Conceptual Doubts in Eletrodynamics


by jaumzaum
Tags: conceptual, doubts, eletrodynamics
jaumzaum
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#1
Mar1-13, 08:07 PM
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I'm having a lot of "simple" conceptual doubts when I try to study eletrodynamics in a qualitative way. Look at the circuit below



First we have a capacitor and a generator of voltage V (A). We wait until the capacitor is fully charged and then disconnect the generator terminals (B). Finally, we connect both terminals of the capacitor together (C). I want to know what will happen and why.

In (A) we have an internal force in the generator (emf=V) that makes electrons flow from the anode to the cathode. This way electrons are deposited in one of the plates of the capacitor and electrons flow from the other plate to the cathode. Right?

In (B) we disconnect the generator terminals and nothing happens, since there is no way for the electrons to flow.

In (C) I really don't know what would happen. I would say there is a force between the plates that can make the electrons stay exactly where they are and cause no current. But there is a force in the thread too right? Is this force (voltage) V too? If so, considering the negative plate, we have a force V to the left and a force V to the right, what would cause an indifferent equilibrium. Can anyone help me?
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BruceW
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Mar1-13, 08:24 PM
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you've got it all right for part (a) and (b). Now for part (c) there is a build-up of charges on the plates. There is no driving force, so the system will simply go to the lowest energy state. The build-up of charges means that there are charges close to each other. Think about the groups of charges on each side of the capacitor. They are closer to each other within their group, than they are across groups. So the force going across the capacitor is the less strong force. Due to coulomb's force, all the charges within group push away from each other. So what do you expect the final state to be? It is fairly intuitive answer.

edit: by 'within group' I mean imagine all the charges that have built up on either side of capacitor. There is a group on each side.
jaumzaum
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Mar1-13, 09:27 PM
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Thanks Bruce. Nice explanation. I would say each plates would become neutral again.

Now another thing came in my mind. How can I explain qualitatively that the emf (drving force) of a battery act through the whole thread even in a great distance? For example, how can the emf of a battery act in a capacitor which is 1 meter far of it, if they are connected only by a thread?

BruceW
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Mar2-13, 02:19 AM
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Conceptual Doubts in Eletrodynamics


oh, I didn't realise it was a thread and a big gap between plates. Hmm, that makes it seem more like an experiment to do with static electricity. Was it actually a battery, or was it a van de Graaf generator?

About your question.. um I am slightly less certain about what happens for a truly large-scale 'capacitor' of width of a meter, as you mentioned. But If you do assume it acts like a usual capacitor, then you can work out how the capacitance depends on the dimensions of the capacitor. Wiki have a short explanation on the 'capacitor' page, under 'parallel-plate model'
jaumzaum
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Mar2-13, 11:21 AM
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Quote Quote by BruceW View Post
oh, I didn't realise it was a thread and a big gap between plates. Hmm, that makes it seem more like an experiment to do with static electricity. Was it actually a battery, or was it a van de Graaf generator?

About your question.. um I am slightly less certain about what happens for a truly large-scale 'capacitor' of width of a meter, as you mentioned. But If you do assume it acts like a usual capacitor, then you can work out how the capacitance depends on the dimensions of the capacitor. Wiki have a short explanation on the 'capacitor' page, under 'parallel-plate model'
Actually the capacitor is normal, Area is much greater than the distance between plates. The 1 meter I mentioned is the thread length. I mean, how can the emf produced by a battery be distributed by the whole 1 meter thread and reach the other end of the thread in witch a capacitor is connected to? How the driving force is permeated through the thread? Ho can I explain it qualitatively?
bgq
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Mar3-13, 04:30 AM
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Hi,
About your second question, I don't claim that I give you a formal explanation, but just simple one that may clarify the situation: The thread is made of up of metal which is able to lose electrons. The repulsive electric force makes an atom in the thread to lose an electron (lets say the first atom); the second atom captures this electron momentarily and then lose it again (because metals cannot gain electrons), now the second atom, by the similar procedure, gives this electron to the third atom, and so on until it reaches the capacitor.

Please note the above process occurs simultaneously through the whole wire. It is somehow similar to put a row of 1000 balls horizontally in a straight line (no spaces between them), now if you push the first ball, the last ball simultaneously goes away.

Have a nice day!
e.chaniotakis
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Mar6-13, 04:59 AM
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If I may say, in the C case, if the wire has resistivity R, then you have an RC circuit. The capacitor is discharged and current flows through the wire, losing part of its energy through R. The capacitor is re-charged with opposite charges and then discharged again with current flowing opposite etc. This way after a few circles, the current will be no more ( all energy will have been consumed by R, and the capacitor will have no charge). How is that?
amos carine
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Mar6-13, 12:55 PM
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Quote Quote by bgq View Post
Please note the above process occurs simultaneously through the whole wire. It is somehow similar to put a row of 1000 balls horizontally in a straight line (no spaces between them), now if you push the first ball, the last ball simultaneously goes away.

