Maker Pro
Maker Pro

Having trouble visualizing Voltage = Joules / Coulombs

wannabegeek

Aug 17, 2011
133
Joined
Aug 17, 2011
Messages
133
Also my thinking is that since magnetic fields induce electric fields, which is the voltage, it makes sense to me that a magnet would allow charge to flow more or less easily.


one more thing, only alternating magnetic fields induce electric fields and vice versa...
the equations that relate them are called Maxwell's equations.

Electric fields are not the voltage... :eek: and magnet's do not allow charge to flow more easily...only super conducting magnets do and they have to be at close to zero kelvin to do that....and only DC currents can super conduct, there's no super conducting radios :D
unfortunately....
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
yes

yes

No, divide both sides by Q ==> V = E / Q (you're taking a step back to the original equation)

And thus E = V * Q (take the same step forward)

I bet it was late when you typed that. I've been down that path before :D

I must admit, it took me a few minutes looking at this before I figured out where the error was.
I guess I did not make it clear. I was showing how it would result in an absurdity if Q was in the numerator in the original equation. So this eq. was supposed to be wrong.

Bob
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
I guess I did not make it clear. I was showing how it would result in an absurdity if Q was in the numerator in the original equation. So this eq. was supposed to be wrong.

Bob

That makes far more sense! :)
 

CluQu

May 22, 2013
14
Joined
May 22, 2013
Messages
14
one more thing, only alternating magnetic fields induce electric fields and vice versa...
the equations that relate them are called Maxwell's equations.

Electric fields are not the voltage... :eek: and magnet's do not allow charge to flow more easily...only super conducting magnets do and they have to be at close to zero kelvin to do that....and only DC currents can super conduct, there's no super conducting radios :D
unfortunately....

Oh darn I was interested in seeing if there was a specific field of research about it. Why do super conductors only work with DC? I'm very slightly familiar with bengzen rings, I think that's how you spell it. If I recall it was the bonds between the carbon atoms that allowed for close to zero resistance when electrons are flowing. I don't see why a change in direction would not work. Very interested in this, post links to places if you could, I would appreciate it. Meanwhile I'll check out Wikipedia. :cool:
 

wannabegeek

Aug 17, 2011
133
Joined
Aug 17, 2011
Messages
133
Turns out I'm wrong about AC SC...there is an ac application, but they don't SC in the strict sense...the resistance increases as the square of the frequency...

To be honest, I didn't have a chance to study SC ( or very much QM ) in my studies...I stuck to mechanic, devices and lab skills....I've never used a SQUID but I know that they are the cat's meow of detectors.

Even though this article mentions an RF application of SC,I 've worked in brain research using MRI and MRI doesn't use SC pickups. They are conventional LC detectors. Only the magnet is SC....

http://en.theva.biz/user/eesy.de/theva.biz/dwn/Superconductivity.pdf

Google Quantum entanglement radio...
 

NuLED

Jan 7, 2012
294
Joined
Jan 7, 2012
Messages
294
Hey you guys are getting me more confused over the issue :-D

Anyway I am still slowly grasping the idea that "charge" does not in itself contain any definition of energy and is dependent upon an external clarification of the quantized energy (in joules) contained "per charge" (which again is confusing since the charge itself is also quantized into discrete units called Coulombs).
 

CluQu

May 22, 2013
14
Joined
May 22, 2013
Messages
14
Hey you guys are getting me more confused over the issue :-D

Anyway I am still slowly grasping the idea that "charge" does not in itself contain any definition of energy and is dependent upon an external clarification of the quantized energy (in joules) contained "per charge" (which again is confusing since the charge itself is also quantized into discrete units called Coulombs).

Well energy isn't really a "thing" more so just an observed behavioral constant. Charge if I recall isn't really a thing either, just another observed constant. These particles do this around others, and so on and so forth. I'm sure there exist a much more formal definition of the terms.

Also what is the point in that RF application where the magnet is SC but not anything else? I'm going to dig into it some more when I have a more appropriate amount of time. :cool:
 

NuLED

Jan 7, 2012
294
Joined
Jan 7, 2012
Messages
294
I understand that the quantization of energy (in joules) and charge (in coulombs) are somewhat arbitrary, for human mathematics (of course, they also follow natural physical relationship with each other).

