# Why is there close to zero potential difference in conductors in a circuit?

1. Feb 19, 2012

### glenn21

This is why I think this is the case, I can't come up with any other reason and have been researching all over the place. I do have a knowledge of circuits, I'm doing an electronic engineering course, but the reasoning has never been explained, as if it's unnecessary to understand why, but I am curious.

So in a simple circuit containing a battery and a resistor the electrons have high electric potential energy at the start due to the protons on the other side of the battery and also due to the repulsion from neighbouring electrons on the negative terminal.

The reason I think there is close to zero potential difference across the conductive wires is that the excess electrons on the negative terminal side of the battery spread out across the surface of the conductor in such a way that as electrons travel through the wire, their potential energy is almost constant at all points in space along it. So in the length of wire which the electrons begin at they have a fixed constant non-zero electric potential energy, but in the length of wire after the resistor connecting the positive terminal the excess protons arrange themselves so electrons maintain zero electric potential energy.

In the resistor such an arrangement is not possible due to the structure being of an insulator in nature and hence all the energy must be lost here.

If my reasoning was not true, then why is it that charges pass through the wire at a constant velocity? As the electrons go through the wire they gain kinetic energy, so they are using up electric potential energy. This would also mean that most of the energy would be lost in the resistor only if it were longer than the wires because more distance means more conversion of electric potential energy.

Is my reasoning correct?

I thank anyone who responds to this in advance.

2. Feb 20, 2012

### davenn

Hi glenn21

welcome to the forums

a few bad assumptions there ;)

Put a Volt meter across the terminals of a good battery and what do you measure ?
the voltage of the battery -- there's your potential difference

take a pair of wires a metre long and connect one end of the pair to the battery terminals and leave the other end open circuit
Put a Volt meter across the open ends of the wire and what do you measure ?
still the voltage of the battery -- there's your potential difference still there

Put a 100 Ohm resistor across the open ends of the wire and then the Volt meter across the resistor and what do you measure ?
still the voltage of the battery -- there's your potential difference still there

dunno where you got that info from, but its incorrect :)
A resistor just impedes the flow of current through it and its assoc circuit and using Ohms law in a DC circuit you can work out using the battery Voltage and resistor value to find out what the current flow is reduced to.

cheers
Dave

Last edited: Feb 20, 2012
3. Feb 20, 2012

### sophiecentaur

The reason that there is almost zero potential drop across a piece of connecting wire is that it takes almost no energy to make the charge flow.
Think in terms of 1 Volt corresponding to 1 Joule of energy being needed to cause a charge of 1 Coulomb to flow. Voltage should strictly be called Potential Difference (difference in energy per unit charge)
i.e. 1V = 1J/C
Resistance is just a measure of the PD required for One Amp to flow. It's an attractive notion and it's frequently stated that resistance is "What tends to stop current flowing" etc. However, if you really want to get the right idea,m you need to appreciate resistance for what it is - the relationship between two electrical quantities. (R = V/I)

4. Feb 20, 2012

### glenn21

Thanks for your replies. Ok I need to make myself more clear.

Electric Potential Energy = Kq1q2/R.

Now each charge in a circuit has electric potential energy due to the charge difference on the plates of the battery/capacitor.

The charge is a constant in the circuit, the distance isn't. So as charges go through the circuit - no matter what they are passing through, they are converting electric potential energy to heat/kinetic etc.

This is just like a ball falling to the ground where gravity is causing the ball to have gravitational potential energy, and as it falls it gains kinetic energy.

Now, I am saying the reason why energy conversion does not occur across the conductors is because excess charge from both terminals spreads itself over the surface of each conductor in such a way their is no conversion of energy i.e. each conductor behaves as almost an equipotential. So that is why most of the energy is converted to heat in the resistors as next to zero energy is lost while propagating through the conductors. Their is a sufficient current such that the resistance causes a voltage drop equal to the terminals across it.

Last edited: Feb 20, 2012
5. Feb 20, 2012

### sophiecentaur

This doesn't seem right. Why do you talk of charge spreading over the surface? With a highly conducting wire, hardly energy is needed for the charge to flow so there will be virtually no PD across it. Current will flow through the complete depth of the wire (for DC), in fact, the Pinch Effect will tend to concentrate the current to the centre of the conductor.
The reason that so little energy is used on the way through a conductor is that the electrons are 'dissociated' and do not interact significantly with the lattice of positive ion cores.

