B Why the work done is negative when bringing 2 opposite charges together?

Hawkingo

we know that if the applied force is in the direction of the displacement then work done is positive.But in case of bringing 2 opposite charges from infinite to a certain distance,the work done is negative even the force and the displacement of the charge is in the same direction.

From mathematical point of view it says the applied force is positive here and the (dr) is negative here.but what's the physical significance of the sign is here?and if so when can we say the applied force is positive or negative(in case of bringing two same charges together)?

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phinds

Gold Member
the work done is negative even the force and the displacement of the charge is in the same direction.
The displacement is the same but the force is opposite. You have to expend force AWAY from the center point to keep them from coming together

• Hawkingo

Drakkith

Staff Emeritus
2018 Award
But in case of bringing 2 opposite charges from infinite to a certain distance,the work done is negative even the force and the displacement of the charge is in the same direction.
The work done by what on what. Always keep those two things in mind. The work done by the attractive force between them is positive, since $W=ΔKE$, where KE is the kinetic energy of the object the force is acting on and this kinetic energy is obviously increasing. However, if you're the one 'holding' each charge and trying to bring them together, you don't actually have to perform positive work since they want to come together anyways. You'll actually have work performed on you in the form of accelerating your hands, so the work you do on the charges is negative. You gain energy instead of expending it.

• Hawkingo

sophiecentaur

Gold Member
The work done by what on what.
It's always a source of confusion but there's no problem if you stick to the rule. The definition involves work done ON a charge in moving it. Hence, for an attractive system, bringing a charge towards another charge involves negative work put in - hence the negative sign when the distance is measured away from the centre. If you over- think this, then you can convince yourself either way so just stick to the definition. It will become second nature to get it right.

Drakkith

Staff Emeritus
2018 Award
The definition involves work done ON a charge in moving it. Hence, for an attractive system, bringing a charge towards another charge involves negative work put in
This is what I was talking about. The work done on the charge by what? An experimenter bringing the charges together would exert a force that is opposite the displacement (after all, if they aren't, then they aren't part of the experiment and it makes no sense to ask what the work done by them is), and so the work would be negative. But the work done by the electrostatic force between the charges is positive, correct?

sophiecentaur

Gold Member
But the work done by the electrostatic force between the charges is positive
If you like, But that's not the definition. I did mention that you can over think this. Work and Energy are like money. Someone gives it and someone receives it but we don't get confused about the sign of a money transaction. In that case, it's somehow 'obvious' - but even with money, it's possible to set up scenarios in which it's not so clear. Con-men exploit this at times.

Drakkith

Staff Emeritus
2018 Award
If you like, But that's not the definition.
Remind me what definition we're using here. I know of several ways of defining work.

sophiecentaur

Gold Member
Remind me what definition we're using here. I know of several ways of defining work.
Work is Force times Distance / (Displacement could be better, perhaps) and that's a vector multiplication.
Work done ON the system by 'the experimenter'. I would say that all experiments involve an input to a system to see the result. Even when we just observe, we identify with part of the system for independent variable and turn it into an experiment.
The work done on a mass as you lower it to the ground is (universally?) agreed as being negative so can't you just take it from there? The force is in the direction of decreasing distance etc. etc.
If you change to asking the work done by the mass ON the experimenter then you get a change of sign and, that would be ok if that were the accepted convention. But I don't think it is.

A.T.

we know that if the applied force is in the direction of the displacement then work done is positive.But in case of bringing 2 opposite charges from infinite to a certain distance, the work done is negative even the force and the displacement of the charge is in the same direction.
You can do positive or negative work on them. It's up to you what force you apply to them.

• sophiecentaur

ZapperZ

Staff Emeritus
2018 Award
As a physics instructor, I often puzzle at why a student cannot make the connection with something that he/she should have learned about and understood from something earlier.

When someone is learning about charges, it is assume that the student has learned basic kinematics, etc. in a General Physics class. So something like this where work done is "negative" should be familiar. For example, lifting a book to a new height means that work is done ONTO the book, i.e. the book gains gravitational potential energy. However, if we simply let go of the book and it free-falls to the ground, the work done ONTO the book is NEGATIVE, and consequently, it loses gravitational potential energy.

This is no different than bringing a negative charge nearer to a positive charge, or bringing a positive charge nearer to a negative charge. Work done by the charge being moved is negative, because it is not doing the work. The field is doing the work, and the field is doing the work ONTO the charge. As a consequence, the charge's potential energy also drops, in this case, becoming more negative (U = qV).

Both situations are analogous conceptually.

Zz.

• sophiecentaur

Drakkith

Staff Emeritus
2018 Award
Work is Force times Distance / (Displacement could be better, perhaps) and that's a vector multiplication.
Okay. Then which part of my posts don't meet this definition?

The work done on a mass as you lower it to the ground is (universally?) agreed as being negative so can't you just take it from there?
It's not just agreed upon, it's required, because the work done by you on the mass is negative since you're applying the force in the opposite direction of the motion.

sophiecentaur

Gold Member
It's not just agreed upon, it's required, because the work done by you on the mass is negative since you're applying the force in the opposite direction of the motion.
We are agreeing here. But the "required" bit is a consequence of how the Potential is defined - i.e., in terms of work done on the system.

