How can you achieve different voltages in an electric circuit?

[Opus_723]

Can you do the integral of 1/r²?

A simple yes or no will let me explain the next bit.

Yes. -1/r + C.

That is mostly correct I believe. Do you have a specific example or question in mind that you would like to ask about? This is a very big subject and there is no way we can explain every situation.

Sadly, I don't think so. I don't know enough about practical electronics to think of anything specific. But if what I said earlier IS correct, then that helps me a lot, I think.

 

[Studiot]

Well I'll try anyway since I want to go to bed.

Isn't that basically what I was saying before? If you separate a positive and a negative charge, then the work done (and therefore potential difference) increases with the distance you separate them by?

The problem with this is the following.

Work = force x distance.

The work done in moving (separating) these charges a small distance dr is

Force x dr

Now the force is given by the inverse square law as inversely proportional to the square of the distance

F = \frac{K}{{{r^2}}}

So if we try to substitute into the expression for work and integrate from r = 0 to r = some distance say a we get.

Edit: Sign correction, work is against force ie negative.

\int\limits_0^a {-Fdr} = \int\limits_0^a {\frac{K}{{{r^2}}}} dr = \left[ {\frac{K}{r}} \right]_0^a = -\infty +\frac{K}{a}

Now can you see why we do not define potential energy this way?

 

[RedX]

There is a point of diminishing returns: you can't just separate two charges forever and increase your voltage to any high value you want.

What's more important than how far you can separate two charges, is how close the positive and negative charges are together to begin with. Then you can define a so-called ionization energy.

But batteries don't really work like this. But ultimately that's where you are going if you insist on thinking of everything in terms of point charges.

 

[Opus_723]

The problem with this is the following.

Work = force x distance.

The work done in moving (separating) these charges a small distance dr is

Force x dr

Now the force is given by the inverse square law as inversely proportional to the square of the distance
So if we try to substitute into the expression for work and integrate from r = 0 to r = some distance say a we get
Now can you see why we do not define potential energy this way?

Alright. I understand that that doesn't work. But I thought potential energy was defined in terms of work, and W = F*d?

What am I missing? Apparently potential energy is not defined in terms of work? Or does work have a broader definition that I'm not aware of?

 

[Drakkith]

Alright. I understand that that doesn't work. But I thought potential energy was defined in terms of work, and W = F*d?

What am I missing? Apparently potential energy is not defined in terms of work? Or does work have a broader definition that I'm not aware of?

(I hope I'm using the right terms here)

Potential Energy and Potential Difference are NOT the same thing. If an electron and a proton are 100 miles apart with nothing in between them, they effectively have Zero voltage, or potential difference. But they both have potential energy, because if I did move them together they would attract. The potential energy is equal to the amount of work required to separate the charges. The voltage is not however.

The closer two charges are, the MORE voltage they have between them because their attractive force increases the closer they are brought together. This is why a capacitor acquires more capacitance the closer you bring the plates together. The charges between the plates attract each other, and the closer you bring them together the more charges you can move between the plates for the same applied voltage.

 

[Opus_723]

I wasn't claiming that potential energy and potential difference were the same. I thought potential difference was the difference between the potential energy of a charge at two points. Is that wrong?

Also, thanks everyone for bearing with me, I know this must be frustrating.

 

[Drakkith]

I think you may be confused because you are looking at a situation between just 2 particles. In reality we don't generally use voltage and such for just 2 particles by themselves.

(I think this is all mostly correct, forgive me if it isn't. This is the way I currently understand it though.)

For example, let's look at a capacitor again. If you connect a conductor between the terminals of a capacitor, you have given the charges on the plates a path to flow back to their source. But let's look at each plate more closely.

On the Negative plate you have a build up of electrons (negative charges) caused by the charging voltage. So that plate has all these electrons that are all exerting a negative charge on each other and pushing them apart. Until the circuit is completed, the electrons have nowhere to go, as they cannot flow through the dielectric and the resistance between the two terminals is much to high for current to flow.

On the positive plate, you have the reverse situation. You have all those electrons, which would usually equalize the charges, no present. This leaves an excess of positive charges on the plate.

Now, when you connect a wire between them, all those electrons which are pushing each other on the negative plate suddenly have the resistance between the terminals removed. So all this negative charge pushes on the electrons in the wire and at the same time you have the positive plate pulling on the electrons in the wire.

So what happens? Since the resistance of the wire is very very low, the electrons in it can very easily move. The potential difference in charge, AKA the voltage, between the plates is enough to force all these electrons to start moving along the wire. The electrons from the negative plate flow into the wire and at the same time the electrons from the wire flow into the positive plate. This continues until the two plates equalize and become neutral.

The VOLTAGE here was the combined force from one plate due to the buildup of like charges in a given area. If we double the size of the plates in the capacitor, then we can transfer double the electrons and have double the power but the voltage would NOT increase.

However, if we increased the charging voltage by double, we CAN transfer double the electrons without having to double the size of the plates. This would result in double the voltage because we have more charges combined in the same area.

 

[Drakkith]

I wasn't claiming that potential energy and potential difference were the same. I thought potential difference was the difference between the potential energy of a charge at two points. Is that wrong?

