1. Oct 9, 2004

### Cyrus

Q: I am reading up on voltage and I want to make sure I am understanding this concept correctly. We say that there is a voltage between two points a and b. Then if I am to put a particle of any charge at point a, and multiply the charge by the voltage, I will obtain the potential energy at point a. Similarly, I will obtain the potential energy at point b. If I subtract the potential energy at point a minus the potential energy at point b, then I will know how much work it will take for me to move that particle of charge from point a to point b. Is this correct? The work in moving from a to b would be,

$$W_a_-_>_b = (V_a * q - V_b *q)$$, where q is any charge I pick.

2. Oct 9, 2004

### Cyrus

Now Is my second question, and Im kind of 'making it up as I go along here' I want to understand this voltage concept in terms of wires. Ok, so,.. Lets say I have a wire with a light bulb attached to it. Lets not worry about where the power is and making it a complete circuit. For now lets just say it gets its power somehow and we dont care how or why. Its my understanding that in order for the light bulb to light up, electricity, i.e. current, i.e electrons, must be moving inside the wire from point a, into the lightbulb, and out the other end at terminal b, and all the electrons must travel in the same direction. Now in order for this to happen, i am also asummue you must have some sort of voltage across each end of the two wires. As the left side, there must be a different voltage than the right side, otherwise the electrons wont "flow" through the lightbulb, or in general any electrical device. So one side has to have more voltage, or potential energy than the other side. Will the electrons want to go to the lowest state of potential energy? I guess this because a weight in a gravitational field will want to fall to the ground, where potential is lower. Will the electrons do the same? Secondly, if this wire has two different potentials, then there has to be more charge on one terminal, lets say terminal a, than there is at terminal b. Does this mean that the wire on the side of terminal a, has more electrons "backed up", and that at terminal b, there seems to be less electrons. In other words, at terminal a, would I find that there is more electrons per unit length of wire, than at terminal b, and if so, is this a requirement for there to be a voltage differential. Or is there the same amount of electrons on each side of the wire, and the voltage differential occurs due to some other reason?

*edit: for now, I guess all I can say correctly is that I can determine how much work it would take to move a charged body from one point in space to another point in space, in the presence of other charges. Thats what my first post tells me. In my second post, I think things will break down and my assumptions might be wrong because we are no longer dealing with going from point a to b in empty space.

*edit2:I also find it interesting that the potential energy does not depend on the mass of the body with a charge q. two particles of different mass and the same charge q have the same potential energy! I suppose you could justify this because the mass has no effect on the electric force. The only way mass would be a factor is if there is gravity to influence the particle in falling down, in which case the electric force would be insignificantly small. Vice versa, if the electric force plays a role, then the gravitational force will be infitesmal, so they wont both be a factor to worry about.

*edit3:I finished reading some more on my physics book. Looks like my assumption was wrong about the electron, it always wants to move to higher potential energy! Thats weard. You would think it would want to go to lower potential energy, but only a positively charged particle will move to lower potential energy!

Last edited: Oct 9, 2004
3. Oct 9, 2004

### Manchot

Electrons move to areas of high potential simply because of the way potential has been defined. It's defined in a way such that it is always relative to positively-charged particles. What's high potential for positive particles is low potential for electrons, and vice versa.

4. Oct 9, 2004

### cepheid

Staff Emeritus
Your first post is correct. The change in potential energy is the negative of the work done by the electrostatic force as the charge moves through the electric field. So if the electrostatic force does negative work (opposes the displacement), the particle's electrical potential energy will increase (i.e. Va < Vb from the formula). This is analogous to a mass being lifted against a gravitational field. If the electrostatic force does positive work, the charge will move in the direction of the electric field lines...it's potential energy will decrease...we say its destination has lower potential than the origin (Va > Vb). This is analogous to the gravitational force accelerating a mass downward..it gains kinetic energy and loses the corresponding gravitational potential energy. By convention, electric field lines point in the direction of decreasing potential for a positive test charge.

Correct. Note that they all move in the same direction continuously for direct current, and all in the same direction but back and forth rapidly for alternating current.

