Coulomb's law, electrostatics?

[Jurtinus]

Equation:

ΣF=Σk(qi)(qj)/(r^2)

Question:

Considering more than a couple of particles. How can a net force on a charged particle be calculated if Coulomb's law is under the restriction of static forces?

Thanks!

 

[Doc Al]

What does the number of particles matter?

 

[Jurtinus]

To imply a generalized approach.

To restate, how can a net force be present under static conditions?

 

[Doc Al]

To restate, how can a net force be present under static conditions?

Perhaps you are thinking of Earnshaw's theorem?

 

[Dale]

To restate, how can a net force be present under static conditions?

You are correct. Statics usually assumes no net force. However:

1) The electric force may not be the net force if there are other non-electric forces involved.

2) The Coulomb force is not “static” but “electrostatic”, meaning that the currents are at least momentarily 0.

 

[Jurtinus]

You are correct. Statics usually assumes no net force. However:

1) The electric force may not be the net force if there are other non-electric forces involved.

2) The Coulomb force is not “static” but “electrostatic”, meaning that the currents are at least momentarily 0.

Assuming no external forces, I suppose Coulombs Law investigates impending motion of charged particles.

 

[ZapperZ]

To imply a generalized approach.

To restate, how can a net force be present under static conditions?

I think you have misunderstood the term "static" in "electrostatics".

It means that any "d/dt" term is zero, or that there is no time-variation in the electric field. In terms of the presence of charged particles, it means that (i) these charges do not move or change position (ii) the amount of charge of each particle does not change over time (iii) you are finding the electric field at a fixed field location. This will then allows you to find the coulomb's force at that field location if you place another charge there.

So yes, there can be a force under static condition. This is not "static" as in "engineering statics", where the NET force on each member is zero.

Zz.

 

[Jurtinus]

I think you have misunderstood the term "static" in "electrostatics".

It means that any "d/dt" term is zero, or that there is no time-variation in the electric field. In terms of the presence of charged particles, it means that (i) these charges do not move or change position (ii) the amount of charge of each particle does not change over time (iii) you are finding the electric field at a fixed field location. This will then allows you to find the coulomb's force at that field location if you place another charge there.

So yes, there can be a force under static condition. This is not "static" as in "engineering statics", where the NET force on each member is zero.

Zz.

In fact, I am taking an engineering statics approach. But statics is statics.

From my understanding of mechanics; d/dt=0 implies a steady state condition, which is not the same as d/ds=0 which implies a static condition, thus no net force. However, the integral of ds is taken over time. So, I suppose this would agree with your argument.

Once again, I’m led to the notion that Coulombs Law defines how a charged particle impends motion on another. Rather than, explaining a static situation explicitly.

Additionally, impending motion is covered in engineering statics with respect to dry friction. Which, nonetheless, can be referred to as Coulombs friction.

 

[Dale]

Assuming no external forces, I suppose Coulombs Law investigates impending motion of charged particles.

Coulombs law certainly applies to situations that violate the no external forces assumption. I would not try to restrict it through unnecessary additional assumptions. It is valid in all “electrostatic” situations, which is a different criterion than “static”.

 

[ZapperZ]

In fact, I am taking an engineering statics approach. But statics is statics.

No, it isn't. That's like insisting "theory" is theory, and using the layman's use of it to apply everywhere. Where is it written that this is the ONLY way to use that word?

If that is the approach you are taking, then this whole discussion is pointless, because you are applying an irrelevant set of rules to something where it wasn't meant to be used. There's nothing written anywhere that "statics is statics"? You made that up, and arbitrarily applying it. Go to your Engineering Statics professors and try to sell your idea to them and see how that goes.

Besides, what makes you think this application is a one-way street? Why can't the "statics" as used in electrostatics be applied to Engineering Statics? After all, if you think about it, the condition used in E&M is a lot more GENERAL, and encompasses what is used in Engineering Statics. What arbitrary rules did you made up to allow for this one-way application?

I see this topic as going nowhere fast.

Zz.

 

[Dale]

After all, if you think about it, the condition used in E&M is a lot more GENERAL, and encompasses what is used in Engineering Statics.

That is an interesting point. I had not considered it before, but that is right. It also incorporates magnetostatics and various equilibrium assumptions

 

[Jurtinus]

No, it isn't. That's like insisting "theory" is theory, and using the layman's use of it to apply everywhere. Where is it written that this is the ONLY way to use that word?

If that is the approach you are taking, then this whole discussion is pointless, because you are applying an irrelevant set of rules to something where it wasn't meant to be used. There's nothing written anywhere that "statics is statics"? You made that up, and arbitrarily applying it. Go to your Engineering Statics professors and try to sell your idea to them and see how that goes.

Besides, what makes you think this application is a one-way street? Why can't the "statics" as used in electrostatics be applied to Engineering Statics? After all, if you think about it, the condition used in E&M is a lot more GENERAL, and encompasses what is used in Engineering Statics. What arbitrary rules did you made up to allow for this one-way application?

I see this topic as going nowhere fast.

Zz.

