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the gravity potential energy to electric potential energy analogy again!!!

Posted on 2010-08-16
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Hi

As mentioned in previous posts electric potential energy (EPE) is always introduced by comparing it to gravitational potential energy (GPE). It goes a bit like this:

work is done on an object to move it against gravity. The work done on the object is equal to the gain in GPE of the object. When the object returns to its lower starting position the GPE is transferred to kinetic energy. Gravity has done negative work on the object (I'd never heard of negative work before today).

I'm trying to now compare this to the EPE example and I am concerned with getting concepts correct as well as the correct wording as I know this matters.

Lets imagine we have a copper wire with potential difference across it because there is a build up of electrons at one end. I don't know how to describe the force/work on the electrons. Is it correct to say the following:

a) the electrons move because there is a pushing force on them because of repulsion from the build up of electrons. So the electrons are moving in the direction of a force but there is no work done. But work is distance moved in the direction of a force. I can see why there is no work done but don't know how to word it. there is a distinction which i don't know how to describe between an existing force that makes something move (e.g. gravity pulls something down) and a force that has to be applied to make something move (i have to lift the box). In this case the electric potential already exists and is making something move. Granted we did work to make the potential difference but once that force is there things can move in the direction of that force without work being done.

b) is something doing negative work on the electrons? I would guess if graivty does negative work on an object then a voltage can do negative work on charge

thanks
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Question by:andieje
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by:IT-Monkey-Dave
ID: 33449324
This is way out of my area of expertise, but I don't think you can move an electron across a wire from Point A to Point B without "work" occurring in the form of the electron overcoming the resistance of the wire.  That resistance, however low, is not zero.  I'm guessing the "work" in this case would be heat generated by the electrons overcoming the resistance of the wire.
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by:d-glitch
ID: 33449402
The analogy works for a charged particle in free space. where the only choices are kinetic and potential energy.
In a wire, things are more complicated, and neither KE nor PE are really relevant.

This is fine:
>> Lets imagine we have a copper wire with potential difference across it
   
This really isn't:
>> because there is a build up of electrons at one end.

a) The electrons move because there is an electric field in the wire.  
    Since the electrons are negatively charged, they move opposite to the direction of the field.
    There is work done on the electrons, but it is primarily dissipated at heat.

>>  Granted we did work to make the potential difference but once that force is there things can move
      in the direction of that force without work being done.

This isn't true.  A charged particle moving in an electric field experiences a force.  If it moves in the field,
there is work.


One small change:
>>  b) is something doing negative work on the electrons? I would guess if gravity does negative work
       on an object then an ELECTRIC FIELD can do negative work on charge
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by:andieje
ID: 33449737
i thought you got a build up of electrons at the negative plate of a battery? Someone explained it to me once that the reason you get a higher current though a wire when there is a higher voltage across it because if you imagined more electrons are the negative end of the wire and then the charge(q) would increase and if you then calculated the force using coulombs law the force on each ecltron would increase
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by:ozo
ID: 33450018
Yes, the electrical force increases when you have more charges, just as the gravitational force increases when you have more mass
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by:ozo
ID: 33450030
Work is done when electrons move in an electric field, just as work is done when masses move in a gravitational field.
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by:andieje
ID: 33450123
This driving me crazy in captial letters. However i know realise what I don't understand which is a relief in itself.

I understand how point charges create electric fields and I understand how the forces vary at different points in those elctric fields. And i understand how the total force on a charged object will depend on the positioning of other charged objects around it. But then when i try and apply this knowledge to what goes on when charge moves along a wire i have no idea.
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by:ozo
ID: 33450188
Can you position masses around at various places and find the G field from them and apply this knowledge to what goes on when a mass moves through a tube?

(that should get you on your way, but in many situations you may also have to remember that the tube is also filled with masses that respond to the field generated by the moving masses)
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by:ozo
ID: 33450332
The main difference between gravity and electricity is that like gravitational charges attract, whereas like electrical charges repel, but virtually all the equations work analogously.
(the fact that gravitons have spin 2, whereas photons have spin 1 has no practical consequences in any of your examples)

The other important difference is that negative charges are a lot easier to find than negative masses,
so that most objects you encounter have an almost perfect balance of positive and negative charges,
with the tiny imbalances being responsible for most of the electrical phenomena you observe.

This also means that it can be tedious to individually sum up all the fields from all the charges and all the motions in response to those fields in order to describe things,
so we often resort to rules of thumb about the collective behavior of the nearly balanced positive and negative charges in materials with various properties.
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by:andieje
ID: 33453554
Ok, thanks. But I am confused now by things other people have told me. Everyone seems to tell you different things where this subject is concerned.

