What is Electric Potential Difference?

Username99

Hi, thanks for all the input regarding my issues with Gravitational Potential Difference. But no sooner has one issue been resolved and I have stumbled across another that I would greatly appreciate some help with. My question is, as the title suggests, What the heck is Electric Potential Difference?

1 Volt = 1 Joule / 1 Coulomb.

So a single volt is the amount of joules * per coulomb of charge.
* emmited, transferred, ...... ???

I am unsure of what to write where the * is above. I do not know what 'happens' the joules of energy.

1 electron has a charge of (1.602 x 10^-19).
There are (6.242 x 10^18) electrons in 1 coulomb of charge.

If another way for describing 1 Volt = Work / Charge, and Work = (Force x Distance), then 1 Volt = (Force x Distance) / Charge.

If I am correct so far can you let me know that I am correct up to the confused face please.

Now here is where I really get lost: If I had a circuit consisting of a battery 2 resistors in series, a conductor completing the circuit and a switch for making and breaking the circuit.

As soon as I press the switch chemical action in the battery causes a seperation of charge. Negative charge at the negative terminal and positive charge at the positive terminal. The free electrons in the valance band of each atom of the conductor break free and drift from atom to atom towards the positive terminal. Opposite charges attract.
As the charge of one electron is too insignificant to talk about I shall follow the path of 1 coulomb as it travels from the negative terminal to the positive
terminal.
So this coulomb leaves the negative terminal due to the force of attraction between it and the electric field of the positive terminal. The voltage is at its highest point between the positive terminal and the first resistor it drops back across each resistor until I reaches 0 Volts between the negative terminal of the battery and the last resistor.

Voltage = Work / Charge, and work is joules.

What causes the work/Joules/energy to rise. My question is why would there be more joules per coulomb in one circuit than another?

Sorry I went around in a roundabout there. Any input at all is greatly appreciated thanks.

riffmongous

You need to think a bit more about what the units stand for, electric potential difference describess the difference in electrical potential energy between two points, so say you had a point with an electrical potential of A, then the electrical potential difference between A and another point B is 1 Volt when you move a charge of 1 Coloumb from A to B and its electrical potential energy changes by 1 Joule.

Username99

I have attached a simple circuit diagram to help me explain my problem a little better.

riffmongous said that 'electric potential difference describes the difference in electrical potential energy between two points'.

From the circuit schematic I have attached to this e-mail and taking the above statement into account, Am I right in saying that the difference in Electric Potential between the negative terminal and the positive terminal is 10V (10 Joules of energy per coulomb of charge)?

So when a charge of 1 coulomb moves from the negative terminal to the positive terminal the electrical potential energy has changed by 10 joules.

But I still can not see what relevance this has to anything. The current is constant throughout the circuit, 1.67A, as it is a series circuit.

If I was shrunk down to the size of a single electron, and if we think of a coulomb as a bucket full of charge. If I hopped into the bucket with all the rest of the electrons at the negative terminal and started moving towards the positive terminal. As I pass from through point 'C' then 'B' then 'A' and finally into the positive terminal, what do I feel at the different points. Another way of looking at it could be what influence does being at lets say point 'B' have on me as opposed to at point 'A'?

I do not move any faster, because the current is constant.

Lbeard

Username99 wrote: »

If I was shrunk down to the size of a single electron, and if we think of a coulomb as a bucket full of charge. If I hopped into the bucket with all the rest of the electrons at the negative terminal and started moving towards the positive terminal. .

Think of it a different way. In the case of a charged plate. Imagine if the electrons were all people. They're very tightly packed together in a room. It's uncomfortable, they're all pressing against it other - that's the negative terminal. There is another room - the positive terminal - that is completely empty. But the door is not open. There is a difference between that room and the negative terminal room. Someone opens the door. People shifted from the negative room to the positive room. until there are the same number of people in both rooms, and in the corridor there is the same density of people - there is no difference anywhere. There is no voltage.

If there is a constant stream of electrons filling up the first room - people realising it's getting crowded will move to the positive terminal (that terminal could be an open door, as in the case of it being ground). The people may have to do a little work to reach the positive terminal - turn a wheel, pull a speaker cone, or push it. But that takes less energy than it would take to remain squashed in the room. In the case of an open door, the electrons will do work until no more electrons are being forced into the circuit. If that happens, anyone (electron) still in the circuit will stop what they're doing - they don't have to work to get out of the crowd.

riffmongous

I hope this is correct since I dont really work with electronics anymore..

but lets look at the circuit math. So the current is 1.67A (V= IR = I*(4+2)).