Mostly correct, but you are mixing up your terminology. A voltage is a difference in potential. In your second sentence, you cannot therefore speak of the voltage at a point. It should say that the electrons move because there is a difference in the potential at point a and the potential at point b. This leads to a potential difference. If the potentials were the same, then there would be no potential difference, i.e. no voltage, across the terminals. In your third sentence, you have also used voltage and potential energy interchangeably. Voltage is difference in potential. Potential is P.E. on a per unit charge basis. So if there is a potential difference between the two points, than any specific given charge's potential energy will change (increase or decrease) as it moves between those two points. Finally, note that when we speak of potential at a point, it is only valid relative to a certain reference at which the potential has been defined to be V = 0 (usually at sea level in a gravitational potential field). Finally, to answer your question: YES...electrons will move to a region of (what is for them) lower potential, undergoing a corresponding decrease in electric potential energy.

I understand what you are driving at, but it seems that you are interpreting a wire as a hollow tube that starts off with nothing in it, until electrons from some voltage source are introduced to it, and propagate along the wire, "spreading out" as it were. This is incorrect. Remember that a wire is a metal conductor...it has free electrons floating around all along its length. So if you apply a voltage to it, would not all electrons at all points in the wire begin to move simultaneously in the same direction? So, I don't think there would really be such a non-uniform charge density as you suggest. If so, where is this "concentration" of negative charge of which you speak? The answer: at the negative terminal (anode) of the battery. Electrons are indeed "bunched up" there, having been liberated from their parent atoms by a chemical (oxidation) reaction. This net charge between the positive and negative terminals is what creates the electric field that drives the electrons through the external circuit, from the negative terminal back to the positive.

I think we ironed out the wrinkles in your second post though eh? Good job though...always try to inquire/work things out the way you have been doing.

Now be specific! The electric potential energy of system is indeed independent of the masses of the charges involved. The gravitational potential energy however, is not. It just usually isn't relevant (see below)

I'm not sure what you're getting at here, but it seems a little messed up. Both gravity and electromagnetic forces have an influence on a particle. In classical physics, they are two of the four fundamental forces. However, the electromagnetic force is an intrinsically stronger force than gravity. Just take a look at the values of the two universal constants...isn't the Coulombic constant something like, TRILLIONS of times, (or more) greater than G? But, you might wish to play devil's advocate and ask why then, do electromagnetic forces not play a great role than gravity in our everday lives? Answer 1 is that electrostatic (coulombic) forces do play a significant role. They hold you and all other matter together, and are the basis of what you experience when you feel/interact with other objects. Most of the contact forces you study in mechanics (friction, tension, whatever) are fundamentally electrostatic in nature. But that's not what you were asking. You were asking why we don't see the super strong electric force superseding gravity and flinging people/objects high into the air? Because all matter we can see is made up of neutral atoms that maintain an exact balance between positive and negative charges (no NET charge). If this balance could be disrupted, the results might very well be catastrophic. But the energy required to unbalance the charge would be equally formidable in the first place.

As I said before, the electron DOES want to move from high potential (negative charge) to low potential (positive). Your textbook is just using a common convention in physics known as Conventional Current, which dates back to the time in which people actually thought (assumed) that the primary charge carriers in a conductor were positive charges. It doesn't matter. You can always just reverse the direction of the arrow shown if you want to see which way the Electron Flow current would be.

Phew!

5. Oct 10, 2004

### Cyrus

Thanks for your help! LOL did not mean to bombard you with so many questions! . I just have a few follow up questions though.

In my physics book, they did a sample problem where the calculated the potential at a point, and the units were V volts, and I confused volts with voltage, now I realize my error.

1.) Im having trouble seeing how the electron actually does go to a lower state of potential energy. I do see how it wants to go to higher potential energy based on how potential is defined, im just not seeing why it would go to lower potential energy.

