Electrons velocity and Potential energy

[tonyjk]

hello all... i have a question: In an electric circuit the velocity of electrons are very small than how they will pass through all the elements of a circuit and gain or/and loss potential energy? for example let's take a basic circuit : An EMF and a resistor. How the electron in a resistor will gain a potential energy through the EMF device if its velocity is small? how can it pass through the EMF device? even in AC circuits the electrons barely move...thanks

 

[hds123523000]

as far as i know, the velocity of the propagation of the electric field is the same as the speed of light so that electrons can sense it quickly and move slowly.

 

[Drakkith]

The electron doesn't gain potential energy by moving. It gains potential energy by having a voltage applied to the circuit. This causes the electron to move and converts potential energy into other types, such as heat or motion.

Remember that voltage is a difference in electric potential in the circuit. When you have a difference in potential, charges want to move to balance out that difference. Applying and sustaining a potential difference is what causes current to flow in a circuit.

 

[tonyjk]

Yeah i know what's voltage than why we say there's
Conservation of energy in a circuit? It does mean the conservation of electron's energy? So when we have 2 resistors the electron does not
Move trough the two resistors and convert its potential energy gained by the EMF to heat?

 

[sophiecentaur]

Forget about the velocity of the electrons - energy is not carried to circuit components in the form of electron KE. The energy moves from source to load by a lot of electrons moving at a very low mean velocity. It isn't the energy conservation 'of the electrons'; the charges, moving through a resistor, for example, lose energy by interacting with the material and this heats the resistor. For two resistors in series, the Potential difference is 'shared' by the two resistors. each Coulomb shares its energy between the two series elements.
Electrons take so long to get anywhere in a normal circuit that it is pretty pointless to describe things in terms of electrons moving. By the time an electron actually might have got round the circuit, the experiment could be finished and the circuit could have been switched off. It's best to talk of Charge - which doesn't concern itself with details like random motion of countless electrons with low mean speeds.

 

[Drakkith]

Yeah i know what's voltage than why we say there's
Conservation of energy in a circuit? It does mean the conservation of electron's energy? So when we have 2 resistors the electron does not
Move trough the two resistors and convert its potential energy gained by the EMF to heat?

Don't think in terms of a single electron. Think in terms of large numbers of charges moving. Current will flow throughout the entire conductor all at once, so current will flow through both resistors and lose energy as heat.
Edit: Sophie, you beat me while I wrote and re-wrote this post a time or two!

 

[tonyjk]

ah yes it's the potential energy of 1 coulomb of charge( meaning several electrons) but even the 1 coulomb of charge will flow all around the circuit to "share" it's potential energy? all what i mean is how those charges will share their potential energy if they don't "interact" with the device(such as resistor)... hope I am not disturbing you with my questions

 

[sophiecentaur]

One Coulomb flows in one end and another Coulombs worth of electrons flow out of the other end. The energy still gets transferred - at the speed of light nearly.
A crude analogue is a bicycle chain, in very high gear. The links move around slowly and the bike moves fast. You don't consider the KE of the links - just the force times distance moved of the chain, when calculating the energy transferred.

"Several electrons" lol. More like 10E23 electrons!

 

[tonyjk]

"Several electrons" lol. More like 10E23 electrons!

haha yeah I'm bad in english :P thank you any way

 

[tonyjk]

Sorry again really I am a little confused... Let's take a circuit that contains an emf and a resistor...a charge in the copper that is after the emf what is its
Potential? This charge(1 coulomb) didnt pass through the emf and gained potential energy it means that no work was done by the emf on this charge..thanks again

 

[Drakkith]

Sorry again really I am a little confused... Let's take a circuit that contains an emf and a resistor...a charge in the copper that is after the emf what is its
Potential? This charge(1 coulomb) didnt pass through the emf and gained potential energy it means that no work was done by the emf on this charge..thanks again

The voltage applied to the circuit causes the EMF in the entirety of the circuit all at once. It doesn't matter where your voltage source is in the circuit. Also, current is flowing through ALL points in the circuit at the same time, not just in one spot.