I think I see what your getting at, perhaps I'm misunderstanding the concept of electrostatics. In an attempt to learn, I'm using concepts familiar to me and trying to form a link to enlighten myself elsewhere. I must be misunderstanding the intent of particular examples problems introducing Coulombs Law. I shall continue to study further.

Although, I believe I have an idea of how Coulomb formulated his Law through his torsion balance experiment, it has been eluding me how it can apply to charges in space. I now see I have been assuming too much, and the example problems I'm using aren't intended to show equilibrium, as such of Engineering Statics.

Thanks for everyones help:)

EDIT: Wikipedia defines electrostatics as "charges at rest." This would agree with what I'm trying to define as "static," or in equilibrium.

 

[Dale]

EDIT: Wikipedia defines electrostatics as "charges at rest." This would agree with what I'm trying to define as "static," or in equilibrium

Even “charges at rest” would still allow non electric forces.

 

[Jurtinus]

Even “charges at rest” would still allow non electric forces.

Yes, I understand equilibrium. But not a net force in equilibrium.

 

[ZapperZ]

EDIT: Wikipedia defines electrostatics as "charges at rest." This would agree with what I'm trying to define as "static," or in equilibrium.

But here again you are confusing "source" charge with "test" charge!

If the source charge, i.e. the charge that is the source of the electric field, are "static" or not moving, then the electric field are also fixed in space at time. This is EXACTLY what I stated in one of my posts!

However, you then claim that this means there's no net force, as in "engineering statics", where everything is not only stationary, but also has net zero force. This is not true in electrostatic, because I can put another charge (the "test" charge) in the static electric field and that charge will experience a net force! This is the coulomb force.

The problem here is that it appears that you have on learned about electric field yet, and how coulomb's law can be obtained by the introduction of a charge into this electric field.

Zz.

 

[Dale]

Yes, I understand equilibrium. But not a net force in equilibrium.

Yes, but the total electric force may be nonzero, which is what Coulomb’s law describes.

 

[Jurtinus]

But here again you are confusing "source" charge with "test" charge!

If the source charge, i.e. the charge that is the source of the electric field, are "static" or not moving, then the electric field are also fixed in space at time. This is EXACTLY what I stated in one of my posts!

However, you then claim that this means there's no net force, as in "engineering statics", where everything is not only stationary, but also has net zero force. This is not true in electrostatic, because I can put another charge (the "test" charge) in the static electric field and that charge will experience a net force! This is the coulomb force.

The problem here is that it appears that you have on learned about electric field yet, and how coulomb's law can be obtained by the introduction of a charge into this electric field.

Zz.

I believe to understand all that you have said, this isn’t my first time dealing with electromagnetism.

If, instead, we drop an actual charge. Would it not either repel/attract from/to the source charge?

If so, than all I’m claiming is impending motion of the test charge, which describes a net force.

If not, than I’m utterly confused.

Thanks:)

 

[Jurtinus]

Yes, but the total electric force may be nonzero, which is what Coulomb’s law describes.

This would then give rise to motion, no?

Thanks:)

 

[Dale]

This would then give rise to motion, no?

No, not if there were other forces.

This goes back to your extra unnecessary assumption that I pointed out earlier

If so, than all I’m claiming is impending motion of the test charge, which describes a net force.

It is only a net force under an additional and unnecessary assumption

 

[ZapperZ]

I believe to understand all that you have said, this isn’t my first time dealing with electromagnetism.

If, instead, we drop an actual charge. Would it not either repel/attract from/to the source charge?

If so, than all I’m claiming is impending motion of the test charge, which describes a net force.

If not, than I’m utterly confused.

Thanks:)

I have no idea what you said here.

The "test" charge that I stated is the charge that you put in an electric field, and which you are finding the Coulomb's force for!

Zz.

 

[Jurtinus]

No, not if there were other forces.

This goes back to your extra unnecessary assumption that I pointed out earlier

It is only a net force under an additional and unnecessary assumption

I have made assumptions to simplify the matter, since the examples I was looking at were introductory (which are never complete in the first place). In fact, I suspect these examples to be the source of my confusion.

For clarity, I’ll post the example problem. Please take it as is, and no more.

Thanks:)

 

[Jurtinus]

The initial problem

Sears & Zemansky's University Physics w/ Modern Physics 13th ed., Young & Freedman, 2012

Thanks:)

 

[ZapperZ]

So what’s the issue here? How does this problem somehow resulted in this thread?

Zz.

 

[Jurtinus]

So what’s the issue here? How does this problem somehow resulted in this thread?

Zz.

There is a net force, the particle isn't in equilibrium. Coulombs law, based on electrostatics or equilibrium (from what I understand).

1) How can there be a net force if the particle is in equilibrium?
2) If the particle is not in equilibrium, how does Coulombs law apply?

These were all my original inquiries, wether or not I made that clear.

Thanks:)

 

[ZapperZ]

There is a net force, the particle isn't in equilibrium. Coulombs law, based on electrostatics or equilibrium (from what I understand).