Have a nice day!
This is a classic physics post, so instantaneous action is permitted, but under a relativistic view, would this newtons cradle action or something like billiard balls lined touching each other, would this transfer of force be limited by c?
Integral
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#9
Mar6-13, 01:39 PM
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Not sure why you think that there will be oscillations. Shorting the cap will cause a current to flow. The voltage on the cap will drop as the current flows, once the cap is back to neutral the voltage and current flow will be zero. There are no forces acting to cause any oscillations.
amos carine
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#10
Mar6-13, 02:15 PM
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Isn't there an electromotive force between all the particles in the wire that is repulsive as the current flows? An oscillation does not happen from the cap to to where it flows, but i'm not ruling out that something i don't understand behaves with oscillations, in certain cases.
bgq
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Mar9-13, 12:45 PM
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Quote Quote by amos carine View Post
This is a classic physics post, so instantaneous action is permitted, but under a relativistic view, would this newtons cradle action or something like billiard balls lined touching each other, would this transfer of force be limited by c?
Interesting question. Relativity states that no matter can move with speed faster than c; yet we may ask can information move faster than c?
DrZoidberg
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#12
Mar9-13, 02:59 PM
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Of course you get an oscillation if you discharge a capacitor through a wire with very low resistance. The wire and also the capacitor have some inductance so you have an LC circuit.
http://en.wikipedia.org/wiki/LC_circuit
It only oscillates for a tiny fraction of a second though and then the capacitor will be discharged.

Quote Quote by jaumzaum View Post
Now another thing came in my mind. How can I explain qualitatively that the emf (drving force) of a battery act through the whole thread even in a great distance? For example, how can the emf of a battery act in a capacitor which is 1 meter far of it, if they are connected only by a thread?
To understand that you need to look at electric fields. When a current is flowing in a metal wire it is always because of an electric field. An E field inside of metal always causes a current to flow. The other way around this means that if there is no current flowing in a piece of metal the inside of that metal must be free of any E field.

There is also a generalized form of Ohm's law that states
J = σE
or J = E/ρ
=> E = Jρ

http://en.wikipedia.org/wiki/Ohms_law

σ is the conductivity of the metal, ρ is the resistivity, J is the current density and E the electric field.
Let's say you connect the two poles of a battery directly with a wire. That wire should have a sufficient resistance not to short circuit the battery. The current will be the same everywhere in the wire. If the wire is equally thick over it's entire length the current density will be the same and therefore the electric field inside the wire will be the same everywhere. e.g. if the wire is 1.5m long and the battery supplies 1.5V the field in the wire would be 1V/m.
Now that seems strange because a battery without an attached wire will have an E field that is stronger close to the battery and weaker further away. Just like the B field of a magnet. So how can the field suddenly be equal everywhere. Even if that wire was several km long, the field would still be equal over it's entire length.
Imagine the wire was somehow brought close and attached to the battery within 0 time, impossible in practice of course but that doesn't matter. In the very first moment after the wire is attached, the field inside the wire would indeed be much stronger close to the battery. That means the current will be much stronger there. Now the current is not the same everywhere in the wire which means electrons start to "pile up". Different parts of the wire receive different amounts of "static" charge. Those charges produce an E field themselves. Eventually (after a tiny tiny fraction of a second) an equilibrium is reached. All E fields combined (i.e. the field of the battery plus the fields of all the "piled up" charges) produce a field that is equal over the entire length of the wire. And now the current is also equal in the entire wire.
BruceW
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Mar10-13, 06:02 AM
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I was a bit put off by the 'thread', but it seems that we can just treat it like a standard conductor. So if this is true, then jaumzaum's question becomes "how can the driving force permeate though a conductor?" Well, the two terminals of the battery have different potential. So in a sense, electrons are always 'wanting' to go from one terminal to the other. (Because they always want to go to a place of lower potential energy). But without a circuit, they can't do this, because the air is difficult to move across. So when you introduce a circuit, all the electrons in the circuit start flowing around it.

Now you might say "but how do they know which way to go, when they are far from the battery?" Well, the electrons near the battery know which way to go, and when they start to move, the other electrons also move because there can be no build-up of electrons.

So now you might say "So that means the electrons near the battery move first, then ones slightly further away, then ones a little bit further away, and so on. Surely this will take a lot of time?" Well the answer is that these little motions are happening very quickly, on a very short time scale. So it does take a tiny amount of time. But this amount of time is so quick, that we can usually just ignore it. (it is only slightly slower than the speed of light).


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