But it just is confusing to me that we do not have energy inherent in charge.

If we have a certain amount of "things" that have a certain amount of "elementary charge" surely there is inherent potential energy in that collection of charged things? Why must we additionally introduce energy into the system, in order to obtain voltage?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
I understand that the quantization of energy (in joules) and charge (in coulombs) are somewhat arbitrary

That's not quantization, they are units of measure.

And whilst they are arbitrary in one sense, they are tied to other measurements of an observable nature. So they form part of a consistent measurement system. Where they have a relationship to constants (for example the speed of light, or absolute temperature, the charge on an electron, etc) then they are tied to physical absolutes.

But it just is confusing to me that we do not have energy inherent in charge.

energy is no more inherent in charge than distance is inherent in time.

energy = voltage * charge
distance = velocity * time

In one case velocity is measured in units of distance per unit time. In the other voltage is measured in energy per unit charge. In both cases they are derived to measure a relationship which is observed in the real world.

To me, both are obvious and intuitive.

If you think of charge in terms of current and time (1C = 1As) then that 1 amp for 1 second can deliver more energy at a higher voltage. That tells you that the coulomb of charge must have more energy. And that energy is described by the voltage.

If you take the time to write down the fundamental units for these things, you'll see they cancel out nicely.

s = ut (s in metres, u in m/s, t in s) we have m = m/s * s

e = VQ (e in kg*m^2/s^2, V in (kg * m^2)/(A * s^3), Q in A*s) we have kg*m^2/s^2 = ((kg*m^2)/(A*s^3)) * (A*s) which also cancels out

For a bit of fun, you can look at the fundamental definition of various units and see where they cancel out. Most often this results in an equation that you are familiar with. However you can also generate equations where you have (say) resistance defined in terms of the universal gravitational constant and other things -- whilst these are true, they may not have any particular use.

If we have a certain amount of "things" that have a certain amount of "elementary charge" surely there is inherent potential energy in that collection of charged things? Why must we additionally introduce energy into the system, in order to obtain voltage?
because, as in the example above, without voltage we cannot relate the two. We have left over units of (in this case) kg, metres, and time. If you look at those leftover units, you'll see mass and distance in the numerator. These tell us that energy is involved (since energy is mass times distance) and the presence of time in the denominator tells us that there is something to do with a rate.

So voltage has something to do with the rate at which we can deliver energy (and that's power).

The rest of the units tell us more about that relationship. (And it turns out that the missing factor is the Amp).

In full, voltage tells us how much power we can deliver at a given current (recall P = V*I).

We know that the coulomb is the unit of charge that is represented by 1 A for 1s, so we can change P = V * I to P = V * (Q/t).

And wasn't Power measured in J/s?. So now we have e/t = VQ/t.

But both of those /t's cancel out, so we are left with e = VQ which is where we started.
 

wannabegeek

Aug 17, 2011
133
Joined
Aug 17, 2011
Messages
133
I understand that the quantization of energy (in joules) and charge (in coulombs) are somewhat arbitrary, for human mathematics (of course, they also follow natural physical relationship with each other).

But it just is confusing to me that we do not have energy inherent in charge.

If we have a certain amount of "things" that have a certain amount of "elementary charge" surely there is inherent potential energy in that collection of charged things? Why must we additionally introduce energy into the system, in order to obtain voltage?

What you are asking is kinda advanced without having taken some course material.

Image a battery, with two wires coming out of making a circuit.
Code:
      _______
      |           |
    _-_         |
      |           |
      | ____ _|
Inside the wires, there is charge Q b/c there are electrons there, spread out evenly on the surface of the wire.

Only consider the wires connected in a loop for a moment. By def, V = int{ E dot dL}. Inside the wire, we draw a rectangle to illustrate the current flow in the bulk of the wire.
Code:
                     --->----
                      |        |
                      ^       v
                      ---<---
the arrows denote the direction of the current along each side ( L). Since they are all opposing each other, the integral equals ZERO.

There is no electric potential in the wire EVEN though there is charge.

Now, the battery provides the force to move the charge around the wires.