2.) In terms of a battery, am I correct to say that this is where the electrons bunch up. If so, when I connect this battery to a simple circuit with a light bulb, will the bunched up electrons proceed to move as fast as they can throug the wire, and then start to "bunch" up at the other end of the battery? (my intuition would tell me that there is some kind of chemistry going on inside the battery so that as soon as the electrons bunch up on the other terminal, they will be chemically altered to go back to the starting terminal, recycling the process.) Would this recycling process end when something causes this reaction to stop. When it does stop, will there no longer be bunched electrons anywhere, but the same amount on both sides, so there is zero potential now.

6. Oct 10, 2004

### Dragon42

I will attempt to answer number 2 from above.

First, you need a little chemistry in order to understand a battery correctly. There are different kinds of batterries, but the basic idea is that a chemical reaction involving metals and acids is taking place. Because of the way these reactions occur, they reach an equilibrium. At equilibruim, you can think of the reactions as having stopped (that is not what actually happens, but it is the simple version). When you connect the terminals of the battery, you disturb the equilibrium. The electrons flow to the lowwer potential, and the reactions start up again trying to reach equilibrium. THERE IS NO RECYCLING OF THESES ELECTRONS. There is just sufficient matterials that this can go on for a prolonged period of time. This process keeps reoccuring :rofl: . As the matterials dwindle, the equilibrium point changes. (thus why you see your electronic gizmo lossing power) Eventually, the matterials run out and the battery is 'dead' :yuck: .

Rechargable batteries are close to the same. What happens is that you plug them in reverse. This reverses the potential to a point that pushes the reactions in reverse. Now you are storing energy back in the battery. This process is not complete. Each time you drain a rechargeable, some of the matterials are permanently changed into unusable forms.

I hope this explains the second well enough.

For the first question, you just need to use your intuition and Conservation of Energy.
Imagine a round stone on a piece of horizantal wood (a charged particle in free space). We now tilt the board (applying an electric field over the space). The stone rolls to the lowwest point possible, ie. it goes to the lowwest potential. (the charge follows the field to the lowwest potential energy for it). Now, this is always the same, you just need to see that the potential for a field may not directly translate sign wise to lowwest potential for the charge. It depends on the sign of the charge.

7. Oct 10, 2004

### Cyrus

I was wondering, is it possible to calculate the potential at a point without reference to another point. It seems so in some cases, like point charges, we can calculate the potential at any point in space by summing the potentials of all the point charges. In other cases, like infinite line charge, it does not seem possible to calculate the potential at a point in space, because the book calculated the potential at one point relative to another point. So it seems that in some cases you can and some cases you can not. Is that true?

I know that Voltage is always refrence to some point, but is potential also referenced to some point?

Last edited: Oct 10, 2004
8. Oct 10, 2004

### cepheid

Staff Emeritus
I think you're still not quite grasping the idea. This sentence^^^ is not quite right. Voltage is a difference in potential (between two points). It is not with reference to anything. But the potentials between those two points must each be stated with reference to some fixed point. An analogy: say you have two points on a mountain, one at a higher elevation than the other. The difference between the elevations of the two points is always the same no matter where your reference is (i.e. where you are measuring the heights from). But the stated value for the elevation at EACH point is meaningless unless it is defined with respect to some reference. In this case, the values given usually represent the elevations above sea level, where height = 0 by definition.

It may not be stated outright in your text, but the electric potential at a point is always given with reference to some other point infinitely far away from the source of the potential (where V can be safely assumed to be ZERO). That should alleviate some of the confusion caused when they started calculating "potential at a point".

For an infinite line charge, you cannot set your reference to V = 0 as r --> infinity because the electric field strength does not fall off as you recede from the source of charge. You must set your reference to something else (whatever you want). But in real life, this is never an issue.

9. Oct 11, 2004

### Cyrus

Thanks that helps alot! One thing im having terrible trouble grasping is that the electric field at the surface of a conductor is always perpendicular to the conductor. The book says that once equilibrium is reached the charge at the surface no longer moves. I know that it can no longer move outwards, but can it move around at the surface. If i have two charges at the surface 180 degrees appart from eachother, can't they move around and around while still being 180 appart?

