# Trouble understanding voltage

1. Sep 9, 2005

### leright

Sorry....I originally posted this in the EE forum, but I realized that for this particular question, I'd want a physicist to explain it, instead of an EE so I moved it to this forum. Sorry for the trouble. A mod can delete the other thread in the EE forum if they'd like.

Ok, I am an EE major that is confused as Hell with the concept of voltage, and I need some assistance, since understanding voltage is extremely important to my major. I am going to explain my understanding of many many fundamental concepts not because I don't think you know what they are, but because I feel it is essential to explaining voltage overall, and a misconception of one of the fundamentals could be throwing me off somewhere and causing my confusion.

In order to truly understand what voltage is physically, one must understand charge, force (specifically, coulomb force), electric field, and energy (specifically, potential energy) and each of these will be explained evne though you are already experts on the subject matter, so bear with me.

First of all, charge is a fundamental physical quantity that cannot be created nor destroyed and accounts for the electrostatic interaction between objects. Charge comes in two types: positive and negative. Two types of charge are needed to account for the TWO types of interaction: attraction and repulsion. Electrons have negative charge and protons have positive charge, so a transfer of electrons between two objects accounts for charge imbalances between protons and electrons, causing opposite charges. Excess of electrons in an objects means negative charge and excess of protons in an object means positive charge. This is very basic but I'd like to put this into my own words.

Second, force needs to be understood. force is determined to be mass*acceleration, which means that a NET force is only applied if a mass is being accelerated. If there is a constant velocity or no motion at all, when a force is applied, then there must be another force cancelling it, such as friction, or a wall exerting a force back. Electrostatic force is a mutual force between two charged objects. This force can be attractive or repulsive. Coulomb's law is a mathematical model for electrostatic force, which is F = [k(Q1)(Q2)]/R^2, where Q1 and A2 are the two charges involved and R is the distance between the two charges.

Electric field is a medium that is used as a model to explain action at a distance in electrostatic interaction. It flows from positive charge to negative charge. Mathematically, electric field is the force a charge could exert on a POSITIVE test charge per unit charge of the test charge (yes, poor wording here). Since the test charge is always positive, the sign of the e-field producing charge determines the direction of the e-field, and thus the direction of the force. Thus, a positive charge produces a positive e-field and a positive force (repulsive) on a positive test charge, and a negative charge produces a negative e-field and a negative force (attractive) on a positive test charge. Basically, repulsion is a positive electrostatic force and attraction is a negative electrostatic force. *whew*

Energy is defined as the capacity to do work. Work is simply a force being exerted through a distance (word = force*distance). The unit for energy is the joule. There are two types of energy: potential energy and kinetic energy. An object has potential energy if a force CAN be exerted through a distance if the force creating an equilibrium environment is taken away. Kinetic energy is another type of energy, and an object with kinetic energy can also do work, or exert a force through a distance. KE = (0.5)mv^2. A mass moving at a constant velocity (0 force on the object moving) has the capacity to exert a force on another mass through a certain distance. The potential energy lost by an object is equal to the kinetic energy gained by an object.

Now, voltage can finally be explained. Voltage is basically the work per unit charge that a given charge can do between a certain distance if it is placed in an electric field and has the freedom to move. This is easy to understand. Since electric-field is simply the force per unit charge we sinple need to multiply by the distance. However, since electric-field varies inversely as the square of the distance, we cannot simply multiply, and we must integrate. So, voltage is minus the integral of E-field times dR. OK, simple. Now, say we would like to determine the voltage across the terminals of a battery. Since the terminals are opposite charges, the e-field is going to be negative, or -kQ/R^2. Integrating e-field with respect to R gives (kQ)/R + C. Since we must multiply by a -1 we get (-kQ)/R + C. Integrating between the two opposite charges (between 0 and R, since the distance between the negative charge and itself is 0, in a sense, and the distance between the negative charge and the positive charge is R) we get [(-kQ)/R] - [(-kQ)/0] = [(-kQ)/R] + [(kQ)/0]. Obviously, this is a problem, since we get a result of infinity. What am I not understanding about the voltage of a battery? In summation, what is the voltage between two charges, such as a battery. I am clearly thinking of the entire concept of the measurement of voltage incorrectly since the integral of the e-field vs. distance graph between a distance of 0 and a distance of R is infinity, since e-field asymptotically approaches x = 0 so the integral is infinty. I think the concept of a GROUND (arbitrary reference point assumed to be a voltage of zero) has to fit in there somewhere but I don't understand where. HELP ME!!!