So there isn't a build up of electrons and negative charge in the wire attached to the negative terminal of a battery?

I've definitely been told on this forum that this is an accumulation of charge at the negative terminal of a battery though I appreciate this isn't the same thing as a charge build up on the wire (though i thought it would be)
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by:d-glitch
ID: 33454340
The analogy between gravitational and electrical PE and KE works fine for particle/masses/charges in free space.
It doesn't work for electrical charges in a wire.

The sort of analogy that might work would involve dropping lots of ball bearings into a tube filled with thick oil.
You would have to think about the BB's bouncing off each other and the wall of the tube, and interacting with the
gravity and the sticky oil.  Such a complicated model is not likely to be helpful.

>>  when i try and apply this knowledge to what goes on when charge moves along a wire i have no idea.

The model that works well for electricity in a wire is current (not charges) and electric field.

Take  12V battery with a 12 meter, 12 ohm cable across its terminals.
The electric field at every point along the wire is 1V/meter.  The current at every point is 1A
Every electron in the wire feels the same force.  The field does work on the electrons.
But the electrons don't accelerate.  They are moving with some spread around an average terminal velocity.
The work done on the electrons is mostly dissipated as heat/friction in the wire.
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by:d-glitch
ID: 33454492
>>  I've definitely been told on this forum that this is an accumulation of charge at the negative terminal of a
       battery though I appreciate this isn't the same thing as a charge build up on the wire (though i thought
       it would be)

This is true, but not really helpful or relevant to the flow of electricity from the wire.

Take a 12 volt battery, with terminals that are 4 x 4 x 0.5 cm  (Wide x High x Thick).
Let the distance between the terminals be 10 cm.

The terminals of the battery form a little parallel plate capacitor.  
You can calculate the charge on each terminal precisely.
If you double the battery voltage, you will double the charge on each terminal.
But this has nothing to do with the flow of electricity once you connect a wire.

The crucial characteristic of the battery is not the charge on the terminals but its ability
to maintain 12V across those terminals even when there is current flowing.

With time and experience, you will figure out which tools/models to apply in different situations.
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BigRat earned 250 total points
ID: 33473435
>> Everyone seems to tell you different things where this subject is concerned.

This is because of the intepretations of the terms in use. You stumbled over "negative work", simply because, I suspect, you equate "work" with the sort of sweaty labour you might do yourself. In fact although the two are related, work in science is a technical term which is BY DEFINITIION the amount of energy transferred by a force acting through a distance. (see http://en.wikipedia.org/wiki/Work_(physics)). So if a force moves an electron so many meters, then work is done. Since force is a vector,  this PRODUCT can be negative, if the force acts in the opposite sense to what is considered positive. Hence negative work.

>>I've definitely been told on this forum that this is an accumulation of charge at the negative terminal of a battery though I appreciate this isn't the same thing as a charge build up on the wire (though i thought it would be)

Accepting for the moment that there is a build up of negative charge on a battery terminal, then any wire attached to it (and nowhere else) will also have this build up of charge. Your question isn't clear as to what this wire is "doing". d-glitch is correct in his capacitance explanation.

In fact what actually happens when an open wire is attached to a battery terminal, is that a wave of energy goes down the wire as if the wire was a closed circuit. This wave of energy will (assuming a perfect switch between the battery and the wire) be a square wave, ie: a voltage jump, and the current will be this voltage divided by the instantaneous resistance of the wire. This is a combination of the normal resistance and the capacitance and inductance of the wire. This will in fact not only attenuate the wave as it proceeds along the wire, but will effect the waves shape. Eventually the wave reaches the end of the wire. The same rules apply, the resistance here is infinte (or virtually so) and the inductance and capacitance fall off almost to zero. the effect is to cause a backwards wave which effectively cancels out the downcoming wave. The net result is a small loss of energy when the wire is attached and a small charge stored in the wire. If you use a high-spped oscilloscope to view this event, you'll see a short pulse with sinusoidal wave on it, which is the wave going down and comming back on the wire.

Incidentally the wave obeys Maxwell's wave equations and you can find out much more on this topic, which is called Transmission Line Theory, in textbooks on Communications.