10. Oct 11, 2004

### ehild

The macroscopic charge density does not change after equilibrium. But the charge carriers - electrons in the case of metal - do move on an irregular way so as the time average of their density corresponds to the observed distribution of charge on the surface.

ehild

11. Oct 11, 2004

### Cyrus

.

Im sorry but im not seeing what your saying. Wont the electon always go to higher potential based on how we defined potential. Isint this analagous to a brick that has negative mass, would fall up when leg go, rather than fall down! Thats neato.

an electron either gets repelled by a similar charge, also negative, and potential energy becomming less negative, or it is attracted to a positive charge, and potential energy increases as they get closer to one another. So in any situation, the electron moves to higher potential. Doesnt it?

Last edited: Oct 11, 2004
12. Oct 11, 2004

### Cyrus

any anwsers?

13. Oct 12, 2004

### cepheid

Staff Emeritus
This guy already explained it to you...

14. Oct 12, 2004

### cepheid

Staff Emeritus
Wrong man! Why would an electron's potential energy increase as it is drawn towards a postive charge? That's where it wants to go, it's going there spontaneously, you don't have to do any work on it to get it there, in fact the electric force does work on it, increasing its kinetic energy (with a corresponding decrease in potential energy...just like a rock falling!).

Yeah sure, so potential has been defined by convention so that areas of postive charge are called "high", but the high has been defined for a postive test charge. This same area is a LOW for an electron. Why is it so hard for you to believe that a direction of increasing potential (towards +) for a +ve charge will be a direction of decreasing potential (still towards +) for a negative charge?

15. Oct 12, 2004

### Cyrus

I think a problem migh be that potential is not the same as potential energy. Is it correct to say that the potential of an electron increases in the presence of any electric field, but the POTENTIAL ENERGY decreases?

Im trying to do an exmaple to see that indeed the potential does decrease as the electron interacts with any electric field. I know from the text that the potential will increase, and the potential is defined as,

$$V = \frac {q} {4pi*e_o*r}$$

I just picked the charge of an electron, -1.60 x 10-19, and I arbitrarily chose the charge of the body producing the electric field to be +5x10-9 C. Lets say the distance at point a is 1m, and the distance at point b is .5 meter.

so the potential due to the body at a distance of 1m is,

Va = +5x10-9 C *9.0x10+9 / 1m = 45V potential at point a

Vb= +5x10-9C *9.0x10+9/.5m = 90V potential at point b,

so far so good, the potential is higher when it moves in closer, but i still want to show that the potential ENERGY is lower,

U(potential energy) = q*V

Ua=q*Va = -1.60 x 10-19 * 45 = -7.2x10-18

Ub=q*Vb=-1.60 x 10-19 * 90 = -1.44x10-17

So then if potential energy is indeed lower, then the potential energy at point a must be greater than the potential energy at point b, Ua-Ub >0.

-7.2x10-18--1.44x10-17 = 7.2x10-18, a positive number! Looks like potential energy is in fact decreasing. Ah now i see silly me, 10-18 is a bigger number than 10-17 because of the negative sign of the number, that was throwing me off for a second.

To do a second test, if both charges are alike, then the potential energy should decrease as they go far appart,

Vb = -5x10-9 C *9.0x10+9 / 1m = -45V potential at point b

Va= -5x10-9C *9.0x10+9/.5m = -90V potential at point a

Ua=q*Vb = -1.60 x 10-19 * -45 = +7.2x10-18

Ub=q*Va=-1.60 x 10-19 * -90 = +1.44x10-17

but now Ub is bigger than Ua, and their difference is also positive, implying that the potential ENERGY is greater at the closer point than the further point. NOW IT MAKES SENSE! I was about to pully my hair out thank you, im sorry I did not see it sooner.

Last edited: Oct 12, 2004
16. Oct 13, 2004

### ehild

Your first sentence is correct and so is the second statement of the second sentence if you meant that an electron would move under the influence of an electric field in such a way that its electric potential energy decrease.

The concept of potential belongs to the field. The static electric field at any point is the negative gradient of a potential function. An electron does not have potential. It has potential energy in the field, and this potential energy is the product of its charge and the potential of the field.

ehild