I know this is a mouthful, but I NEED some help. I think I would be ok if I just seeked a superficial understanding of voltage across two charges, but I am not that type of person.

I think this confusion is causing me to really be confused with circuits in general and it is making me sad.

edit: Also, can someone explain the meaning of the sign on a voltage? I assume that the sign is negative simply because the e-field is negative when the charge emitting the e-field is negative, which would in turn translate the e-field vs. distance graph about the x-axis, causing a negative voltage, correct? So, negative e-field emitting charge and a positive test charge means a negative voltage, and a positive e-field emitting charge and a positive test charge means a positive voltage. Thus, attractive force means negative voltage and repulsive force means positive voltage, right?

THANKS A TON FOR YOUR HELP!! My first foray into Electrical Engineering has not been a pleasant one.

Last edited: Sep 9, 2005
2. Sep 9, 2005

### teclo

potential is energy per unit charge. it's the integral of the energy field over the distance you've chosen, regardless of the path.

think of a positive charge. if you let is go, it will start to move in the direction of lower potential - the direction of the energy field. if you push it against the energy field you are moving in the direction of higher potential.

it's easy to get potential energy from the voltage.

$$\Delta U_e_l = q\Delta V$$

an increase in $$q\Delta V$$ is associated with a decrease in kinetic energy, where a decrease is associated with an increase in kinetic energy.

does this help?

3. Sep 10, 2005

### teclo

also -- from what i learned, it would be best to think of an object with a positive charge as something that has lost electrons. if you start messing around with protons you're getting all up in the strong nuclear force. it's kind of tough to rip a proton out of an atom, but it's relatively easy to get electrons out.

also, as far as the negative sign on potential -- recall:

$$\Delta V = -\int\vec{E}\bullet d\vec{l}$$

the sign just determines whether a particle will gain or lose energy, depending on what way it goes.

when dealing with this stuff it helps to consider what's going on at the subatomic level. for instance, when electricity is flowing, it's really just electrons flowing through the conductor, following the path determined by the sign of the potential (from negative to positive - a path towards a region of higher potential).

again, i hope this sheds some light on your quandry

4. Sep 10, 2005

### leright

I actually already understood that and explained that in the post above, but it might not have been clear since I don't know how to use latex. I am having trouble understanding HOW to get voltage between a point in space and a point charge when the integral of e-field with respect to distance is infinity, between 0 and R, where R is the distance between the point charge and the point in space. I think something in my math is just incorrect. I get infinity for the integral because as the e-field function approaches x=0 e-field approaches infinity...therefore the integral between 0 and R is infinity...

edit: I can get it to work out right if integrate between like 0.000001 and R instead of 0 and R. That way I don't get a voltage of infinity.

Last edited: Sep 10, 2005
5. Sep 10, 2005

### leright

I honestly already understood all of that, as I explained it in the above post. You may not have caught this, since the post is so insanely long, but I explained this. It might be easier to understand my explaination if I knew how to use latex.
:tongue2:

6. Sep 10, 2005

### leright

Ok, let me restate my problem in a more simple way. I am trying to determine the voltage at all points due to a positive point charge, for instance. so I am going to integrate -integral(E*dR). Integration is needed to find the total work a positive test charge can do per unit charge because e-field varies inversely with the square of R, so I cannot simply multiply E by R. However, e-field varies inversely with the square of distance, so its integral from 0 (distance from reference positive charge) to R (distance to positive test charge) is INFINITY. What am I doing wrong???? HELP ME PLEASE!!

I've looked for some help on the internet, but every website simply MULTIPLIES e-field by distance to get voltage, but obvious you cannot do that, since e-field varies inversely with the square of the distance and you must integrate.

