- #1

FeDeX_LaTeX

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**Voltage is the "Push on Electrons"?**

Hello;

This is the definition that I have always been given. However, what exactly pushes these electrons? What is the force that is causing this?

Thanks.

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- Thread starter FeDeX_LaTeX
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- #1

FeDeX_LaTeX

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Hello;

This is the definition that I have always been given. However, what exactly pushes these electrons? What is the force that is causing this?

Thanks.

- #2

collinsmark

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Hello;

This is the definition that I have always been given. However, what exactly pushes these electrons? What is the force that is causing this?

Thanks.

Well I can give you some equations, but you may not find this a satisfactory answer.

The force which actually pushes them apart is the Coulomb force.

[tex] \vec F = \frac{1}{4 \pi \epsilon _0} \frac{q_1 q_2}{r^2} \hat r [/tex]

Where

You can break this down into field theory by defining the electric field associated with a given charge,

[tex] \vec E = \frac{1}{4 \pi \epsilon _0} \frac{q}{r^2} \hat r [/tex]

And then,

[tex] \vec F = q \vec E [/tex]

to find the force on some other charge.

You can find the electric potential (aka voltage) by integrating over some path.

[tex] V(b) - V(a) = -\int _a ^b \vec E \cdot \vec {dl} [/tex]

It is often the custom to express potential difference in terms of a known "ground"; a common reference potential . And when dealing with point charges, a good choice for this ground potential is infinity. So the equation for potential (Voltage) of a point charge becomes

[tex] V(r) = -\int _{\infty} ^ r \frac{1}{4 \pi \epsilon _0} \frac{q}{r'^2} \hat r \cdot \vec {dr'} = \frac{1}{4 \pi \epsilon _0} \frac{q}{r} [/tex]

Thus the

---------------------------

But I'm guessing that these equations are not quite what you are asking. I'm guessing you might be asking what is that causes this electrostatic force, or the corresponding electric field

The answer is somewhat of an embarrassment in the world of physics, in that nobody really knows. Physicists know that the field is there, and they know how it behaves, and they know how to use these models and equations to make amazingly accurate predictions. But why is electric field there in the first place? I honestly don't know. What is "charge", really? Where does it come from? I don't know that either.

[Edit: Quantum electrodynamics (QED) deals with the forces using force carrying particles (photons essentially). Yet, in the end, they leave the same questions unanswered. I.e. What is charge really? Why are there these forces (force particles) present at all? I don't know.]

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- #3

rcgldr

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The push on electrons is electrical force, and doesn't have a special name. Voltage is a potential, the potential energy per unit charge, and equal to the negative of the work done per unit (positive) charge when a charged object moves from one point to another point within a field.

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Hello;

This is the definition that I have always been given. However, what exactly pushes these electrons? What is the force that is causing this?

Thanks.

See replies below for description of the mystery of charge and electrical forces.

However, strictly speaking, voltage is not the "push on electrons". You'd have to say that voltage is related to the "push on electrons". A "push" is what in physics is termed a force. The force on an electron is given by the Electric Field it resided in time the charge of the electron. Now a voltage is a potential and the potential related to the work it takes to move a charge (like an electron) from point A to point B in an electric field. Thus to call voltage a "push" is to say that work is a force which is comparing apples to oranges. But there obviously is a relationship between the two.

- #5

sophiecentaur

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A voltage of100,000V across two plates, separated by 1km, will not provide much of a force on a charge because the Electric Field is only 100V per metre (that is the 'gradient' of the energy). The same voltage, applied across two plates separated by 1m, will provide a force 1000 times greater because the field will be 100,000V per metre.

So the Voltage has to 'be there' but it's the Volts per metre that provide the Force- not the Volts. The posh maths in the earlier post says the same thing, btw.

Voltage relates to Energy, more directly, than to Force, in the same way that the height of a hill tells you more about the energy you use getting up it, rather than how hard you need to pedal on the way.

- #6

FeDeX_LaTeX

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The force which actually pushes them apart is the Coulomb force.

[tex] \vec F = \frac{1}{4 \pi \epsilon _0} \frac{q_1 q_2}{r^2} \hat r [/tex]

Whereqand_{1}qare the charges of each point particle. Substitute -6.02 x 10_{2}^{-19}C for each charge if you are dealing with electrons. [tex] \epsilon _0 [/tex] is the permittivity of free space, 8.85 x 10^{-12}F/m.

Okay, I understand. So what is the r with the bracket over it?