My first encounter with this was in the sixties. If your television antenna is in direct sight of the transmitter you'll get a perfect picture. But if there is some large object, like a gasometer, say off to the right, which reflects the signal to you, then your net signal will consist of the original signal plus a slightly delayed and weaker signal. The effect is a "ghost" on the screen. If you splice a bit of coax cable into your antenna feed which is open at the other end, the main signal will go down the wire and come back slightly attenuated and negative. If the wire is cut to the correct length, the returned signal will coincide with the ghost and cancel it out.

The same effects can happen with current switching logic, and is one of the reasons why SCSI devices need to be properly terminated. One often sees "terminating resistors" which do this sort of thing on logic boards.
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by:andieje
ID: 33487890
dglithch this is obviously where I am going wrong:

"The terminals of the battery form a little parallel plate capacitor.  
You can calculate the charge on each terminal precisely.
If you double the battery voltage, you will double the charge on each terminal.
But this has nothing to do with the flow of electricity once you connect a wire."

I'm trying to apply what I know about charges in free space to what is going on in a wire and like you said it doesn't apply. However I'm still left not really knowing what happens in a circuit/battery.

Obviously from bigrat's explanation its not simple!
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by:andieje
ID: 33488420
Is it useful to think of electrons in the wire moving because of the electric field generated by the potential difference between the 2 plates of a battery? I've been thinking in terms of the force on an electron in a wire based on its position in the electric field between the terminals of a battery and I'm not sure that way of thinking is helping me understand current electricity
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by:d-glitch
ID: 33488734
>> Is it useful to think of electrons in the wire moving because of the electric field generated by the potential
     difference between the 2 plates of a battery

Yes, because the electric field in the wire is uniform along the entire length.
 If the battery is 12V and wire is 12 meters, the electrical field strength is 1V/meter everywhere.


>> I've been thinking in terms of the force on an electron in a wire based on its position in the electric field
     between the terminals.

Since the field in the wire is uniform, the position of an electron in the wire doesn't matter.  All the free/outer/
conduction electrons feel the same field and respond (statistically) in the same way.

The flow of electrons (number and velocity) is uniform everywhere in the wire.
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Author Comment

by:andieje
ID: 33489553
Do you mind if I ask some more questions? I will open another question but it can be hard when too many people contribute to a question as you end up going off on too many tangents
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by:andieje
ID: 33489566
I'll try and break everything down into little yes/no questions
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Author Comment

by:andieje
ID: 33489589
Is it correct that there is an electric field around the terminals of a battery. I've done a little diagram (its good i know) supposed to show some of the field lines around a battery.
field.jpg
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by:ozo
ID: 33489666
yes
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by:phoffric
ID: 33489716
I didn't see the word battery in the OP so is this another question? But here is one thought about batteries.

If you have a 9 volt battery that you use to light up a flashlight (and you verify that it works fine), and then you use that same battery to use in your electric piano, maybe you'll get a note out, maybe not.

The problem is that the charge build-up on the terminals cannot be maintained by that little flashlight battery to sustain the power requirements of the electric piano. The piano consumes the energy available faster than the battery can deliver it, so the charge on the battery terminal decreases and the voltage of the battery decreases (while the piano is still on and attached to the battery), and the electric field generated by the battery decreases, and the piano doesn't work.

What you need is a battery whose internals are such as to be able to sustain the charge buildup on the terminals to maintain the sufficient electric field necessary to supply the piano with as much energy it needs per second. That is, you would need a battery which can sustain the power requirements of piano. (A good specification for a battery might be "volts, amps, hours". I'm sure there are more extensive specifications for more general purposes. This specification says that this battery can sustain V volts at A amps for H hours.)
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by:andieje
ID: 33491742
no the question didn;t start off about batteries but i have now understood where the source of my misonceptions are.
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by:andieje
ID: 33491769
I was confused because i was trying to imagine the forces on electrons in the wire (e.g. e1 and e2 in my diagram below) based on their position in the electric field. I thought e1 was closer than e2 so should feel more force. d1 and d2 represent distances on the daigram

However i hope I'm correct in saying that this was an unproductive way to think because the electrons aren't free to through space along the lines d1 and d2 so it doesn't work like that.

If you agree (ozo, dglitch or phoffric) then i shall try and empty my head of all these unhelpful thoughts and try and look at the subject again from a different perspective
img057.jpg
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by:ozo
ID: 33491945
the wire is full of charges that affect the field and are affected by the field
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by:andieje
ID: 33492009
ok, but the point i was trying to make is that the charges aren't in fact affected by the field inthe way that i thought they were because they can;t move directly through space.