Last edited: Sep 10, 2005
7. Sep 10, 2005

### teclo

i think i understand your problem. if you want to find the potential at a certain point in relation to the positive charge. if you set your lower bound to the location of the particle, it's going to take an infinite amount of energy to move the other positive charge. if it's a negative charge, the kinetic energy would approach infinity. i suspect that quantum mechanics would need to be invoked to explain something along those lines. i'm only a sophomore, just starting electrodynamics - so i could be off on something.

what you can do, is say what is the potential at some point around this source charge. in other words, what would it take to move a particle from a distance so far away that the efield for the other charge is zero to a certain point nearby. unless you want a vector answer, it's only the magnitude of the distance that matters. let's consider the potential at a point around the source charge, at some distance r away.

$$\Delta V = -\int_i^f\vec{E}\bullet d\vec{l}$$

$$\Delta V = -\int_\infty^r\frac {1} {4\pi E_0}\frac {q} {r^2}$$

$$\Delta V = \frac {q} {4\pi E_0}\int_r^\infty\frac {1} {r^2}$$

$$\Delta V = \frac {q} {4\pi E_0}\left[ -\frac{1}{r}\right]_{r}^{\infty}$$

$$\Delta V = \frac {q} {4\pi E_0}\left[ -\frac{1}{\infty} - -\frac{1}{r} \right]$$

leaving us with the formula for the potential due to a point charge

$$\Delta V = \frac {1} {4\pi E_0} \frac {q} {r}$$

i hope maybe this helps?

oh and as far as latex goes, i still dont' know it, but i'm getting there. you can click on any latex that's done to see what the code is, and there's a post in the general physics forum with a lot of examples. epsilon is still out of my reach, however :(

8. Sep 10, 2005

### leright

This is EXACTLY my problem! Good job for understanding my cryptic statements.

I understand that you can determine the voltage between a small positive test charge and a point near the source charge, but not between the positive test charge and the actual source charge, since you get a voltage of infinity, which is no good. :tongue:

Now, what about a voltage across the two terminals of a bettery. This is a question that I feel is important to get answered, since I am an EE major. In this case, let's take the negative terminal to be the e-field source charge, and the positive terminal to be the positive test charge. We are presented with the same problem we had before, getting a voltage of infinty. For this reason, I do not understand what is meant by the voltage ACROSS the terminals of a bettery. What does the voltage specification of a battery mean? I understand voltage in its physical sense, but I do not understand what is being described with a battery's voltage....

I wish I would have kept my physics book, because this circuit analysis and design book is doing a very poor job of explaining voltage, and this gives me a very poor omen about EE as a whole....

Also, why are some websites just saying V = E*R to get voltages, where E is the e-field at that distance R? Obviously, this doesn't work. If it did work, I wouldn't be having this problem.....

Last edited: Sep 10, 2005
9. Sep 10, 2005

### leright

Someone please help me. This is driving me insane, and I am mad at how the professor and textbook are explaining these things.

10. Sep 10, 2005

### teclo

you have to understand the question you need to ask. if you are looking at it at from the particle physics point of view, it will take an infinte amount of energy to move a point particle of the same charge to the exact location of another point particle. try to see what the potential is at an extremely close location -- it's insanely high. the reason for this is what you're looking at is a force that approaches infinity as the seperation approaches 0.

if you want to calculate the potential for a two particle system, are you asking the right question? the potential is how related to the energy it takes to move another, 3rd, particle from a certain distance to another distance due to the charge configuration.

have you taken an electrostatics class? E*R means that the electric field in the region is uniform. in many macroscopic situations you can make some simplifying assumptions to get this. it's not precise, though.

concerning a macroscopic systems:

the voltage of a battery would be well described as emf

$$|\Delta V_b_a_t_t| = E_C s = \frac {F_N_Cs} {e}$$

when you are looking at a wire connecting these two, you can consider the electric field the same in each section. this gives you the E*R formula. the electric field, in reality, is produced by surface charge configurations that are established very quickly after the current is started.

your question regarding the integral form for potential is a bit off. i would suspect that it's impossible to most two protons or electrons to the exact same position. and even then, quantum mechanics does take over and you aren't really sure about where these things are.

11. Sep 10, 2005

### leright

ok, so the answer to my question is that for the purpose of batteries (voltages between two terminals), the e-field can be considered to be close enough to being uniform regardless of the distance, and e-field can be considered constant, so the formula V = E*R is acceptable? This is very unsatisfying to me, but if that is the real explaination, then I will just have to live with it.

12. Sep 10, 2005

### teclo

that's how you can consider it if you connect a circuit to the battery. the produced electric field is uniform throughout the wire, unless there are other elements to change this.

as far as a battery, you can think of it perhaps as a capacitor. the charge seperation produced is chemical in nature and i'm no chemist. the reactions give up or take electrons, and this flow of electrons is put out through the posts. the potential difference depends on the type of battery and how it's constructed.