As I commented earlier - it is possible to over think this when it's just a matter of accepting the convention.

Drakkith

Staff Emeritus
2018 Award
We are agreeing here. But the "required" bit is a consequence of how the Potential is defined - i.e., in terms of work done on the system.

As I commented earlier - it is possible to over think this when it's just a matter of accepting the convention.
I don't know, Sophie. I don't even know what we're arguing over anymore since we seem to be half agreeing with each other, lol.

• sophiecentaur

A.T.

if we simply let go of the book and it free-falls to the ground, the work done ONTO the book is NEGATIVE, and consequently, it loses gravitational potential energy.
But the book gains kinetic energy. Applying the definition of work to the gravitational force and displacement of the falling book will yield a positive value. So what is doing negative work on the book here?

sophiecentaur

Gold Member
But the book gains kinetic energy. Applying the definition of work to the gravitational force and displacement of the falling book will yield a positive value. So what is doing negative work on the book here?
That's just generating more confusion about which argument / definition to use. In the end you have to choose which way round to specify how Potential is defined. You and I know the answer to that. The fact that the book would accelerate means that Work is done on it. The book takes the place of 'the experimenter', perhaps and the book does negative work. (Nothing is doing negative work ON the book)
Or maybe you wrote your post in order to make the student think?? Drakkith

Staff Emeritus
2018 Award
That's just generating more confusion about which argument / definition to use.
I don't see how. It matches the definition I asked for, and you gave, earlier.

In the end you have to choose which way round to specify how Potential is defined.
This has nothing to do with choosing how potential is defined. It has everything to do with explicitly stating which forces are involved in the question and which ones your interested in.

In fact, the relationship between work and potential energy perfectly supports this. A book that starts with 20 units of potential energy and ends with 10 units (i.e. it's falling) will have W=-ΔU=-U2-U1=-(10-20)=-(-10)=10 units of work performed on it by gravity. Raise it by 10 units of PE and gravity does -10 units of work, just as it should be given the other definitions of work.

The crux of the OP's question is answered, in my opinion, by recognizing that some of these questions and scenarios are vague and not quite technically accurate (such as what does negative work on a free-falling object solely under the influence of gravity). You can choose to say, "Well, that's just convention" and be done with it, or you can acknowledge that it doesn't actually make sense when you try to figure out what's actually happening.

Some questions have implicit and unstated conditions that aren't immediately obvious. For example, if I hold a book in my hand and raise it by one meter, then I've done work on the book obviously. But what if I lower the book by one meter? That implies that:
A.) I let the book free fall that one meter and stop looking at the question once the book hits the one meter mark.
B.) I slowly lower the book, exerting a force the entire way to keep it from free falling.
C.) I let the book free fall, and quickly 'catch it' so that it comes to a halt at the one meter mark.
D.) Some combination of the above.

In scenario A, I would argue that nothing has done negative work on the book. The only force here is gravity, which is certainly doing positive work on the book given every definition of work I've seen.

In the other scenarios, it is easy to see what is doing negative work on the book. You are. Or whatever is exerting the force that slows the book down. Exactly how much work is done on the book depends on if you completely stop the book's motion or not at the one meter mark.

So how does this generate more confusion? It doesn't in my opinion. I think it removes a great deal of the confusion. I think the confusion comes primarily from poorly worded questions, and that explicitly stating which forces are at work and elaborating on what is happening to the object in question would help a great deal.

The book takes the place of 'the experimenter', perhaps and the book does negative work. (Nothing is doing negative work ON the book)
I can't see how the book could do any work on itself.

• A.T.

sophiecentaur

Gold Member
in my opinion, by recognizing that some of these questions and scenarios are vague and not quite technically accurate
That's right.
As far as I can see, there is a mix up between discussing two objects on their own and two objects with an experimenter.
I can't see how the book could do any work on itself.
The book is gaining KE so it is having work done on it as with my example of money, it can be regarded as doing negative work which is the same as receiving energy / work (which is not "doing negative work on itself"). There would be no work done at all if it were not for the existence of the gravitational field; that's where the energy comes from.
So how does this generate more confusion?
The fact that we are still discussing it is a good indication that confusion exists. I don't see why there should be any and it appears that you don't either but this basic question comes up frequently and my point is that there is a disconnect between intuition and the results of following logical steps in the argument. If you accept the conservation of energy thing that we all learned at school and if you apply it rigorously - with the proper use of signs - it's all perfectly consistent. Just like with money transactions.
Work on and work by have just different signs in the same way that going forwards and going backwards can be easily dealt with by using different signs and not different words; the Maths sorts it out.

