Joules per Coulomb and the Volt

[nDever]

Hey,

A coulomb is the amount of charge that passes a point through a wire carrying one ampere for one second. Voltage is a measure of electrical potential energy in units of volts or joules per coulomb (energy/charge). Then 1 volt means 1 joule per coulomb; 2 volts mean 2 joules per coulomb, and 5 volts mean 5 joules per coulomb.

But what is meant by 2 or 5 volts? If one coulomb contains a set amount of electrons, how can one coulomb have more energy or potential to do work than another coulomb? What is different?

 

[cmb]

But what is meant by 2 or 5 volts? If one coulomb contains a set amount of electrons, how can one coulomb have more energy or potential to do work than another coulomb? What is different?

That's what voltage means. It's like asking why a sack of potatoes hits the floor harder if dropped from 10m than if it is dropped 1cm - the 'height' is the analogy of 'volts' here; the more volts (or height) the electrons (or potatoes) transition through, the greater the energy exchange is involved in them transitioning those volts (or height).

 

[nDever]

I understand the potato analogy. The higher the height, the greater the momentum. But why does a potential of 2 volts do more work than 1 volt? Is it the amount of the excess/deficiency of electrons that determines how much energy is expelled?

 

[cmb]

why does a potential of 2 volts do more work than 1 volt? Is it the amount of the excess/deficiency of electrons that determines how much energy is expelled?

To paraphrase your terms; the "energy [that is] expelled by the excess/deficiency of electrons determines the volts".

In effect; to determine the voltage, you perform a 'test' on some of the electrons to see how much energy they release when they undergo a transition to a known [or given/datum] potential. A bit like if you don't know the height you are at, so you pull a potato out of the bag, and time how long it takes to fall from your height to wherever you're dropping the potato to.

 

[nDever]

So then when we consider current and resistance, 2 V means that in a wire with 1 ohm of resistance, at a certain point in the wire, there will be 2 coulombs of charge passing that point every second.

 

[cmb]

Yes, that's right. Was there a question? Each coulomb that passes through a resistor with a 2V potential drop across it will release 2J into that resistor (because 2 V = 2J/C). Two Coulombs will mean 4J of energy is released, through a 2V potential drop.

In fact, it is, again, somewhat the other way around - the resistor, R, is such that it will cause J Joules to be released when C Coulombs pass through it, according to J=C^2 x R

 

[nDever]

I see, I think I understand now. So the volt is a standard? 1 V means the energy or work that that the charges have the potential to do through a wire of 1 ohm of resistance, in other words, the amount of energy used up through the resistance.

So when we buy a 9 V battery from the store, this means that the battery is somewhat "guaranteed" to do 9 joules of work through a resistance of 1 ohm and at a rate of 9 coulombs per second...?

 

[cmb]

Almost. There is a ^2 term in there. I think I'm over-stretching the analogy to explain it, but think of the potato and that the wind resistance is a ^2 term of the motion of the potato falling through the air.

The 9V battery denotes that One Coulomb of electrons passing from one terminal to the other can nominally perform 9J of work. Across a 1 ohm resistor there would be 9 Coulombs of charge per second, so that'd be 81 Joules per second.

 

[nDever]

That's where I get stuck. I get the calculation of power. With 9V, 9 coulombs/second with each coulomb being able to do 9 joules of work computes to 81 joules/sec (as you said). I don't get what being done to get the electrons to do more work per coulomb, to get more joules out of each coulomb.

 

[cmb]

I don't get what being done to get the electrons to do more work per coulomb

The value of volts specifies precisely, and linearly, how much work is done per Coulomb as the charge changes voltage, but you get to double-count the volts because it also determines not only how much work a Coulomb can do, but also how many Coulombs will flow across a linear (ohmic) load.

