# F = mass * gravity

1. Jun 23, 2006

### meldave00

Hi I'm back. Sorry for all the threads. I hope I'm not disrupting the normal thread protocol of this forum. However, there are some things that I really been wanting to understand for some time now.

I was talking about capacitors in series, but this thread sort of ties into it but does deserve its own thread.

I read and hear the same analogy over and over again. Voltage is potential energy analagous to rolling a boulder up a hill. There is work required to roll a boulder up a hill. So as you move a boulder up a hill it gains potential energy. This is exactly what happens in an electric field. As a positive point charge is brought closer and closer toward another positive point charge there is work required to move it closer to another point charge. It gain potential energy.

However, there is a big difference between the boulder on the hill and the positive point charge in the electric field. The boulder's potential energy may increase, but the force remains constant no matter how high you bring it up the hill. F = MASS * GRAVITY (ACCELERATION).
For the positive point charge, the potential energy increases as you bring it closer to another positive point charge, but also the force exerted on the positive point charge increases exponentially as you bring it closer to another positive points charge. F = (q0*q)/(r*r). Where r = distance between the charges. Notice that its squared. So I don't think a hill analogy is adequate in describing voltage.

I believe that voltage is like water pressure in pipes. That is, as more water is packed into the same pipe volume then the water pressure goes up. I believe that as more charge is packed into a conductor, that more force is exerted on each individual charge and hence the whole conductor has more charge pressure. This pressure is what pushes charge thru a circuit.

I am looking for an open ended discussion on this matter. Please let me know if my thoughts are correct. As I am assuming this to be the case and don't know for sure.

2. Jun 24, 2006

### capnahab

These are good questions of someone trying to understand electrical also known as the FM principal. To use the water pipe analogy:
Think of electrons replacing the water molecules in the pipe.
The amount of electrons is called Coulombs (bad spelling I think).
The pressure is Volts.
The amount of electrons (one coulomb) that pass a given point in one second is one amp.
The resistance is the length of the pipe and the diameter that is slowing down everything.

For capacators and inductors remember "Eli the ice man". E (volts) leads current (i) in an inductor (l) and current (i) leads voltage (e) in a capacator (c). Hows that for a start?

By the way the M in FM principal means magic and I'll let you figure out what F means.

3. Jun 24, 2006

### Danger

I've never heard of the boulder analogy; the water one is the way that I was taught to envision it.

4. Jun 24, 2006

### Staff: Mentor

The water pressure analogy still won't give you what you are looking for, Dave. Water is an incompressible fluid, and electron density can vary. Probably a better analogy to add into your mental picture is isolated masses on springs. You mentioned how the force between two positive charges increases as you push them together (and they gain potential energy) -- you might use the spring analogy to help you feel better about that picture.

I'd also suggest that you try to build a few kit projects on your own as time allows (are you out of school for the summer yet?). You are asking good questions as you learn the fundamentals of EE, but it will help you a lot to get some practical experience with real electronic components. I'd also suggest that you get a copy of "The Art of Electronics" by Horowitz and Hill. Read it cover to cover over the summer, and you will be in great shape for the fall term.

Some kits like these, for example: http://www.transeltech.com/kits/kits1.html

5. Jun 25, 2006

### meldave00

Voltage

Berkeman and company,

Sorry I don't have time to mess around with Kits. My wife would not understand why I am not paying attention to my kids when I am sittin in the garage geekin out with kits. LOL :)

Anyway, the reason I am debating the hill analogy is because someone tried to explain to me that voltage is all about potential energy. The higher up the hill you go, the higher the potential energy an object has. I believe that this analogy is incorrect as most of you out there tend to agree. I think the water pressure analogy is the way to go. I would agree that the isolated charges with outwardly attached springs is the best thought process for the thought on voltage. I just needed consensus and if anybody out there agrees, still send more consensus my way because I still am not fully confident on this yet.

