# Theoretical electrodynamic cycle 1

1. Jul 5, 2011

### Ambforc

I am casually working through Introduction to Electrodynamics, 3rd ed. by David J. Griffiths and created some theoretical situations to test my understanding of the subject matter. I am having difficulty with some, and will appreciate input.

Consider the following theoretical cycle (see the attached figure):

A positively charged mass starts from state 1 and falls under the influence of gravity through a horizontally orientated electrical field. Since the acceleration can be broken up into independent orthogonal vectors, the acceleration due to gravity and the acceleration due to the force on the charged mass due to the electric field can be considered independently. The path A the particle is to follow between states 1 and 2 is subject to vertical as well as horizontal acceleration.

Upon reaching state 2, the positively charged mass exits the horizontally directed electrical field. Since the mass is no longer subject to a horizontal force, horizontal acceleration will cease and the path B the mass follows to state 3 is subject to vertical acceleration only. (I assume it is this point that causes me to err, see the question below.)

At state 3 there is a theoretical transducer that can convert the total kinetic energy of the mass into another form for later use.

The mass is moved from state 3 to state 4 in a way that requires no work input (frictionless surface, and all work required to accelerate the mass to get it moving is recovered when it is decelerated to a standstill at state 4, for example).

Next, the mass is moved up against the force of gravity to state 5, its initial height. The work required to do this can be considered to be equal to the vertical component of the kinetic energy only that was collected at state 3, for this theoretical reversible lifting process.

Path E between states 5 and 1 can be considered to be without work, same as path C.

It appears that, if all the assumptions are valid, the horizontal component of kinetic energy collected at state 3 is excess. This is not supposed to happen, and I would appreciate input as to where there is a mistake. Here are some of my assumptions I am suspicious of, as well as possible workarounds:

• The charged mass exiting the electrical field is subject to some attraction force to the negatively charged plate. This will cause it to experience a force in the vertical direction upon exiting the horizontally directed electrical field. With luck, this force will be exactly equal to the horizontal kinetic energy of the mass after leaving the electric field. This may possibly be overcome by letting the whole falling procedure finish closer to the positively charged plate than the negatively charged one. It may perhaps help to complete paths B and C inside a Faraday cage.
• The field lines cannot be assumed to be completely horizontal near the edges, due to fringe effects.
• There is an accompanying magnetic field due to the moving charge that will cause it to have a back magnetic motive force on some sort, and the error arises by analysing an electrodynamic system with electrostatic arguments. This may be overcome by letting the charge complete its path at a constant slow speed inside the electric field, gradually collecting energy from the forced doing work instead of collecting it all at once at state 3.

I will appreciate comments as to what I miss from an electrodynamic perspective. The theoretical system appears to gain due to the fact that it can move the positively charged mass to its original horizontal position without doing work against the force that caused it to move to the right, as the restoration movement occurs outside the “force field”.

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2. Jul 5, 2011

### ehild

Do you mean that the charged particle gained some KE in the electric field from nothing?
You can set up a much simpler example. You throw a stone horizontally. When it falls down it has both a horizontal and vertical component of velocity and contributions to the KE. Then you can do the same process as in your example to move the stone back to your hand. And you would say that the universe gained the horizontal "component" of KE as excess energy. Did not you (as part of the universe) do work when having accelerated the stone to its initial velocity?

ehild

3. Jul 6, 2011

### Ambforc

If I were to throw a stone horizontally, I will be inputting a horizontal component of force for the time the object is accelerating, so simply attempting to regain that will not help, agreed. However, in this example, the charged mass is simply dropped, its support against gravity alone is removed. Also, the only force against it in its initial position is gravity, since it is still outside the electric field. Once it enters the electric field, it is expected to have a horizontal component of force acting against it as well, which will cease once it leaves the field again. As to your opening line, that is my question and reason for posting this in the first place, since it does not make sense.