The issue is, in effect, the 'resistor' value that makes this look more complicated than you are seeing it to be. The resistance of an electrical element to the flow of electrons is analogous to the resistance of air to an object passing through it - resistance through air is a v^2 term. I'm trying to draw out this analogy (which I don't think is actually a particularly good one! :shy:) only because I can see you are wanting to conceptualise why the power delivered into a resistance is a function of volts^2. I'm not sure I have a better picture to paint for you than that. Maybe someone else will chime in with a better way to provide a conceptualisation of why this is, without simply pointing to the maths of the thing, but maybe you are better just to stick with the maths, and the reality of those equations:

R = V/I
Energy = V.C
Power = Energy/s = V.C/s = VI
so
Power = V^2/R = R.I^2

 

[Pengwuino]

That's where I get stuck. I get the calculation of power. With 9V, 9 coulombs/second with each coulomb being able to do 9 joules of work computes to 81 joules/sec (as you said). I don't get what being done to get the electrons to do more work per coulomb, to get more joules out of each coulomb.

The height analogy cited above is probably the best way to understand this. If an object has to fall through a larger height difference, the force of gravity will have to act on the object for a longer time, thus giving it more energy. The same can be said of electrical potentials. Suppose you have two charged objects, one positively charged and one negatively charged. Let's say that the positively charged object is fixed and the negatively charged object is allowed to fall in. For the negatively charged object to fall inwards, the positively charged object must pull on it via Coulomb's Law, thus doing work. If you wanted to have the charge fall even further, more work has to be done to bring the object closer. This need to do more work is quantified by potentials and voltage where the change in potentials tell you how much work per coulomb of charge is needed to do that work.

So if you want to relate the voltage on a battery to the charged particles falling inward, the voltage basically tells you, in analogy, how far the charged particle can be pulled inwards and the further you pull it in, the more work done per Coulomb of charge. A battery's voltage basically tells you how capable it is of pulling a current through a circuit that has some resistance.

 

[nDever]

Maybe all of my confusion will be eliminated if someone answered this. Quantitatively, does the positive terminal of a battery have the same tendency or force to receive electrons as the negative terminal who is trying to give them away? In other words, does the negative terminal have as many excess electrons as the positive has a deficiency?

 

[cmb]

does the negative terminal have as many excess electrons as the positive has a deficiency?

The answer to that ["it might do, but likely not"] is only likely to confuse you, and nor do I think it would do anything to address your original question.

'Net neutrality' (or otherwise) does not define 'voltage'. You are best to think of 'voltage' as explicitly the relative potentials between two conductors, i.e. as purely a conceptual quantity of charge, for the purposes of your question.

Back to the height analogy again - what is important when you fall off a wall is the height of the wall, not your height above or below sea level.

 

[nDever]

While thinking about it, I may have answered my own question.

The voltage between two objects is their measured difference in electric potential. So for example the 5V battery means that the difference in the electric potentials is 5V. The potentials themselves can be anything so long as their difference is 5V. The potential of the cathode could be 8V and the anode could be 3V but the difference is still 5V, so, hence the 5V battery.

The electric potential (in volts) is the ratio of the electric potential energy (in joules) and the charge (in coulombs).

Am I on the right track?

 

[Pengwuino]

While thinking about it, I may have answered my own question.
The voltage between two objects is their measured difference in electric potential. So for example the 5V battery means that the difference in the electric potentials is 5V. The potentials themselves can be anything so long as their difference is 5V. The potential of the cathode could be 8V and the anode could be 3V but the difference is still 5V, so, hence the 5V battery.

The electric potential (in volts) is the ratio of the electric potential energy (in joules) and the charge (in coulombs).

Am I on the right track?

Yes. Potentials are relative. In fact, the potential you see of a point charge as $V = {{kq}\over{r}}$ is with the (arbitrary) assumption that the V = 0 potential is at infinity. If I were to drop a rock from 1000m to 950m, it would gain the same energy as a rock dropped from 100m to 50m. The absolute heights are unimportant just as the absolute potentials are unimportant.

 

[nDever]

I think I'm beginning to see the picture. So, everything has a force due to electric charges just as everything with mass attracts everything else with mass due to gravity, and you can't really diminish the electric potential to zero. Is this correct as well?

 

[jim hardy]

Think of volts as pressure on the electrons,,, potential energy not kinetic.

I would be careful with the potato analogy because it leads one's mind naturally to imagine their velocity when they hit the floor.
Electrons in a wire have quite low velocity, but can do work by pushing against resistance to make heat, or against a magnetic field to make force as in a loudspeaker or electric motor.

It's natural to think of parallels with Bernoulli , but be careful. They're pretty near massless compared to real fluids.

old jim