The reason I ask this is because I am interested in how capacitors work. In order to understand how capacitors work. I need to understand voltage and charge. As I think about voltage more and more there seems to be two different types: The individual pressure of electrons within a conductor and the collective pressure that electrons bestow upon electrons on the other plate of a capacitor. Respectively, conductors deal with individual charges on another and capacitor plates deals with a collection of charges that pushes electrons away from another plate. Does anyone understand where I'm coming from on this? If not, I can explain more.

Even if you don't understand what I am saying. Please reply with your thoughts. I hope to explain my thoughts with more clarity with further discussion.

regards,

David

6. Jun 26, 2006

### capnahab

7. Jun 26, 2006

### meldave00

Voltage

Nrged and whomever may be interested,

I acknowledge your statement that the slope of the hill may change. But you said that the hill's slope may change steeper or less steep and go flat. However, I think that the slope of the hill would have to increase exponentially in order to be a good analogy to electric charge. For two point charges F= (qo*q)/(r*r) where r is the distance between point charges. That is, the force increases exponentially as two point charges get closer and closer to each other. (That is the hill get steeper and steeper as the boulder goes up the hill).

This statement could be false: "Since Force increases exponentially as two point charges get closer than the E field intensity increase exponentially. Then the votlage increases exponentially. This is all the case because of the r square term in the denominator.

I also agree that potential difference is relative to where you are comparing it too. One node could be at 1 million volts and another node could be at 1 million + 5V. The difference between the two nodes is 5V.

What I think is happening is that electrons (charges) at higher potential are experiencing less charge to charge force than electrons (charge) at a lower potential. When I say this, I am talking relatively and not absolutely. Another way to look at it is that electrons that are at 1 Million volts are going want to migrate toward a node that is at 1 million and 5Volts. (And it would if it had some sort of path to do so.) This is because the electrons at the lower potential are experience more charge to charge pressure than the node at higher potential.

Let me know what you think.

Again, I am not completely sure of this. I still need to refer to what you said more and try to understand. I'm just wondering what I am saying here makes any sense to you?

8. Jun 28, 2006

### nrqed

Let me start by including my previous post on this subject

First, the word "exponentially" in physics and maths has a specific meaning: a quantity must follow an exponential! This is not the case here. we say that the force follows a *power law* (more specifically an inverse square relation). I know that in every day usage the expression "increasing exponentially" is used in all sorts of contexts but in science it has a very specific meaning.

Anyway, yes, the force goes like 1/r^2 so the potential goes like 1/r for a point charge. That does not invalidate the "hill" analogy at all.

Have you ever seen those pictures of the potential of a point (positive or negative) point charge that looks a bit like those embedding diagrams of a black hole (for a negative charge) or an inverted black hole (for a positive charge). One can really imagine placing a test charge somewhere on those embedding diagrams and imagine it "rolling" downhill.

But there is a contradiction in your two sentences. Your first sentence *does* talk absolutely!! Since you say that a charge AT a higher potential is experiencing less force than a charge AT a lower potential!! My point is that you CANNOT say this! Just knowing the potential at which the two charges are does not tell you *anything* about the force they experience. You have to specify the potential at the position of the charge AND the potential at all points infinitesimally close to the charge.

I am not sure what "charge to charge pressure" means. This is a nonstandard expression. The only relevant thing is the electric force felt by the charge at each point. And that depends on the derivative (actually, the gradient, in 3 dimensions) of the electric potential at that point. The electron will feel an electric force on the direction in which the electric potential goes up the fastest and will move in that direction unless other forces are preventing it.
I hope what I wrote makes sense.

Patrick

9. Jun 28, 2006

### meldave00

Voltage

Patrick and whom ever may be interested,

O.K. Let me put it another way. Imagine this mechanical circuit.

"Output of Air pump 1 > copper pipe1 > closed stop valve > copper pipe2> back into input of Air pump 1"

The system is air tight and does not let air escape. I turn the air pump on and it pushes additional air into copper pipe1 and pulls a little bit of air out of copper pipe 2. The air pressure in copper pipe1 increase because the stop valve between copper pipe 1 and copper pipe 2 is closed and does not let any additional air to flow into copper pipe 2. Now the air pump will eventually stop pushing air into copper pipe1 because it isn't strong enough to keep on pushing air into copper pipe 1 and pulling air out of copper pipe 2, but it was able to push a little more air into copper pipe 1 and pull a little bit of air out of copper pipe 2. This little bit of more air in copper pipe 1 causes the air pressure in copper pipe 1 to be greater than copper pipe 2.