There is a danger that you are using a sort of circular argument in your explanation and you have to be precise in describing each step.

6. Feb 20, 2012

### glenn21

If this were not the case, as charges flow through a wire they convert electric potential energy to kinetic energy mainly through conductors. Why do you speak of hardly any energy? Electric potential energy is a function of distance (charge is constant in this circuit), and as such, as particles traverse through it - no matter if they pass through a resistor or a conductor - they convert electric potential energy.

So if the wire is comparatively longer than the resistor in the circuit, more energy will be converted to kinetic than thermal energy throughout the circuit, as electrons don't collide as much.

Compare this with a hypothetical example involving gravity: A ball is falling down and their are rocks everywhere on the way down just stationary in midair, as the ball passes through these rocks most of their gravitational potential energy is converted to thermal energy. If their was more free space than rocks on the way down, more energy would be converted to kinetic energy.

What I am suggesting is that some of the excess charges spread over the surface of the conductor in such a way that as an electron travels through the wire - they are not converting electric potential energy to kinetic energy, instead they are converting no energy because the distribution keeps the electric potential energy constant.

7. Feb 20, 2012

### sophiecentaur

Kinetic energy does not come into it. The drift speed is about 1mm/s and the electron mass is 10e-5 ish of the mass of the metal. Don't make unwarranted assumptions. It is risky.

8. Feb 20, 2012

### glenn21

Ok I agree, but then the energy has to be converted into some form as the electron travels through the wire. What is the energy converted to then, it wouldn't be heat because there are almost zero collisions inside the conductor?

9. Feb 20, 2012

### sophiecentaur

If there is no mechanism then why should it be transformed? You may be thinking in mechanical terms but we are talking of charge transfer and not massive rocks falling. The analogy fails.
mgh is not qV.

10. Feb 20, 2012

### glenn21

But V=Ed and E is a constant. So U=qEd, that is analogous. That is the mechanism. You could similarly say U=mV. Where V is J/kg.

How is charge transfer different to mechanical transfer? Both occur due to a force created by a field.

Thanks for putting in the time to help me so far.

Last edited: Feb 20, 2012
11. Feb 20, 2012

### rcgldr

This the EPE for a charged particle in a field. The EPE in a capacitor can be thought of as stored in it's field. In the case of a circuit attached to the capacitor, assuming the circuit is outside of the field of the capacitor, the electron flow doesn't occur within the field of the capacitor. A different type of logic is needed to describe EPE in a circuit.

How much potential difference there is in a conductor depends on the ratio of resistance in the conductor to the lowest resistance path in a circuit. If the only component in a circuit is an energy source and a conductor, then all of the potential difference occurs in the conductor and due to the low resistance, a lot of heat is generated (including the energy source depending on it's internal resistance, like a battery). If the circuit contains a relatively high resistance, then most of the potential loss occurs across that resistance and not with the conductor.

Last edited: Feb 20, 2012
12. Feb 20, 2012

### glenn21

Actually I am the one who is correct, the formula you have quoted is the formula for electric force - Coulomb's law. If you integrate force with respect to distance you arrive at the formula I stated.

I thought the electrons on the negative terminal of the capacitor experience the field of the capacitor through the capacitor's field going through the circuit from positive to negative terminal. The charges experience a field through the capacitor, but they can only move through the field they feel from the charges at the end of the circuit. Not from the field which tries to pull them through the dielectric.

Last edited: Feb 20, 2012
13. Feb 20, 2012

### rcgldr

Sorry, I was distracted. I was thinking of starting with force and deriving EPE for an analogy with a circuit, but abandoned it and didn't clean up my post. I corrected my previous post, please read it again.

Last edited: Feb 20, 2012
14. Feb 20, 2012

### glenn21

That's ok, that's understandable.

What is it that drives the electrons through the circuit if it isn't the electric field from the capacitor? I thought they felt an electric field due to the protons at the other end of the capacitor (or rather the lack of electrons). And hence why I think of their energy as being converted to other energy forms as they pass through the conductors.

15. Feb 21, 2012

### rcgldr

I'm not sure of a good analogy for fields and forces in a circuit. The EPE for a capacitor is related to position between the plates, while the EPE for a circuit is related to the amount of remaining resistance at some point in a circuit versus the total resistance of the circuit. For a simple steady state DC circuit, the convention for "remaining resistance" would be relative to the negative terminal of the energy source, so maximum at postive terminal, zero at negative terminal.