Int{ E_total / Q_total dot dL } = int{ E_wire dot dL_wire } + int{ (F_bat / Q_bat) dot dL}

Now, F / Q is force per unit charge. Recall that F_electric = q E. The battery dL is the distance between the electric field lines drawn
between the wires. Not shown...

Delta V := - int{ E dot dL} from a point r to a ref point.

Satisfied...? :eek: This is hard to explain simply...
 
Last edited:

NuLED

Jan 7, 2012
294
Joined
Jan 7, 2012
Messages
294
I appreciate you guys' help on this. Again let me ponder all this in detail for a bit. This is the kind of learning "disability" that has always been an obstacle for me in school (although I got a high GPA because I just kept at it until I understood). Strangely enough some things I grasp immediately, before anyone else, and yet some of the most basic things that everyone gets, I don't. I can't explain it.
 

wannabegeek

Aug 17, 2011
133
Joined
Aug 17, 2011
Messages
133
I am def not one of the "smart" people I went to school with...those people have advanced degrees and work for Harvard, etc...

I took a lot of course work on this stuff and had hands on labs and all that...so I can accept a lot of it without remembering how it really works...

Most of the time, it's best to just get a working definition....work with the stuff, come back to the theory and repeat.....

cheers
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Again let me ponder all this in detail for a bit

Assuming you can accept P = VI, the following is about as simple as I have been able to put it:

voltage tells us how much power we can deliver at a given current (recall P = V*I).

We know that the coulomb is the unit of charge that is represented by 1 A for 1s, so we can change P = V * I to P = V * (Q/t).

And wasn't Power measured in J/s?. So now we have e/t = VQ/t.

But both of those /t's cancel out, so we are left with e = VQ which is where we started.

Essentially e = VQ is the same as P = VI, except that the former describes total energy, where the latter describes instantaneous power.

For instantaneous power we need to know how fast the charge is moving (the current), but for total power we simply need to know how much charge flowed (Q) because we don't care how long it took.
 
Last edited:

NuLED

Jan 7, 2012
294
Joined
Jan 7, 2012
Messages
294
OK, so thanks for trying to help me out. I know it must be sort of like drilling blood from a rock. LOL

So here is how my crazy brain works. I went and googled around and ended up studying the electrostatic field and how it is all calculated using Coulomb's law about electrostatic force between charges.

So in THAT I sort of finally got the distribution of force over a quantity of charge. That is, newtons per unit charge (within a field).

SO... if force can be divided amongst coulombs, then the analogue to that for voltage is the energy being divided per unit charge (instead of force being divided).

Now you might be thinking this is obvious but I think part of the "endless loop" in my thinking was that I was always harping on about why there isn't any "energy" inherent in charge, and in fact the "thing" that I was looking for can probably be answered by the fact that there IS a certain "thing" inherent in charge, and that is the natural force it exerts over distance, in an electrostatic field, and that is something already intrinsic to it.

So now, back to voltage, I am close but still have not fully completed the linkage to the energy being distributed per unit charge, but at least I know it is something now external to the charge quantity (as opposed to the natural field force). You can say that in my brain, it is a different kind of "energy" being discussed in the context of voltage.

I guess most people will think this is rather strange to go all around just to get to this point but, unfortunately for me, this is how my quirky brain works. Again I don't know why and certainly I don't really like it whenever this kind of learning hurdle happens for me (it only happens for certain things and I don't know what is the common denominator amongst all those things).
 

NuLED

Jan 7, 2012
294
Joined
Jan 7, 2012
Messages
294
I would like to report that OK I finally got it!

Thanks to everyone who helped me get it into my brain.

I really got it now. I got it after waking up this morning.

In fact now everything you guys said makes sense.

I also came up with a theory about this which has nothing to do with electronics so you can skip the rest of this post if you don't need to read it; it has to do with my theory about brain chemistry.

I think there is a genetic predisposition in my brain for the following effect:

There is a concept called neuroplasticity. And what that means is our brains are not static, and they are actually changing all the time, even all through life. It is a misconception that our brains are very slow in learning when we are older compared to when we were kids. While it's true children learn super fast (e.g., language) in fact our neurons are very much like muscles and need to be exercised (via learning new topics) or they will atrophy. The catch with neuroplasticity is that there is evidence the neurons actually PHYSICALLY change as a response to new learning stimuli (and this is NEW learning, not just reinforcing existing knowledge). It can get harder as you get older (much like our muscles are not as agile) but it doesn't HAVE to be that way.