Also, thanks everyone for bearing with me, I know this must be frustrating.

Hrmm. That looks correct to me.

 

[Drakkith]

Alright. I understand that that doesn't work. But I thought potential energy was defined in terms of work, and W = F*d?

What am I missing? Apparently potential energy is not defined in terms of work? Or does work have a broader definition that I'm not aware of?

See http://en.wikipedia.org/wiki/Potential_energy for more info on potential energy.

 

[Studiot]

You really want to get an understanding of this?
That means that you must just stick to the definitions of all the quantities involved and not try to jump to your own conclusions about the relationships between them. Maths is everything in this basic electrical theory. The analogies just lead you astray.

SC is right ! Ok so posts #29-32 establish that you have the necessary mathematical apparatus to do the job properly.

First an apology for an error which I take it you noticed. In my integral I inadvertently missed a negative sign, so the signs are the wrong way round. However the underlying principle is the same so I have just corrected post#32.
Don’t forget that what I did is a gross simplification in order to highlight a particular point.

To start at the beginning. –Empty space.

If we introduce a neutral object there is no ‘field of electric force’

If we introduce one single charged object, of charge q1, there is still no field of force, since there is nothing for it to act on.

If we now introduce a second charged object, of charge q2, there is a mechanical force acting between them given by the inverse square law.

F = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}

Where r is the distance between them.

It is important to realize that the existence of this force require both charges and is dependent upon both.
As a consequence.
The introduction of the first charge takes no energy ie no work is done.
This is because no force is acting or the force is zero.
As another consequence the law implies that when the charges are infinitely far apart the force between them is zero. This coincides with common sense.

So the work required to bring the second charge from infinity to some point a distance a away from the first charge is given by the integral

W = \int\limits_\infty ^a { - Fdr = \int\limits_a^\infty {Fdr} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}}}\left[ {\frac{{ - 1}}{r}} \right]} _a^\infty = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}a}}

Which we can evaluate as I have done above.

Since work has been done in assembling this simple system, it contains that work in the form of potential energy.

So this is a definition of the potential energy of the system. Within the modern MKS system the units have been arranged for this energy to be measured in Joules.

The thing to note here is that the fundamental quantities are Force, Distance and Charge.

continued in next post

 

[Studiot]

The potential energy is not fundamental, but relates only to the system concerned. (Two charges in our case). There is no ‘absolute PE’.
If you think about it to try to establish this would involve calculating the PE due to the interaction of every charge in the universe.

So PE is a derived quantity.

We can adopt a different viewpoint and derive another quantity.

The force and PE are due to the presence of both charges which both appear in our equation.
If, however, we regard q1 as fixed and divide the expressions by q2 we get the force per unit charge or the PE per unit charge.

We call the PE per unit charge the potential and give it the units – volts.
Potential is a derived quantity.

Potential = \frac{{{q_1}}}{{4\pi {\varepsilon _0}a}}

It is this process that gives rise the confusion that you and others experience.

Note this is called the potential of q2, not of q1, although the potential is due to the existence of both charges q1 and q2.

If we have q1 alone there is no potential.

We can use it to consider the effect on other charges than q2.

As already noted in the last post, trying to establish an absolute PE or potential is unrealisable.

So we introduce yet another derived quantity the potential difference or p.d.

This is a useful quantity because (in our case) it depends only on position (distance) and can be calculated in an absolute manner. This is similar to the difference between definite and indefinite integrals where the definite integral has no abitrary constant.

The potential difference is the difference in potential in moving q2 from a distance a from q1 to a distance b from q1.

p.d. = \frac{{{q_1}}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{a} - \frac{1}{b}} \right]

Since this is a simple subtraction sum it is also measured in volts.

Finally this brings us to the formula

Energy (Joules) = pd times charge (q2) = volts times coulombs.


Hope this helps.

 

[sophiecentaur]

(I hope I'm using the right terms here)

If an electron and a proton are 100 miles apart with nothing in between them, they effectively have Zero voltage, or potential difference. But they both have potential energy, because if I did move them together they would attract. The potential energy is equal to the amount of work required to separate the charges. The voltage is not however.

This is just not correct. The potential between the two charges increases as distance increases. The "volts" between these well separated objects (the PD) would be exactly the work (per unit charge) needed to take one charge (just the charge on one electron) from where it is to meet the proton. So, by definition, the voltage is exactly what you'd expect.
The problem is when you try to relate this to the situation in a circuit or when a charge is moved about in the presence of a field which has been set up by moving a lot of other charges about in the first place - for instance, if the charges have been pushed to the terminals of a battery. The first case involves just two charges but the usual case involves discussing a charge being moved about in the presence of a set of fields due to all the charges in a wire, battery, bulb situation.

But, in both cases, the definition applies but the two charge situation may seem counter-intuitive.

[Edit: I suppose I should add that in the case of the proton and electron, there is a minimal potential which is the ground state of a Hydrogen atom, which you could look upon as 0V]

 

[olivermsun]

Voltage is not analogous to water pressure. It is potential energy so, in the water analogy, the energy contained in a mass of water at a given height over the pump.

