exact shape of the field lines at the edge is such that it will acquire a greater horizontal increment to the left when it leaves.]

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Last edited: Oct 20, 2012
4. Oct 20, 2012

### AlchemistK

What if we compare two instances when the particle is well inside the capacitor? There will be gain in vertical velocity (Keeping your diagram as reference) due to the acceleration it faces by the electric field.

Philip Wood's diagram accurately represents my problem (minus the fringe effect, it doesn't really have to be two capacitor plates,any uniform field parallel to the particle will do, I just though of those because maybe the source of the field will be effected)

(And TL;DR is an acronym for Too Long; Didn't Read but its actually used to summarize a big text)

5. Oct 20, 2012

### Q-reeus

There is a net gain - although you are right in respect of that having shifted vertically, particle horizontal velocity vi is slightly altered by reason of asymmetrical fringing fields even though symmetrical horizontal location applies. That's something I missed. However the degree is highly dependent on capacitor aspect ratio, time of flight (therefore initial velocity), strength of field E between plates etc. At any rate, in a typical scenario where plate separation is much less than plate lateral dimensions, horizontal motion will be negligibly perturbed during capacitor transit proper, and thus a net gain in KE almost entirely given by to qEdx - where dx is the change in transverse displacement during cap transit.
These horizontal component incremental balances will be highly dependent on exact initial transverse displacement, cap geometry, particle initial speed, cap field E etc. And as said above, generally negligible in terms of net KE gain during cap transit proper.

6. Oct 20, 2012

### Philip Wood

If there is a net gain in the particle's KE (and I'd be the first to admit that my edge-effects (fringing fields) argument is open to challenge) where would the particle's extra KE come from?

7. Oct 20, 2012

### AlchemistK

That is what I'm concerned about, the energy lost by the field as qEdx, what happens to compensate that? Please neglect all imperfections including fringing.(Unless that is whats really causing it)

8. Oct 20, 2012

### Q-reeus

Same story as in #1, #2 really, only difference here is there is no appreciable fringing of field E between plates. So there is a gain in KE equal to minus the drop in field energy = charge PE = qEdx - terms as per #5 entry.
In that case scenario is exactly as described in #2, #5.
OK thanks for that clarification. Are you happy with explanations given?
Stop Press: just caught your #7:
Well it's there in the added KE owing to added transverse velocity! As discussed! Is this now clear?
@Philip Wood in #6 - hope above clarifies your question.

9. Oct 20, 2012

### AlchemistK

No, no! Let me start over. Electric field has energy, particle comes through, and electric field does work on it, particle gains kinetic energy. Electric field lost energy, but then, something must happen to change something about the electric field, if the magnitude of the electric field remains constant, its energy remains constant.

10. Oct 20, 2012

### Staff: Mentor

It came from the particle's electric potential energy in the field, which in turn came from the work done on the particle in the process of putting it into the electric field in the first place.

11. Oct 20, 2012

### AlchemistK

Just wandered in with a constant velocity perpendicular to the field lines. (somehow dodging fringing?) It wouldn't have potential energy if the thing that put it there didn't have to do any work, right?

12. Oct 20, 2012

### Q-reeus

So you are worried about exactly how the cap E field changes during charge transit to account for charge q's energy gain? Well if you want the exact detail there's nothing for it but to do a detailed computer simulation of the entire field throughout all space - not a trivial task. Or you can just take it on faith that conservation of system total energy holds at all times. Since for uniform E case (well inside capacitor) horizontal initial motion is really irrelevant, why not just consider where a single charge q leaves one plate and travels to the opposite plate? This 'easy' way of looking at it then has that a partial discharge - by a single charged particle, results in a net drop in cap field E in the ratio dE/E = dq/Q, where Q is the magnitude of initial charge on each plate. Accounted for in that the discharge 'current' owing to q moving across the plates represents a gain in KE energy of qEx in so doing.

13. Oct 20, 2012

### AlchemistK

Or could it happen that the plates move towards each other and reduce the volume of the space containing electric field, ultimately reducing electrostatic energy?

14. Oct 20, 2012

### Jasso

Let me elaborate a little on what jtbell said.

As the particle passes through the capacitor (let's assume that it enters at the exact midpoint between the plates), it will be deflected to the side. As it moves closer to one plate, the electric potential will drop because it will be farther from the repulsive plate and closer to the attractive plate. This drop in potential is where the horizontal velocity comes from.

15. Oct 20, 2012

### Q-reeus

This represents a different scenario to #1 where we assume particle is in free flight. If the particle is now constrained - on rails in effect, then situation energy wise is effectively reversed to what you asked in #4. Now the sole opportunuty for q's energy gain/loss lies in influence of any horizontal component of fringing fields. If rails are exactly mid-way between the plates, there is no horizontal fringing field component acting on q, hence zero net energy change. The field pattern will alter as particle moves through, but not net field energy. Are we now done?

Stop Press: @13:
Not really relevant - will depend on mechanical stiffness of capacitor supports and can be taken as negligible effect.

16. Oct 20, 2012

### AlchemistK

By the statement that the particle comes in perpendicular to the fields, I only meant the initial condition, not that it was constrained throughout its motion.

17. Oct 20, 2012

### Q-reeus

I really believe all your questions have been addressed multiple times (with repetition by some respondents of points already made). What is still not clear or doubtful in your mind?

18. Oct 20, 2012

### Staff: Mentor

I seriously doubt that it's possible to ignore fringing here. Consider that in order to define electric potential (and therefore electric potential energy), we must have $\nabla \times \vec E = 0$. If the field between the plates is strictly perpendicular to the plates and uniform in magnitude, and the field outside the plates is zero, then $\nabla \times \vec E \ne 0$ along the boundary between the two regions. If you let the magnitude of the field decrease continuously through a "transition zone" between the inside and outside, while keeping the direction perpendicular to the plates, then $\nabla \times \vec E \ne 0$ in that region. In either case, you cannot define a difference in potential energy between the inside and outside.

19. Oct 20, 2012

### Philip Wood

In my opinion, the interesting case is when we consider the moving charge when it is a very long way from the capacitor, (a) before it has gone between the plates, and (b) afterwards. In that case the charge's potential will be the same at (a) and (b), because at large distances (compared with the capacitor dimensions) the capacitor's potential will fall dipole-wise to zero at all angles. The capacitor's energy will also be the same, as the charges on either plate will be the same. So the system's PE will be the same for states (a) and (b), and so the moving charge will be moving at the same speed at (b) as at (a).

The overall effect, then, is that the particle has been deflected without change of speed (like an alpha particle deflected inelastically by a nucleus).

Since the moving charge has acquired a velocity component perpendicular to the plates, it has lost velocity parallel to the plates, which, can only, I believe have occurred through the action of the fringing field. Don't forget that if (a) and (b) are a very long way away from the capacitor, the fringing field, though weak, acts through a long distance.

If we bring (a) and (b) close to the capacitor, then the potentials at (a) and (b) will be different, and the KE of the charge will be different, and there is no paradox.

Last edited: Oct 20, 2012
20. Oct 21, 2012

### Staff: Mentor

You don't need a rearrangement of charges on the plates in order to conserve energy. Just remember that the system consists of the plate charges plus the mobile charge. When the mobile charge is in the high PE location its field acts to increase the total field between the plates. When the mobile charge is in the low PE location its field acts to decrease the total field between the plates. Decreasing the field decreases the energy stored in the field, and the various conservation laws ensure that the amount it decreases is equal to the mechanical work done.