1) How can there be a net force if the particle is in equilibrium?
2) If the particle is not in equilibrium, how does Coulombs law apply?

These were all my original inquiries, wether or not I made that clear.

Thanks:)

And again, it is a misuse of the term "static" in this case. It does NOT mean it is in equilibrium. It means that the electric field that the center charge sees does not change with time at that location, thus a "static", non-time varying field.

Static in "engineering static" means not moving. "Electrostatic" here means it is not time varying.

If you have such inflexibility in adapting to the various different rules and definitions, you will have a rough time in school.

Zz.

 

[Jurtinus]

And again, it is a misuse of the term "static" in this case. It does NOT mean it is in equilibrium. It means that the electric field that the center charge sees does not change with time at that location, thus a "static", non-time varying field.

Static in "engineering static" means not moving. "Electrostatic" here means it is not time varying.

If you have such inflexibility in adapting to the various different rules and definitions, you will have a rough time in school.

Zz.

Why then, does Wiki and Young & Freedman define electrostatics as “charges at rest”?

Thanks:)

EDIT: Perhaps, the source charge is at rest?

 

[Dale]

There is a net force, the particle isn't in equilibrium. Coulombs law, based on electrostatics or equilibrium (from what I understand).

Indeed, the problem is one where Coulomb’s law is not clearly applicable. I assume that in the broader context of the chapter it is obvious, but I agree with your objection in the specific context of this problem.

Often simplified laws are applied even when the exact assumptions are violated. In such a case it is important to understand when you can get away with the approximation. Chapter 3 below is a good explanation

http://web.mit.edu/6.013_book/www/book.html

 

[ZapperZ]

Why then, does Wiki and Young & Freedman define electrostatics as “charges at rest”?

Thanks:)

Those are the SOURCE charges. They are the two charges on the two ends. The TEST charges is the one in the middle, because that is the charge where you are calculating the force for!

The two source charges are NOT MOVING, resulting in electric field that is STATIC WITH TIME atthe middle test charge. The middle test charge still has a NET FORCE. Nothing is moving with respect to time in that problem.

When will this sink in?

Zz.

 

[Jurtinus]

I'm terribly sorry, but I've made a mistake. The entire problem wasn't shown...

Sears & Zemansky's University Physics w/ Modern Physics 13th ed., Young & Freedman, 2012

Dale: I'm unsure wether your reply would change. Thanks for the book! Although, it seems too advance at first glance. But I hope to get to this point.

ZapperZ: My mistake aside, from the FBD alone, the particle would have to be in motion or impending motion. This is where you lose me.

Thanks again:)
Sorry for the mistake:(

 

[Jurtinus]

Indeed, the problem is one where Coulomb’s law is not clearly applicable. I assume that in the broader context of the chapter it is obvious, but I agree with your objection in the specific context of this problem.

Often simplified laws are applied even when the exact assumptions are violated. In such a case it is important to understand when you can get away with the approximation. Chapter 3 below is a good explanation

http://web.mit.edu/6.013_book/www/book.html

New to PF, not sure if you've been notified. Check above.

 

[Jurtinus]

Those are the SOURCE charges. They are the two charges on the two ends. The TEST charges is the one in the middle, because that is the charge where you are calculating the force for!

The two source charges are NOT MOVING, resulting in electric field that is STATIC WITH TIME atthe middle test charge. The middle test charge still has a NET FORCE. Nothing is moving with respect to time in that problem.

When will this sink in?

Zz.

New to PF, not sure if you've been notified. Check above.

 

[Dale]

Dale: I'm unsure wether your reply would change.

Yes, it does change. The additional information clearly specifies that the charges are at fixed locations.
Coulomb’s law clearly applies as specified.

The FBD seems incomplete.

 

[Mister T]

There is a net force, the particle isn't in equilibrium.

True.

1) How can there be a net force if the particle is in equilibrium?

The particle is not in equilibrium.

2) If the particle is not in equilibrium, how does Coulombs law apply?

Because the particles are not moving.

 

[Jurtinus]

Yes, it does change. The additional information clearly specifies that the charges are at fixed locations.
Coulomb’s law clearly applies as specified.

The FBD seems incomplete.

What if it is not. Could Coulombs Law alone suggest the impending motion of a particle at an instant, without the need of additional conditions? Such as Coulombs dry friction, covered in engineering statics.

Thanks:)

 

[ZapperZ]

ZapperZ: My mistake aside, from the FBD alone, the particle would have to be in motion or impending motion. This is where you lose me.

Thanks again:)
Sorry for the mistake:(

The FBD has nothing to do with the particle being in motion or not. The FBD shows the forces acting on the particle at that particular instant at that particular location.

If you let the situation continues, then the particle will move, and now, we no longer have the same situation as before because at a different location, the forces will be different. Furthermore, a moving change in a changing electric field (the electric field can be different at different locations) has other complications added to it that is NOT covered here.

You will deal with uniform electric field soon enough, and you'll see the case of charged particle moving in such a field. In that case, then you have the same kinematical equation that you had when dealing with particles in a uniform gravitational field.

Zz.