16. Feb 21, 2012

### glenn21

Thanks for your help rcgldr, I really appreciate it.

Yeah, I see what your talking about, it just doesn't make sense to me that no energy is converted through conductive sections. In what way is a circuit different than if a wire was short circuited between the capacitor terminals? Would the EPE then still be related to position between the plates?

17. Feb 21, 2012

### sophiecentaur

@ glenn21
Perhaps you could use your original 'rolling downhill' analogy but in a different way.
You have a car going down a mountain and you want to use its GPE in order to heat up water, using its brakes. You want it to be going at uniform, low, speed all the way down.
On the steep bits, you will have to apply the brakes hard and you will be converting energy at a fast rate (big force X speed), on the nearly horizontal stretches, the brakes will be off so there will be no energy conversion (zero X same speed).
If you ignore the fact that, in this analogy, someone has to be controlling the brakes to maintain this uniform speed then you have the following parallels. In the sections where there is a big Potential Drop will be where there is a lot of energy transfer (brakes get hot), in the sections where there is little Potential Drop there will be little energy transferred. So we have energy transfer per metre proportional to the slope in potential. Also, of course, the force will be proportional to the Slope - which ties in with the fact that Field is proportional to the differential of Potential (basic rule about field and potential).
The above analogy is not based on KE; it works as speed approaches zero. In the case of electrical conduction, the reason that such a lot of energy is transferred is that there is a huge amount of Charge moving (one electron per atom). This would be the equivalent of 10^23 cars going downhill very slowly. The difference in the scale of numbers is due to the vast difference between Electric and Gravitational forces - which we already know about.

N.B. The situations of Electrons moving through a Conducting Metal and a Vacuum are very different. When the electron is not interacting with a medium, it WILL gain considerable KE, all in one go, as it loses PE on the way between Cathode and Anode.

18. Feb 21, 2012

### rcgldr

That is true for idealized circuit diagrams, but (other than super conductors), there is some energy converted to heat even in a conductor.

I mentioned this case before. All of the energy would be converted to heat (and light if glowing) in the wire (ignoring issues like a battery where there is internal resistance, or the equivalent in terms of the rate of conversion of potential chemical energy into electrical energy and heat). In this case, the EPE would be relative to position in the wire, assuming uniform resistance.

You also need to keep in mind that for a given potential (voltage), the higher the resistance of the total circuit, the lower the current and the lower the amount of energy converted into heat. So if there is a high resistance component in a circuit that is otherwise all conductor, almost all of the conversion of energy to heat takes place in the resistor, but the power (rate of energy conversion) is much less because of that resistor.

19. Feb 21, 2012

### pbuk

Isn't all of this overcomplicating things? The original question was "Why is there close to zero potential difference in conductors in a circuit?" The answer to this is that conductors have close to zero resistance. For instance a typical 1mm2 stranded wire has resistivity of c.0.0175Î©/m. Carrying a current of 1A the PD along 1m of wire is therefore 0.0175V, and the heat loss along the wire is a negligible 0.0175W.

Low resistivity is the definition of a conductor so you can't really ask 'why does a conductor have low resistivity', but it does make sense to ask 'what are the characteristics of a conductor that make it have low resistivity'. The answer is that electrons can move freely along the conductor.

Considering your falling ball analogy, a conductor is a bit like a dust storm - the ball is impeded slightly by the dust, but not enough to notice. A resistive component however is a bit like the 'cloud of rocks' - the ball collides with the rocks and transfers most of its kinetic energy to them. However it is very important to realise that electrical energy is not the kinetic energy of electrons (which is tiny by comparison), any more than the energy equivalent of a rest mass given by E=mc2 is the kinetic energy of a falling ball.

20. Feb 21, 2012

### sophiecentaur

But the Kinetic Energy analogy is not appropriate. There is so little, compared with the total energy transferred that it just doesn't count. It's like saying the kinetic energy of your bicycle chain is what gets energy from your feet to the wheel. If there is any mechanical analogy it's that of Work (force times distance) which corresponds to Charge times PD.

But I agree that, too many times, people seem to demand an over-familiar (and often metaphor-laden and circular) treatment of new concepts rather than accept and use them. After a very short while, with the latter approach, what was unfamiliar and difficult becomes familiar and reasonable. Remember, everything was unfamiliar once - we weren't born with any of this knowledge.