Now, in my case, I think my brain chemistry must have something that causes some neurons to form a kind of endless loop, and I think that because there is no common denominator among the topics that I have trouble with (and it is not even that frequent, but when it does I am truly stuck) then this brain chemistry must be a random occurrence. And due to the neuron restructuring, the more I think about the problem, ironically the more I make it worse because I am reinforcing again and again the new (erroneous) structure. This is why practicing something makes us more competent (including reviewing of learned material). So for me to break out of the rut, I need to reinforce OTHER stimuli that will somehow, eventually be able to replace the erroneous formed memory structure. The analogy might be replacing a wrong logic gate with a proper one.

This neuron rewiring happens more often during sleep. When we learn material, often our brains use sleep as the reformatting stage, to consolidate the information (maybe it needs to do that while taking some other processes offline, and is the reason we need to sleep, which is a very dangerous thing to do in nature if you think about it, so it must be absolutely necessary otherwise it would have been naturally selected out).

So after countless bombardment of the erroneous learning (thanks to all your efforts) and some copious surfing around, last night finally the blockage must have been removed, and today I totally get it.

The funny thing is, I have no recollection why I did not get it, because it seems so obvious to me now, and the only proof that I had a problem with this concept is in these messages.

Anyway again I sincerely thank all of you for helping me out.
 

wannabegeek

Aug 17, 2011
133
Joined
Aug 17, 2011
Messages
133
Good work....I hope my post helped...

I've been trying to tell you something and I think you just haven't heard it yet.
From Griffiths, Introduction to Electrodynamics, chapter 2.3 Electric Potential:

"The name 'potential' is a hideous misnomer because it inevitabley reminds you of 'potential energy'. This is particularly confusing, because there is a connection between 'potential' and 'potential energy'. The best I can do is to insist once and for all that potential and potential energy are completely different terms and should by all rights, have different names."

You keep using the word energy. This is energy: U = int{ F dot dL} or the negative work, - W, done by an electric force on a charge.

Electric Potential which can also be called a Potential Difference which is really DELTA V is often shortened to just V because we assume that the second ref point is 0.

FACT: The electric field produced by a charge element is ALWAYS a collection of straight lines with no curves EVER.

Therefore the integral around a curve has a special outcome. That is, only the field components which lie along the direction of integration will have non zeros values.

V = - int{ E dot dL} is electric potential. Maybe you don't know what a dot product is..? A dot product is the projection of a vector onto the one axis. (in past replies I left off the negative sign on the int )

So a diagonal E vector projects |E| cos ( theta ) amount on to the x axis. Or, |E| cos ( theta) onto the unit vector along the dL line element. Where theta, is the angle between an axis of reference and the dL unit vector.

Electric potential is a scalar description of the SUM of electric field lines that lie along a direction. Only the straight line path between those points has any contribution to the integral.

Going from that to wires across a load is a bit funny....
 
Last edited:

wannabegeek

Aug 17, 2011
133
Joined
Aug 17, 2011
Messages
133
I'm happy for you....it feels great to "get" something like this...

and I know how bad it feels to 'not get it'....I'm a slow learner myself, thanks to family with learning issues...a big part of enjoying a pursuit like this is to relax....

If you can ever take an introductory course in physics, you'll learn some of this pretty damn well. You don't need calculus either. The problems can be stated in geometries that just need trig and that's how it's taught for biology students and the like.

Happy 4th for all....
wbg
 

NuLED

Jan 7, 2012
294
Joined
Jan 7, 2012
Messages
294
Happy 4th to everyone as well and also to our friends abroad; it is a celebration of freedom of thought for all humanity.

(I am reading tons of books and websites these days; part of my "bombardment campaign" to reinforce the right neurons and replace the wrong ones. That's basically how I've been learning stuff over the years. I guess I am fortunate in that I enjoy the learning experience and also there is a nagging motivation when I don't understand it, so I don't just give up. Of course, the Internet helps a lot and I am fortunate to have websites like this one populated with helpful members).
 
Top