 

[tonyjk]

i know that's there's a flow of charges all around the circuit and about the emf causing the move.. I'm talking about the gain of potential: here the figure:
https://www.circuitlab.com/circuit/ha4c48/screenshot/540x405/
let's say at t=0+(after the closing of the switch) the charge at A what is it potential energy? and the charge at B what is its potential energy? ( referring to the ground(V=0)) thanks again

 

[sophiecentaur]

Here's your problem. You need to understand that Potential Energy is something that the source of emf (battery / capacitor / generator) gives to the charges, in the same way that the elevator gives you Gravitational Potential by taking you up to the top of the building. When the charges (and you) have that potential (maximum at the top of the building and at the +12V battery terminal), the energy can be transferred from them (and you) as they move to a point with lower potential. On the way round the rest of the circuit the electrons lose potential energy as it's transferred into heating / light / sound / movement in the circuit components. There is no Potential Energy left when the charges have made it to the other terminal of the battery (@V=0).

 

[tonyjk]

really i know that its similar to the gravitational potential but here what i mean: the elevator is TAKING ME to another potential but the charge at point A didnt pass through the battery to gain it's potential... i hope you understand my point and sorry for any disturbance again

 

[ModusPwnd]

Its the the electric field strength at the location of the charge that determines its potential energy. It need not ever have gone through the battery. Similarly, its the gravitational field strength at your location high up in the building that gives you gravitational potential energy. You need not have ever been in the elevator to have gravitational potential energy.

 

[sophiecentaur]

If you put an arrow into a taught bow, it has potential without actually being moved back with the string. There is no need for the charge to 'move' for it to have potential. You don't even need to move, yourself, to gain Gravitational Potential - you could have been held in position by a sky-hook and the Earth moved away from you. (Unlikely but, in principle . . . .)

 

[tonyjk]

ok thank you all

 

[tonyjk]

can anyone give the name of a book that goes into details about all what you talked about.. thanks

 

[ModusPwnd]

That depends on how much math you know. The standard introductory text is by Griffiths, it requires a knowledge of vector calculus though. If you don't have that much math you will have to check out the electromagnetism sections of a broad introductory "college physics" or "university physics" textbook.

 

[sophiecentaur]

It's pretty standard stuff that you should be able to find in any A Level Physics book. Else, look around on the hyperphysics site. That may not be 'chatty' enough for you but it has facts and formulae on it and is excellent revision-type material.

 

[tonyjk]

actually i have a lot of ebooks from fundamentals of electromagnetism and physics but i was asking if you knew better ones any way...
here's a quote from a book called fundamentals of physics by Walker and Halliday:

The emf E is the energy per unit charge transferred to the moving charges BY the battery. The quantity iR is the energy per unit charge transferred from the moving charges to thermal energy within the resistor. Therefore, this equation means that the energy per unit charge transferred to the moving charges is equal to the energy per unit charge transferred from them

What i don't understand this amount iR is the energy lost by 1 coulomb of charge when this charge pass through the entire resistor but the resistor has a length about 0.5cm for example so how this charge will pass through this resistor with its velocity? even in AC circuits the charge almost don't move so the IeffectivexR is what? thanks

 

[sophiecentaur]

Re-read the bicycle chain analogy and you will see that the speed of the things that move is not relevant to the energy transfer. All the charges in the resistor at anyone time are dissipating some of their Potential energy - the process goes on over the whole length of the resistor. 'Fresh electrons' arrive with all their PE and 'spent electrons' (god how I hate those words - but they are sort of explanatory) leave with less PE.
It's a bit like the old canal boats carrying coal. It didn't matter how fast they traveled as long as the rate in and out of the length of canal was fast enough to run the factories.

 

[tonyjk]

but when we say R is proportional to the length of the resistor and the potential lost when a charge pass through the resistor is iR so the "spent electrons" that they loss their energy didnt pass through the entire resistor so how come they lost iR energy?