Drakkith

Staff Emeritus
2018 Award
The book is gaining KE so it is having work done on it as with my example of money, it can be regarded as doing negative work which is the same as receiving energy / work (which is not "doing negative work on itself"). There would be no work done at all if it were not for the existence of the gravitational field; that's where the energy comes from.
I suppose you can say that the book does negative work, but I think it just adds confusion since the book isn't a force and is the object having work performed on it. Is this the convention you were referring to?

Work on and work by have just different signs in the same way that going forwards and going backwards can be easily dealt with by using different signs and not different words; the Maths sorts it out.
My entire point here is not that something with the math or the definition needs to be changed, it's that it's necessary to understand which forces are involved and which ones you're interested in. Because if you don't understand this, then you can't really hope the math will work out since you haven't set the math up correctly in the first place.

• A.T.

A.T.

However, if we simply let go of the book and it free-falls to the ground, the work done ONTO the book is NEGATIVE
Nothing is doing negative work ON the book
You both seem to be contradicting eachother.

sophiecentaur

Gold Member
You both seem to be contradicting eachother.
@ZapperZ is implying that a negative displacement times a negative force produces negative work so I don't agree with him.
But this is getting daft. We've made so many excursions around this that it's very easy to make a slip in a comment. I already gave warnings about over-thinking this.

rcgldr

Homework Helper
The work done by the attractive force between them is positive, since $W=ΔKE$
Or using a frame of reference with its origin at the center of mass of the two body system, and with both objects on the x axis, then the attractive force on the left object is positive and the displacement is positive, while the attractive force on the right object is negative and the displacement is negative, so both components of work done by the attractive force are positive.

rcgldr

Homework Helper
The displacement is the same but the force is opposite. You have to expend force AWAY from the center point to keep them from coming together
Only if the force slows down or stops the rate of approach due to the attractive force. However, if the external force acts on each object in the same direction as the attractive force, the work done by that external force is positive, and the charges accelerate towards each other at a greater rate.

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phinds

Gold Member
Only if the force slows down or stops the rate of approach due to the attractive force. However, if the external force acts on each object in the same direction as the attractive force, the work done by that external force is positive, and the charges accelerate towards each other at a greater rate.
Fair enough.

rcgldr

Homework Helper
Only if the force slows down or stops the rate of approach due to the attractive force. However, if the external force acts on each object in the same direction as the attractive force, the work done by that external force is positive, and the charges accelerate towards each other at a greater rate.
Fair enough.
Even if the force opposes the attractive force, the magnitude is also a factor. Assume an initial state where the objects are a finite distance apart (not infinite as in the OP). If the opposing force is equal in magnitude to the attractive force, then no work is done. If the opposing force is greater than the attractive force, positive work is done (and the attractive force component is negative work).

So whether or not a net force in addition to the attractive force performs negative, zero, or positive work depends on the direction and magnitude of that force.

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vanhees71

Gold Member
I think the problem with this thread, as many in this forum, is that there are too many words rather than a handful of formulae. Math provides clear definitions for the quantities at hand. Let's do the most simple example: consider a single particle moving in some external force field. Then you have an equation of motion, which is usually of the form
$$m \ddot{\vec{x}}=\vec{F}(t,\vec{x},\dot{\vec{x}}).$$
Now it is sometimes of advantage to consider kinetic energy,
$$T=\frac{m}{2} \dot{\vec{x}}^2.$$
Now let's see how it changes with time:
$$\dot{T}=m \dot{\vec{x}} \cdot \ddot{\vec{x}}=\dot{\vec{x}} \cdot \vec{F}(t,\vec{x},\dot{\vec{x}}).$$
In the last step I have assumed that $\vec{x}(t)$ is a solution of the equation of motion. Then you can integrate this equation over $t \in [t_1,t_2]$,
$$T_2-T_1=W_{12} = \int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \vec{F}(t,\vec{x},\dot{\vec{x}}).$$
$W_{12}$ defines the quantity "work", and it's clearly defined as a line integral along the actual trajectory of the particle under the influence of the force. The above equation is known as the "work-energy theorem".

So far, it's not so clear, why this might be of some use. Now for many forces in non-relativistic physics the force is of much simpler structure than assumed above. It may only depend on $\vec{x}$ (the location of the particle) and neither explicitly on time nor on velocity. It then further often can be written as a gradient of a potentil,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Then work becomes independent of the trajectory of the particle! It only depends on the values of the potential at the endpoints of the tractory you integrate over. Thus you know the work done without any reference to the equation of motion or its solution, and the work-energy theorem provides a "first integral":
$$T_2-T_1=W_{12}=-(V_2-V_1),$$
which you can reformulate in terms of the energy-conservation law
$$T_2+V_2=T_1+V_1.$$
This again holds for all solutions of the equation of motion, but you don't need to know any solution to state it. It simply helps to find a solution. Since you can choose $t_1$ and $t_2$ arbitrarily you can simply write
$$E=T+V=\text{const},$$
and call $E$ the total energy of the particle (with $T$ the kinetic and $V$ the potential energy).

With these clear mathematical definitions, there's no more doubt about the sign of work or potential energy differences anymore.

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• sophiecentaur

"Why the work done is negative when bringing 2 opposite charges together?"

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