You can find the electric potential (aka voltage) by integrating over some path.

[tex] V(b) - V(a) = -\int _a ^b \vec E \cdot \vec {dl} [/tex]

Can you give me an example calculation? I know how to integrate, I just don't understand what you mean by "integrating over some path"... by V(b) - V(a), are you saying you subtract one voltage from the other?

It is often the custom to express potential difference in terms of a known "ground"; a common reference potential . And when dealing with point charges, a good choice for this ground potential is infinity. So the equation for potential (Voltage) of a point charge becomes

[tex] V(r) = -\int _{\infty} ^ r \frac{1}{4 \pi \epsilon _0} \frac{q}{r'^2} \hat r \cdot \vec {dr'} = \frac{1}{4 \pi \epsilon _0} \frac{q}{r} [/tex]

Thus thepotential energyof some other charge,q, brought into this electric field isqV.

Can you show me an example calculation please? Just to make sure I understand...

---------------------------

But I'm guessing that these equations are not quite what you are asking. I'm guessing you might be asking what is that causes this electrostatic force, or the corresponding electric fieldin the first place.

Yes, but the equations were interesting too, since I'd never seen them before. Thanks!

A voltage of100,000V across two plates, separated by 1km, will not provide much of a force on a charge because the electric field is only 100V per metre (that is the 'gradient' of the energy). The same voltage, applied across two plates separated by 1m, will provide a force 1000 times greater because the field will be 100,000V per metre.

Because voltage splits in Series, right? Okay, I understand.

- #7

collinsmark

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Okay, I understand. So what is the r with the bracket over it?

The [tex] \hat r [/tex] is the unit vector. It's nothing more than a way to keep track of where things are pointing. It's just a way to keep track of direction. By definition, unit vectors have a magnitude of 1. Direction is the information that they contain.

[tex] \vec F [/tex] is a vector, and [tex] \vec E [/tex] is a vector field. Vectors specify both magnitude and direction. Here, [tex] \hat r [/tex] specifies the direction, which always points away from the point charge (or in the case of [tex] \vec F [/tex] points from the first point charge to the second point charge, indicating the force on the second point charge -- switch point charges around to find the opposite; the force on the first point charge).

Anyway, don't put too much concern into unit vectors, they're just a way of keeping track of direction.

Can you give me an example calculation? I know how to integrate, I just don't understand what you mean by "integrating over some path"... by V(b) - V(a), are you saying you subtract one voltage from the other?

The equations in my last post only applies to point charges. If you have a different configuration, such as a charged plate, or a charged wire, you'll end up with different equations.

But let me elaborate on the V(b) - V(a) part. As others have made it clear, Voltage (electric potential) is a measure of

For example, suppose you are to measure the potential energy of car on a roller coaster at any given instant in time. A typical choice of the reference potential energy is the surface on which the coaster stands. That way, you can use PE = mgh, where h is the height of the car with respect to the ground/surface. But you could choose some other surface too such as 1000 meters below ground or 10000 feet above the ground. Or if you wanted to crazy and measure the potential energy with respect to infinity (pretending that there is no Sun or other celestial objects besides the earth), change the potential energy equation to PE = -G(Mm/h) where G is the gravitational constant , M is the mass of the earth, and h is the height above the earth's center.

Anyway, the point is that there is the assumption that electric potential is always in reference to something -- i.e. it's essentially a potential difference. If you measure the "voltage" of a battery, you are measuring the potential of the positive terminal

And like I mentioned before, electric potential (aka Voltage) is a measure of

Can you show me an example calculation please? Just to make sure I understand...

(a) How much energy is required to bring and electron from infinity to a distance of 1 meter to another electron, assuming that there are no other charges involved?

[tex]

V(r) = \frac{1}{4 \pi \epsilon _0} \frac{q}{r}

[/tex]

[tex] V(\textbox{1 \ m}) = -\textbox{1.44 \ nV}. [/tex]

[tex] W = qV = (-1.602 \times 10^{-19} \ C) (- \textbox{1.44 \times 10^{-9} V}) = 231 \times 10^{-30} \ J [/tex]

---------------------------

Because voltage splits in Series, right? Okay, I understand.

Force and electric potential (voltage) are related. But the relationship is very configuration dependent. So the statement that "Voltage is the push on electrons," is sort of misleading, but probably might suffice for the lay person. While that statement might be okay for the lay person, one needs to make the distinction between force and Voltage when pursuing the details of electrostatics.

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