I'm hoping i was wrong because that would explain why its always confused me
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by:phoffric
ID: 33492088
Yes, there is not just a charge build-up at the terminals. There is a charge gradient across the wire, and each charge effects the electric field. Review this post to see if it helps:
    http://www.experts-exchange.com/Other/Math_Science/Q_26288244.html#a33079582


For your light reading entertainment, here is a "mad", but true theory to "answer the question: 'How does energy really flow in electric circuits?":
     http://www.furryelephant.com/content/electricity/visualizing-electric-current/surface-charges-poynting-vector/
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by:phoffric
ID: 33492184
>> I thought e1 was closer than e2 so should feel more force.
The force on e2 comes from every net charge in the wire, not just in the battery terminals.
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by:andieje
ID: 33493266
It was that very post that helped me find my misconceptions phoffric. It's so good i've printed it out!

So you're agreeing with me to ditch those misconceptions?

You can only look at that furry elephant site for 10 mins (though i definitely had less than that) without a subscription!!!
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by:phoffric
ID: 33493379
Which misconceptions? IMO, after you see a consensus of viewpoints in this thread, you should then reread a physics text or papers on the subject and see if this read makes more sense. If there still appears some issues, then certainly bring them up.
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by:d-glitch
ID: 33495555
There are several things to consider concerning your second picture, with the long
curvy wire between the battery terminals:

The electrons in the wire are constrained to stay in the wire.

The presence conductors (carrying current or not) can the electric field configuration.

In your drawing, the wire could be a complicated 3D curve.  
The complete description of the free space electric field would be horribly complex.

But the electric filed in the wire is still the simple volts/length in magnitude
and parallel to the wire in direction.
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by:andieje
ID: 33526297
What i mean phoffric was to ditch my misconceptions about how the electrons in the wire are affected by the electric field on the battery because I was treating the electrons in my mind as charged particles that can move through free space which they can't
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by:phoffric
ID: 33526336
In general, ditching misconceptions is a good thing only after asking how one came to believe in these misconceptions; otherwise, history will repeat itself without improvement in the process.

>> charged particles that can move through free space which they can't
It sounds like you have answered this particular issue of electrons in  wire under normal usage conditions not being able to move through free space. However, I imagine that possibly there may be less usual conditions where charged particles could leave a wire and travel through free space.
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by:ozo
ozo earned 250 total points
ID: 33526438
I think the error in predicting the motion of the electrons is less the assumption that they can move through free space,
than it is ignoring the field generated by all the charges in the wire, and the response of those charges to that field.
(which also act to confine free electrons to the wire)
Of course, since there are so many charges in the wire, summing the effects of all of them can be tedious,
and since the effect of all the positive charges and the negative charges come so close to canceling out,
even a small error can make a big difference.
Which tends to make it more practical to analyze electric potential in wires using rules of thumb like
Kirchhoff's and Ohm's laws than from first principles and analogies with gravitation (although that is possible)
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by:BigRat
ID: 33529686
>> I imagine that possibly there may be less usual conditions where charged particles could leave a wire and travel through free space

The only charged particles we are talking about here are electrons, since the atomic nuclei are strongly bound in the metal lattice (after all metals, which conduct electricity well, are very hard materials).

Conductors, as I have mentioned before, are materials whose atoms have outer electrons shielded from the nuclei which are able to move. These are call band electrons and it does not take much energy to shift them into a (further) outer level. Moving through the metal lattice is difficult, since the lattice is electron rich, so electrons tend to move along the surface of a conductor (this is called the skin effect).

The actual electrical energy travels down the conductor according to Maxwell's Wave Equation, which means that the energy travels quite near the speed of light. On printer circuit boards this is around 1 nano second per foot. The electrons however DO NOT more at this speed, they just drift around, and this drifting can cause a random noise in conductors - and more strongly in semi-conductors - and is called shot noise. The actual energy transfer is actually due to the interaction of electrons which exchange virtual photons.

http://www.tpub.com/neets/book1/chapter1/1n.htm
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by:phoffric
ID: 33541445
>> I imagine that possibly there may be less usual conditions where charged particles could leave a wire and travel through free space

    Here is what I was thinking. If incorrect, please correct. If the wire had high resistance per meter, and the wire was very large, and if the "battery" were actually a DC generator, and if the generator produced a very large voltage across the terminals, then I thought that if the total resistance of the wire, very big R (v-b-r), was greater than the electrical resistance of air, then I thought there might be a static discharge from one terminal to the other, or possibly to another highly charged object through the air.
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by:BigRat
ID: 33541633
OK.
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Author Closing Comment

by:andieje
ID: 33681605
thanks
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