# I Electric potential and work

1. May 26, 2018

### mk9898

Hello,

I have a question on electric potential, potential energy and work.

$\Delta V = \frac{\Delta U}{q_o}$
$W = q_o\Delta V$
$W = -\Delta U$

Example:
A point A that is closer to the source charge than point B, then the electric potential difference is negative ($\Delta V = V_B - V_A$)
When we now have a negative test charge that moves from A to B. If a negative test charge is moved through a negative potential difference, the potential energy is positive. But then that means that the work is both positive and negative given the two equations above. Can someone explain the problem here?

2. May 26, 2018

### Staff: Mentor

Your equations are not self consistent. Can you spot the inconsistency?

3. May 26, 2018

### mk9898

I just looked at the equations in University Physics and Serway's book and they both have these equations.

4. May 26, 2018

### Staff: Mentor

It doesn’t matter where you got them. They are inconsistent. Can you see why?

Last edited: May 26, 2018
5. May 26, 2018

### mk9898

Yea I can. But why the inconsistency? Which are correct?

6. May 26, 2018

### Staff: Mentor

It is very common for W in one equation to mean the work done by a system and for W in another equation to mean the work done on a system. In this case, the second equation is the work done on a charge, and the third equation is the work done by the charge.

7. May 26, 2018

### mk9898

Ah got it thank you. Is there a rule of thumb of how to remember these? I had the same problem learning thermodynamics and it was never really explained thoroughly. Especially when some exercises don't necessarily specific which work they are speaking of. Is there a "go-to" case or should I always try to understand from what perspective the work is being done?

8. May 26, 2018

### Staff: Mentor

I think that you need to look at it in each case. Assume that W is positive and then look if the energy of the system has increased or decreased. If it increased then W is work done on the system. If it decreased then W is work done by the system.

9. May 26, 2018

### mk9898

Wait so now I'm confused. I see now four cases:

1. Work done on a charge
2. Work done by the charge
3. Work done on the system
4. Work done by the system

Work can also increase when we consider the charge so which ones of the cases are the same?

10. May 26, 2018

### Staff: Mentor

Think in terms of systems (so you can delete 1 and 2). You can make the system be just the charge, but you can make the system be something more complicated if needed.

11. May 26, 2018

### mk9898

Ok so when work is done on the system, then we consider the test charge and work by the system we consider the source of the electric field.

12. May 26, 2018

### mk9898

Per the photo: so could we also see this as the charge does work on the electric field when we consider the perspective of the charge? My apologies if I am being slow here but this topic of work and negative and positive signs have been confusing me for quite some time and I really want to put the nail in the coffin. If you or anyone knows where I can find a detailed explanation, I would be indebted if you could share it.

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13. May 26, 2018

### Staff: Mentor

There isn’t any secret here. This is completely general.

Simply define what your system is. Write your equation W=whatever. Assume that W is positive and see if the energy of your system increases or decreases. If a positive W makes the system’s energy increase then W is work done on the system.

It doesn’t matter if the system is a charge, a field, a car, or a pot of water.

14. May 26, 2018

### mk9898

Makes sense thanks. So if I write this:

$W = -q_o\Delta V$
$W = \Delta U$

Would the change in the signs change the system I am referring to?

For the example a ball thrown up in a gravitational force, we consider the the earth as the system and the force doing work on the ball. So would the system earth be $W=Fds$ and the system of the ball would be $W = -\Delta U$?

15. May 26, 2018

### sophiecentaur

There is a secret! And the secret is to go over it so many times in your mind that it works and makes sense, whatever the wording that you find in the various sources. I don't think there is any hope that one can write out a set of rules about which way round applies where.

16. May 26, 2018

### Staff: Mentor

No. It would only change whether W represents work done on or by the system. There is no need to change the system either way.

It seems like you have a misunderstanding of how to use the concept of a system. The system is the thing you are interested in. You are free to choose its boundaries, so you are free to choose what you are examining. Once you do that you apply the laws of physics and keep track of what forces or energy or heat or matter crosses the boundary.

So usually the ball would be the system because usually that is what you are interested in. You keep track of work done on the system and its energy and so forth. You typically don’t calculate the energy of the earth, although you could if you really wanted to.

Last edited: May 26, 2018
17. May 26, 2018

### Mister T

You are using two different definitions for $W$. A very common source of confusion.

18. May 28, 2018

### mk9898

And is there some resource that alleviates this confusion? Or do all textbooks just skip over this and leave students not really understanding it.

19. May 30, 2018

### Mister T

Ahhh... Well, the big challenge (as far as I'm concerned) is making the definition of $W$ that's used in mechanics consistent with the one that's used in thermodynamics. Some of the college-level introductory physics textbook authors ignore the issue and some bugger it all up (again, as far as I'm concerned) trying to address it.

Notice, in particular, the way friction is dealt with. Sometimes authors will claim that when the mechanical energy of a system decreases it's because negative work is done by friction, but that use of $W$ is not consistent with the way $W$ is used in thermodynamics.

Take, for example, a block sliding across a level table top at a steady speed. There is no change in kinetic energy. Someone is pushing on the block with a force of 10 N while the block moves a distance of 1.0 m. All will agree that the person does 10 J of work on the block. But how is one to explain that the block's kinetic energy doesn't change? One way is to say that friction does -10 J of work on the block, so the net work done is zero. This is a perfectly valid way to handle the dynamics of the situation, but it cannot be a valid way to handle the thermodynamics of the situation. After all, the internal energy of the block-table system increases by 10 J (the two surfaces get warmer at the locations where they make contact) and that energy has to come from somewhere. It does indeed come from the 10 J of work done by the person! So, from a thermodynamic perspective, the friction does no work at all.

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