4. Jul 6, 2011

### ehild

As in case of the stone, your body loses energy when you accelerate it. This energy comes from burning some sugar, so chemical energy is transformed to mechanical one. The energy of the universe does not change.
In case of the particle in the electric field, there must be a source - a battery for example, which produces the field. The moving particle interacts with the charges accumulated on the plates. The charges on the plates will rearrange when the particle gets closer to one plate and farther from the other one. What happens in the source while it keeps the potential difference constant between the plates? You can calculate the electric work done while the particle moves in the field - it is provided by the source, on the account of its internal energy.You have to take this work of the source into account.

ehild

5. Jul 6, 2011

### Ambforc

According to my understanding, a source is only needed to separate the charges initially. After the charges have been separated and provided there is no leak to the environment such as the atmosphere, the accumulated charge on the plates is going nowhere and the electric field will stay intact. After the potential difference source has done work to separate the charges, it can be disconnected from the plates, and the plates will stay charged, as in a normal capacitor. Even if the source where to remain connected, it will not do work as there can be no flow of current once the voltage over the plates is equal to the voltage over the source, as the circuit is not closed looped.

Note that the positively charged mass never touches the negative plate physically. Although there may be slight rearrangements of the charge on the plate, this will return to normal once the positive charge exits the electric field and enters the Faraday cage or simply just gets an arbitrary large distance away from the plates.

Are you proposing that, if the plates are charged and the source disconnected, and the charged mass is allowed to fall through repeatedly (without touching either of the plates to allow for charge transfer), that the charge on the plates will eventually neutralize, eliminating the electric field? How can this be without the charges having a conductive pathway to travel between the two plates?

Thank you for your input so far.

6. Jul 6, 2011

### ehild

Well, I need to present a new theory if those plates are not connected anywhere.

You agree that the charged particle has a varying electric potential energy while it moves between the plates don't you? If we assume that nothing happens on the plates, the charges are fixed, there are no radiation effects, the KE of the particle gained from the field equals to the loss of its electric PE.
When the particle falls out of the space between the plates, it does not get into zero electric field. Call it "edge effect". Such electric field as shown in the figure does not exist . The negative charges on the plate nearer to the particle will attract it stronger than the positive charges on the other plate. As a result, it will loose KE, and in case the ground is very far away, it loses all what it gained from the field. If not, you need to do work against the coulomb forces along the way back to the positive plate.

ehild

7. Jul 6, 2011

### Ambforc

Absolutely. I thought that the problem would be with the fringe or edge effect usually ignored in the textbooks since they assume that the ratio of the distance between the plates to the edge of the plate is such that fringe effects can be ignored for the situations studied. However, in my original post I proposed two workarounds for this:

1. Let the whole procedure finish on the left side (to the positive side) with respect to the middle between the two plates. Then the charged mass will still be closer to the positive side than the negative side upon "exiting" the main horizontal portion of the field. Prevent it from moving to the other half by stopping it before it does. It will now experience a net repulsive force as it exits. Getting around the positively charged plate can be done infinitely far away if required, or done in a Faraday cage (see next point).
2. Let it fall into a Faraday cage, shielding it from all external electrical fields. It is possible that the field will be such around the opening into the Faraday cage that it will still retard it somewhat, but will this necessarily be exactly that amount required to stop all it gained? Can this be proven by a vector field description of the electric field around the opening into the Faraday cage?

Also, I will appreciate it if you can comment on a second proposed scenario in the post Theoretical electrodynamic cycle 2, I have not received any reply to it as yet.

8. Jul 6, 2011

### ehild

When you perform thought experiments on electrodynamics you have to accept Maxwell's equations. One of those is that the work done by the electric field along a closed path is zero, if the path does not enclose external currents and the time dependence can be neglected.

The change of KE of the particle is equal to the work done on it. In a complete cycle, the electric work is zero, therefore the electric contribution to the kinetic energy is also zero.

The charges on the plates interact with the moving charged particle both in between the plates and when it falls out from or moved back into the space between the plates again. You can not gain energy.
Calculating the work done in your set-ups might be tedious, but it can be done applying Coulomb's law.

ehild