Next at each resistor there will be a voltage drop equal to V=IR
for R1 V=1.67*4= 6.67V
for R2 V=1.67*2= 3.34V

The power dissipated by each resistor is P=V*I
for R1 P=6.67*1.67=11.14W
for R2 P=3.34*1.67=5.57W

The total power dissipated by the circuit is then 16.71 for 1.67A of current, so for 1 amp of current it would be 10.01W
Watts are joules per second, so each second we send 1.67A of current through the circuit we use 10.01*1.67 joules of energy. The electrical potential energy of the charges is converted to different kinds of energy and lost by the resistors as heat.

I sincerely hope the reasoning is all correct, I havent done this in years!

Username99

Lbeard wrote: »

Think of it a different way. In the case of a charged plate. Imagine if the electrons were all people. They're very tightly packed together in a room. It's uncomfortable, they're all pressing against it other - that's the negative terminal. There is another room - the positive terminal - that is completely empty. But the door is not open. There is a difference between that room and the negative terminal room. Someone opens the door. People shifted from the negative room to the positive room. until there are the same number of people in both rooms, and in the corridor there is the same density of people - there is no difference anywhere. There is no voltage.

If there is a constant stream of electrons filling up the first room - people realising it's getting crowded will move to the positive terminal (that terminal could be an open door, as in the case of it being ground). The people may have to do a little work to reach the positive terminal - turn a wheel, pull a speaker cone, or push it. But that takes less energy than it would take to remain squashed in the room. In the case of an open door, the electrons will do work until no more electrons are being forced into the circuit. If that happens, anyone (electron) still in the circuit will stop what they're doing - they don't have to work to get out of the crowd.

Thanks Lbeard, I like this example. So the voltage is like the desire of the people (electrons) to move from the negative terminal to the positive terminal. The more packed in the people (electrons) are at the negative terminal, the greater their desire to get to the positive terminal. And they have to do work on the way through the corridor connecting the two rooms.

To go back to the maths for a second, In the circuit schematic attached to my last post. There is an accumulation of electrons (net negative charge) at the negative terminal and a net positive charge at the positive terminal. The electric field set up due to the positive charge interacts with the free electrons 'sucking' them towards the positive terminal.

So the 'suction' at one end and the discomfort of the electrons at the other end is basically the function of the battery. And even though the electrons have to do work to get from one terminal to the other there would be a greater amount of work involved in overcoming the forces of repulsion at the negative terminal.

This whole idea of WORK is catching me out though, it was fine for gravitational potential energy because: Work = (Force)(Distance). The work done in bringing a mass a given height relative to the ground is equal to the change in kenitic energy.

But in this case, distance does not seems to come into play, if opposite charges attract, the electrons are attracted to the positive terminal. If we negate the resistance in the conductor and if there were no resistors present there would be no voltage, therefore no work would need to be done moving electrons from negative to positive?

But in our case there is no potential difference between the negative terminal and R2, but there is a potential difference between point 'B' and point 'C', therefore the work done per unit charge to get through the resistor is 3.33J?

I'll leave it there for now or this post will be too long. Thanks for all the help so far, I really do appreciate it.

riffmongous

Username99 wrote: »

This whole idea of WORK is catching me out though, it was fine for gravitational potential energy because: Work = (Force)(Distance). The work done in bringing a mass a given height relative to the ground is equal to the change in kenitic energy.

But in this case, distance does not seems to come into play, if opposite charges attract, the electrons are attracted to the positive terminal. If we negate the resistance in the conductor and if there were no resistors present there would be no voltage, therefore no work would need to be done moving electrons from negative to positive?