13. Sep 10, 2005

### LeonhardEuler

I'll take a crack at this:
Ok. Voltage is a potential energy. The baseline you choose for zero PE is arbitrary. Just like for gravity, where you can say PE=mgh, and choose h=0 wherever you want, it ia only the change in PE that is ever important. Now in order to define a potential, the vector fied of forces you are integrating must be conservative, or else the value would depend on the path chosen. It is convenient to define the potential as zero infinitly far away from any charges. So the expression for the voltage is:
$$V=\int_{r=\infty}^{R}\vec{E}\cdot d\vec{s}=\frac{kq}{R}$$
Now the question is how to find the potential difference between two points. The potential willl be undefined at the exact locations of the charges, but not near them. Remember that the vector field is conservative, so all loop integrals vanish. Consider a triangle with sides consisting of the two points of interest and the point at infinty. The integral over this path is the sum of the integrals along each side of the triangle, and should be equal to zero. The question of finding the potential between point 1 and point 2 is the same as finding the integral along the trialgle leg from point 1 to point 2. Keeping track of orientation in the triangle and setting the loop integral to zero gives:
$$\int_{\infty}^{1}dV+\int_{1}^{2}dV+\int_{2}^{infty}dV=0$$
$$\int_{1}^{2}dV=\int_{\infty}^{1}dV-\int_{2}^{infty}dV$$
$$V=\int_{\infty}^{1}dV-\int_{2}^{\infty}dV$$
So it is a matter of choosing a path where the integral is easy to deal with.

Last edited: Sep 10, 2005
14. Sep 10, 2005

### leright

explain what you mean when you say the electric field is uniform throughout the wire. Also, is it possible that the charges just don't meet one another, and become 'one', and there is in fact a separation of charge? This would also take care of the infinity in the integration....but the notion that e-field is uniform confuses me.

15. Sep 10, 2005

### teclo

say you have a wire that is bent into a rectangle. the efield points from the positive to the negative. the magnitude of the efield in each part of the wire is about the same. if your wire is 1 meter long, and there's a 1.5V potential, then the efield is 1.5 v/m in each part of the wire pointing in the direction of the wire towards the negative terminal. this field is because of the charge configuration (shortage/overage) of electrons on the surface of the wire.

no, two protons can't merge and become one. neither can two electrons. there is a massive force pushing them apart that increaces to infinity as they their distance approaches zero.

16. Sep 11, 2005

### lightgrav

First of all, Voltage "between" locations A and B
(or Voltage from A to B, or across AB) is the
difference in Electric Potential: Delta V.
While the E.Potential "V" is zero at infinity,
its DIFFERENCE is V_B minus V_A .

Next, the E-field is a VECTOR , so it has DIRECTION,
not just being positive or negative. At any location,
E is the VECTOR sum of contributions by +Q and -Q .

Chemical reactions occur inside a battery cell
until enough electrons accumulate at the cathode
to effectively stop the reaction. So batteries
make the potential difference between terminals
"constant" - they are Voltage sources.

A real conductor (terminal) has nonzero surface area,
so the local E-field is essentially 4 pi k Q/A ,
pointing perpendicular to the surface (out if Q >0).
The E-field at the other terminal is similar, but inward.
The E-field "lines" look like that of a dipole, weakest
in the center between + and - . It is nowhere infinite,
because the charges that cause it are spread out.
For a car battery post, Q is about 1/4 picoCoulomb.
When you integrate this E-field from + to - , you get
the Voltage of the battery.

You might worry that batteries with smaller terminals
would have higher Voltage? It doesn't work that way,
because the reaction would've stopped long before.
That is, a smaller battery terminal has less charge on it.
Only as the electrons are removed into the circuit
do the reactions begin to supply more to replace them.

A point charge has an electric potential V = kQ/r .
But V at the surface of a conducting battery post
is caused by +Q that averages some 8 mm away,
and -Q that's about 200 mm away.

Lastly, the E-field in a wire will be uniform if the wire
connects the + to - terminals (as a "short"). But this
is because the charges being supplied by the battery
brings PE which is uniformly converted to thermal Energy.
Conducting wires don't have excess nor deficit electrons
on the outside of the wire. They are neutral metal,
except VERY near the terminals or other irregularity,
like where it joins to carbon (in a resistor).