More generally, pressure differential in water pipes is analogous to potential difference, aka voltage, in an electrical circuit. In a water pipe, a certain pressure differential between the ends of a pipe is required to make a certain water flow through the (restrictiing) pipe. In the same way, a certain voltage is required across the ends of a resistive load to make a certain current flow.

As HallsofIvy mentions, the height of a water column (in gravity), is a particularly good way to create a (known) pressure, and two columns of differing heights can create a known pressure differential, etc. (although if you connect them you can't expect the differential to last forever!).

Water is not the way of madness -- one just has to make the right analogies. :)

 

[sophiecentaur]

Water is not the way of madness -- one just has to make the right analogies. :)

The water analogy is fine - looking back at it from a position of already understanding the electrical 'reality'. I have heard people make so many howlers when using it that I really think it's not worth going with. For a start, people don't even know what they mean by water pressure half the time.

 

[RedX]

For analogies, see this table:

http://en.wikipedia.org/wiki/Bond_graphs

Electromotive force is to current as pressure is to volumetric flow rate, and torque is to angular velocity, and temperature is to entropy flow rate.

 

[Studiot]

Hey, guys, I really would lay off the water pressure analogy as it is so easy to find exceptions that it is just a waste of time, especially when you could be using the real thing.

The OP here is trying to understand voltage not water pressure.

 

[Drakkith]

This is just not correct. The potential between the two charges increases as distance increases. The "volts" between these well separated objects (the PD) would be exactly the work (per unit charge) needed to take one charge (just the charge on one electron) from where it is to meet the proton. So, by definition, the voltage is exactly what you'd expect.
The problem is when you try to relate this to the situation in a circuit or when a charge is moved about in the presence of a field which has been set up by moving a lot of other charges about in the first place - for instance, if the charges have been pushed to the terminals of a battery. The first case involves just two charges but the usual case involves discussing a charge being moved about in the presence of a set of fields due to all the charges in a wire, battery, bulb situation.

But, in both cases, the definition applies but the two charge situation may seem counter-intuitive.

[Edit: I suppose I should add that in the case of the proton and electron, there is a minimal potential which is the ground state of a Hydrogen atom, which you could look upon as 0V]

Hrmmm. Wouldn't that be the potential energy, but not the volts that is increasing? I don't think there is any voltage between just 2 particles at a large distance from each other.

By your definition, the voltage would be almost 0, as it would take just the slightest push to make an electron move to the proton. (Given that both are in space with nothing between them) Shouldn't the "electromotive force" increase as the 2 particles got closer, as the attraction grows stronger?

Or am I just getting confused here?

From wikipedia's reference on voltage:

To find the electric potential difference between two points A and B in an electric field, we move a test charge q0 from A to B, always keeping it in equilibrium, and we measure the work WAB that must be done by the agent moving the charge. The electric potential difference is defined from VB − VA = WAB/q0" Halliday, D. and Resnick, R. (1974). Fundamentals of Physics. New York: John Wiley & Sons. p. 465.

Would 2 isolated particles provide the requirements for defining voltage?

 

[sophiecentaur]

I agree that there may be confusion when applying the definition to just two particles. But it is still consistent.
Yes "a gentle push" but a push is not energy. There is always a finite force of attraction. Btw, electrons in a wire only need a "gentle push" to move from place to place; that's what Low Resistance means. They can still have a high PD wrt another part of the circuit.
If one argues that the Volts are Zero at a distance, one would have to ask "when does the trend change direction" as the volts clearly increase when they're close together?

 

[Studiot]

Or am I just getting confused here?

From wikipedia's reference on voltage:


To find the electric potential difference between two points A and B in an electric field, we move a test charge q0 from A to B, always keeping it in equilibrium, and we measure the work WAB that must be done by the agent moving the charge. The electric potential difference is defined from VB − VA = WAB/q0" Halliday, D. and Resnick, R. (1974). Fundamentals of Physics. New York: John Wiley & Sons. p. 465.

Would 2 isolated particles provide the requirements for defining voltage?

Yes you are getting confused.
My posts do explain all this.

Note the following

1) The force between charges can be attractive or repulsive. If it is repulsive the force gets larger and larger as they get closer and closer ie it gets more and more negative. Something getting more and more negative is technically decreasing.

2) I did warn against considering the energy of separation and later showed that it leads to an impossible calculation ie infinity. That is why we do not do it this way.

3)

The word potential is used in three different ways in electrics. Sometimes people mix these up.

Potential energy - measured in Joules
Potential or electric potential - measured in Volts
Potential difference - measured in volts

Notice that two of these are measured in volts. A recipe for confusion.

4)Where does the electric field in your example come from?
It can only come from another charge, since charges are the source of electric field. It is possible to postulate an electric field and use this to derive the same expressions as I did.
I avoided this since I think it only adds an extra layer of complexity for beginners any my way preserves the direct links between the mechanical effects (work and force) and the electric effects withouf intermediate calculations.
That means I am working from what most people already know (the mechanical stuff) to the new (electrical) stuff, rather than presenting a new purely academic construct.