What you describe there though is an unconnected battery In that case you can't neglect the resistance in the conductor I think

And the concept of work is still the same using distance and forces, just an electric force and charge instead of a gravitational force and mass, its just not very convienient to talk about work like that for electrics. One flaw with your analogy is that the current doesnt move from terminal to terminal like a water flow you just turned on, you can actually do the calculation but the speed of an electron moving through a wire is actually very small, what happens is that all the electrons in the circuit start to move (iirc), if the electrons were coming from the battery terminal you could be waiting a while for them to show up at the end terminal, which isnt the case. The electric field however move at the speed of light and that causes the movement of the electrons in the circuit (iirc)

Username99

riffmongous wrote: »

What you describe there though is an unconnected battery In that case you can't neglect the resistance in the conductor I think

Ha, sorry I ment to say short circuit the resistors.

riffmongous

Well with a short circuit you still have to have a conductor of really small resistance joining the terminals, V=IR so a small resistance means large current, large current means large power needs to be dissipated and your conductor will probably melt and your battery explode

I also added in an edit to my previous post to try to explain whats going on a bit better

Username99

riffmongous wrote: »

Well with a short circuit you still have to have a conductor of really small resistance joining the terminals, V=IR so a small resistance means large current, large current means large power needs to be dissipated and your conductor will probably melt and your battery explode

Oh I am aware of this, I was not proposing the application of the situation rather the theory behind it. And I was neglecting the resistance of the conductor as this was a hypothetical situation. I feel like my head is going to explode with this voltage problem to be honest:D

riffmongous

Username99 wrote: »

Oh I am aware of this, I was not proposing the application of the situation rather the theory behind it. And I was neglecting the resistance of the conductor as this was a hypothetical situation. I feel like my head is going to explode with this voltage problem to be honest:D

Yeah I know its just a hypothetical, but if you neglect the resistance well you get an infinite current then, not 0 voltage I guess.

This page here sums up a lot of what we were discussing I think
http://www.physicsclassroom.com/class/circuits/u9l1c.cfm

Lbeard

Username99 wrote: »

So the 'suction' at one end and the discomfort of the electrons at the other end is basically the function of the battery.

No, there is no suction. Electrons all have the same charge, like charges repel, the electrons are trying to get away from each other. It takes energy to force them together - when they move apart they release that energy.

This whole idea of WORK is catching me out though, it was fine for gravitational potential energy because: Work = (Force)(Distance). The work done in bringing a mass a given height relative to the ground is equal to the change in kenitic energy.

You can think of work done in an electric circuit in kinetic terms. As the electrons move through the circuit they bang off atoms. the atoms move, they jiggle. This movement results in light and heat. If they move through a circuit and there is an electric motor, they have to mechanical turn the motor wheel to get through that bit of the circuit. When they're trapped in a confined space, they don't do any work because they can't - but they have the potential to do work.

Username99

Lbeard wrote: »

No, there is no suction. Electrons all have the same charge, like charges repel, the electrons are trying to get away from each other. It takes energy to force them together - when they move apart they release that energy.




You can think of work done in an electric circuit in kinetic terms. As the electrons move through the circuit they bang off atoms. the atoms move, they jiggle. This movement results in light and heat. If they move through a circuit and there is an electric motor, they have to mechanical turn the motor wheel to get through that bit of the circuit. When they're trapped in a confined space, they don't do any work because they can't - but they have the potential to do work.

Alright so I have thought about it some more. So through chemical action inside the battery a separation of charge occurs, positive charge at the positive terminal and negative charge at the negative terminal. The bigger the separation the bigger the 'pressure' inside the battery.

So if you have two circuits consisting of exactly the same elements the only difference being that circuit 'A' has a 5V battery and circuit 'B' has a 10V battery, the pressure inside of circuit 'B' is twice that of 'A'.

This means that the electrons in circuit 'B' will do more work to get to the positive terminal as opposed to the electrons in circuit 'A' who will do less work. And WORK is the transfer of energy from one form to another, so the amount of WORK that will be done is the amount of energy that will be transferred from kinetic energy into other forms such as light and heat when a coulomb moves from positive to negative.

We take advantage of this fact by placing elements such as light bulbs in the circuit and when the electrons pass through, if more work is being done in circuit 'B' then the emission of light will be greater.

Lbeard

Username99 wrote: »

So if you have two circuits consisting of exactly the same elements the only difference being that circuit 'A' has a 5V battery and circuit 'B' has a 10V battery, the pressure inside of circuit 'B' is twice that of 'A'.
.

Yes you could say pressure - the term current comes from the fluid analogy. Current as you'd have in a sea.

You should also think what pressure is. If you have two tanks - say glass jars - connected by a pipe, but the pipe is closed. In one tank you have a pressure of 10 atmospheres, and the other is just 1 atmosphere, the same as the room. There is a difference between each tank - since the pipe isn't open, it's a Potential Difference. Open the pipe a little - and there will be an air current.