 

[tonyjk]

Re-read the bicycle chain analogy and you will see that the speed of the things that move is not relevant to the energy transfer. All the charges in the resistor at anyone time are dissipating some of their Potential energy - the process goes on over the whole length of the resistor. 'Fresh electrons' arrive with all their PE and 'spent electrons' (god how I hate those words - but they are sort of explanatory) leave with less PE.
It's a bit like the old canal boats carrying coal. It didn't matter how fast they traveled as long as the rate in and out of the length of canal was fast enough to run the factories.

so the iR is the amount of energy lost by all the charges that are inside the resistor and its also the loss of energy of a charge that(virtually) pass from one terminal to another?

 

[sophiecentaur]

Jeez, you are making me THINK!
Right. The charges near the bottom at switch on only have a small potential. As you go up,
the charges will have more etc. So charges with all values of PD will be there. When you switch off, the same situation obtains. Energy is transferred uniformly all the time the volts are applied. Good enough?

 

[tonyjk]

what i understand from you:P : the delta V is between the two terminals of the resistor and it's equal to the potential that a charge of 1 coulomb will loss if it passes across the resistor but in reality its equal to the the loss of potential of all the charges inside the resistor?

 

[tonyjk]

delta V = -integral of the electric across the length of the resistor and it's also the work done by the electric force(-q*E) inside the resistor to get a charge of 1 coulomb pass across the resistor so actually they are saying that this charge passed the resistor and lost its potential but in reality this charge didnt pass through the entire resistor... hope you understand me:P

. If a charge
is moved from a through the battery, its electric potential energy is increased
by dU=Dq*DV . On the other hand, as the charge moves across the resistor, the potential
energy is decreased due to collisions with atoms in the resistor

http://web.mit.edu/viz/EM/visualizations/notes/modules/guide06.pdf

 

[sophiecentaur]

The charges that pass through the whole of the resistor lose all that PE. The ones that only pass through a part of the resistor (this will be at switch on and switch off time) will only dissipate a proportional part of the PE because the potential difference was only there for part of the possible journey through the resistor. They are moving soooo slowly that they are only propelled through a part of the resistor.

 

[tonyjk]

The charges that pass through the whole of the resistor lose all that PE. The ones that only pass through a part of the resistor (this will be at switch on and switch off time) will only dissipate a proportional part of the PE because the potential difference was only there for part of the possible journey through the resistor. They are moving soooo slowly that they are only propelled through a part of the resistor.

ah ok so the charges can pass through the whole resistor

 

[sophiecentaur]

It's perfectly correct in the steady state. But none of Kirchoff is correct for the first few ns, in any case. Are you trying to find 'loopholes' or what? DC is DC - that means (in Engineering / Science) 'after a long enough time'. This what I (and others) am getting at when I /we say that considering the speed of charges doesn't help you in this stuff.

Would you like to have the whole of the state of current Physics dumped on you all in one go? Could you cope?

 

[tonyjk]

thank you sophie really now i understand it :D

 

[sophiecentaur]

You're a hard man to convince but no worries, I enjoyed having to justify my answers!

 

[tonyjk]

Just a last question:p what do you mean by charge? Is it consisting of large numbers of electron? So how we won't look at drift speed of electrons?( i was re-reading your first post:p)

 

[sophiecentaur]

Because the drift speed is irrelevant to the situation. Have you not read any of this thread or any textbooks about this? By "Charge" I mean exactly what all textbooks mean - standard definition in a proper textbook and not some noddy thing aimed at 'lower school pupils' - that will not help you at all. Whatever happens to be the carrier of charge, the equations all apply the same and do not include drift velocity. Why do you think that would be?

 

[tonyjk]

I'm confused about when we say the power is equal to the product of UxI but I is for a cross sectional area and U is the electric potential between 2 terminals and by coulomb of charge what i mean is those charge dq = Ixdt how they will pass both terminal?