# A confusion in work and potential energy

1. Mar 24, 2013

### ehabmozart

This is an abstract of my book " Potential energy is always defined relative to some reference point where U=0. Therefore U represents the work that would be on the test charge q0 by the field of q if q0 moved from an initial distance r to infinity. If q and q0 have the same sign, the interaction is repulsive, this work is positive, and U is positive at any finite separation. If the charges have opposite signs, the interaction is attractive, the work done is negative and U is negative" ... All right, let's take this into small parts. I agree that U is zero at infiinity following the formula U = kq1q2/r .. Now, we know Work done is the negative of delta. U. Taking the second sentence, shouldn't work be done by a force and the electric field does no work. Now, with the first condition when both charges are positive. which work is positive. Which work is actually referred in this text. I assume it is the work done due to the field of let us say q on q0. If so, why is U positive then. When they have opposite charges, I don't get why is the work negative. I need clarification for this confusing text. I am sorry it is a bit long but try being patient with me and thanks a lot to whoever shares his ideas.

2. Mar 24, 2013

### Simon Bridge

If both charges are the same, you need positive work to bring them closer and negative work to separate them. If the charges are opposite, it is the other way around.
Since the force is proportional to the field, it does not matter if we say the work was done by a force or a field.

In the book's somewhat sloppy abstract, q0 is a small test charge and q is the charge whose potential we are investigating. By convention, the test-charge is taken to be positive.

If q is negative, it attracts the test charge, and you have to do work to get it to infinity.
Since the potential energy at infinity is zero, then the potential energy close to q must be less than zero for final-initial to give a positive number.

If q is positive, it repels the test charge: you'd have to do positive work to bring the test charge closer. This reverses all the signs in the above discussion.

3. Mar 26, 2013

### e.chaniotakis

From the definition I know, when the work is done by the field spontaneously then it is positive, else if we do it it is negative.

Therefore for to same charged particles to approach, work is negative (we give our own muscles' energy) and this energy is absorbed by the field as the potential energy between the two particles.

The test particle goes from infinity to a distance R from the source particle. Thus its potential energy goes from 0 to U , leading to -ΔU=0-U=-U
From energy conservation : W=-ΔU =-U <0

If you take it the other way round,
a positive test particle coming from infinity to a distance R from a source negatively charged particle.
The field of the source particle attracts the test particle .
In infinity , U is zero
In distance R, U is negative (attractive potential). We can write U(R) as -|U(R)|

Thus from energy conservation W = -ΔU = 0-(-|U(R)|) = |U(R)| >0 , work is positive.

(That's the way I teach it at least:-) )

4. Mar 26, 2013

### ehabmozart

I just wanted you to identify to me what reference are you taking. You say at the beginning work done. Can you be more specific. Work done from which point to which point?.. Thanks!

5. Mar 26, 2013

### e.chaniotakis

Sure!
Suppose that you have a source particle at x=0
You have your test particle in x=>00 and you want to move it from there to x=R from the source

6. Mar 26, 2013

### ehabmozart

I am sorry but this somehow added to my confusion. Can you specify when you say positive work. Consider two positive charges. By nature they would repel and displacement is away. To keep them together you need to do work against displacement which is negative and not positive as you mentioned!?

7. Mar 26, 2013

### jbriggs444

If you are keeping them together then displacement is zero and work done is zero.

If you are bringing them together then displacement is measured from the starting position to the ending position. This may be either positive or negative depending on how you choose to lay out your coordinates. When you consider the direction of the force, you must adhere to that same chosen sign convention.

8. Mar 26, 2013

### Staff: Mentor

To bring those two charges closer together, the force you must exert is inward (towards the other charge) and so is the displacement. That means the work you must do to bring them closer is positive, which results in a higher potential energy.

9. Mar 26, 2013

### e.chaniotakis

Take the interpretation I gave you and consider that the work one needs to do in order to move a positive charge from infinity to some distance from the source (for two same charged particles) is positive. Else it is negative

10. Mar 26, 2013

11. Mar 26, 2013

### Simon Bridge

Take care to notice who is doing the work.

To move the particles closer together, when they repel, you have to apply a force in the same direction as the displacement. Therefore you are doing positive work.

i.e. a point charge fixed at x=0, the same sign charge at x=a requires a force to move it to x=b<a. The force is in the -x direction and so is the displacement, therefore the work (done by that force) is positive.

However, ΔU is negative - because that is the work done by the field.