Now if I open up the stop valve a little bit so that air can leak into copper pipe 2. This release in air pressure of copper pipe 1 and increase in air pressure of copper 2 allows the air pump to begin pushing air into copper pipe 1 and pulling air from copper pipe 2 again. The air pump still causes air pressure in copper pipe 1 to be greater than the air pressure in copper pipe 2. The air is now flowing thru this mechanical circuit at a rate that the slightly opened stop valve will allow.

Relation:
Air pump > Voltage Source
stop valve > resistor
copper pipe 1 > ground
copper pipe 2 > Node equal to Voltage Source Value
air > electrons

What I'm saying is that the air is pushed thru the stop valve because the air pressure in copper pipe 1 is greater than the air pressure in copper pipe 2.

Hence, electrons at ground migrate to higher voltage because there is more electron to electron force at ground than there is at the voltage source node and the electrons want to migrate to the votlage source because there is less electron to electron force or pressure at the node at the voltage value.

Let me know what you think and if this makes sense. Everyone's thoughts are welcome.

Last edited: Jun 29, 2006
10. Jul 7, 2006

### meldave00

Voltage

Any feedback on my last entry will be much appreciated. Anyone?

11. Jul 7, 2006

### Valhalla

I don't know if this helps but that is a completely false statement. Examine the actual form of gravity and you will find that it looks very similar to an electric field. Force due to gravity is actually:

$$F_g = \frac {G*m_1*m_2}{r^2}$$

where G is the gravitational constant and r is the distance from the center of the first object to the center of the second object. So technically as you push the boulder up the hill it follows the similar inverse square law found in the electric field.

If we examine the potential energy due to gravity, with some integration we end up with the following formula.

$$U_g = \frac {-G*m_1*m_2}{r}$$

So the potential energy due to gravity also varies inversly.

Now lets look at the force due to an electric field from a point charge:

$$F_E = \frac{k*q_1*q_2}{r^2}$$

This looks almost exactly the same! They both follow an inverse square law relationship!

If we examine voltage (electric potential) we find that:

$$V = \frac {k*q}{r}$$
$$V = \frac {J}{q}$$
(note: voltage is joules divided by columbs)
It behaves mathematically equivalent with the exception of a constant!

I don't know if this helps or not. I only used the water through a pipe example when I am discussing current or the flow of charge. When I studied the electric field I related it to gravity because gravity acts very similar on a larger scale (Notable exception: electric fields have sinks and sources, gravitional fields only have sinks..unless im missing something please correct if I am!)

So I guess my point is the boulder example could work, if you pushed the boulder from the Earth's surface to the moon. It just takes a much larger scale.

Last edited: Jul 7, 2006
12. Jul 7, 2006

### meldave00

Voltage

Thanks for your response. I do now believe that voltage potential energy is analogous to the boulder up a hill. There is one caveat that I have. Imagine a large electronic board with a complex circuit on it. Node A is 2 ft away from Node B and NodeA and NodeB are 2ft away from ground. Both are referenced to the same ground. Node A = 5V and Node B = 3V. Since they are 2ft from each other and ground the Efield between them is neglible. My guess is that the reason one node A = 5V and nodeB = 3V that means that there less electron to electron force at NodeA than at Node B. That is Node A has less electron pressure then Node B. Let me know if this make sense. The boulder analogy seem to work for a point charge to another point charge. But is does not explain circuit nodes that are far away from each other and their common ground. Please let me know if the electron pressure is a good way of thinking about it in this case.

regards,

David

13. Jul 7, 2006

### meldave00

Voltage

Correction. That is Node A has less electron to electron force than Node B with respect to Ground's electron to electron force. When I say electron to electron force, I mean the force between all the electrons at a given node. Sort of like pressure.