How Do Electric Potential, Potential Energy, and Work Interact in Physics?

Click For Summary
SUMMARY

This discussion focuses on the interaction between electric potential, potential energy, and work in physics, specifically addressing the equations ##\Delta V = \frac{\Delta U}{q_o}##, ##W = q_o\Delta V##, and ##W = -\Delta U##. Participants highlight the inconsistency in definitions of work (W) across different contexts, such as mechanics and thermodynamics. A clear understanding of the system boundaries is emphasized, with the recommendation to define the system and analyze whether work is done on or by the system based on energy changes. The conversation concludes that consistent definitions are crucial for resolving confusion in physics problems.

PREREQUISITES
  • Understanding of electric potential and potential energy concepts
  • Familiarity with the equations of work in physics
  • Knowledge of system boundaries in physics
  • Basic principles of thermodynamics and mechanics
NEXT STEPS
  • Study the differences between work done on a system versus work done by a system
  • Explore the implications of system boundaries in physics problems
  • Learn about the relationship between work, energy, and friction in mechanics
  • Investigate how thermodynamic principles apply to mechanical systems
USEFUL FOR

Students of physics, educators teaching mechanics and thermodynamics, and anyone seeking clarity on the concepts of electric potential, potential energy, and work interactions.

mk9898
Messages
109
Reaction score
9
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?
 
Physics news on Phys.org
mk9898 said:
Can someone explain the problem here?
Your equations are not self consistent. Can you spot the inconsistency?
 
I just looked at the equations in University Physics and Serway's book and they both have these equations.
 
mk9898 said:
I just looked at the equations in University Physics and Serway's book and they both have these equations.
It doesn’t matter where you got them. They are inconsistent. Can you see why?
 
Last edited:
Yea I can. But why the inconsistency? Which are correct?
 
mk9898 said:
Yea I can. But why the inconsistency? Which are correct?
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.
 
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?
 
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.
 
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
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
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
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.
 

Attachments

  • Bildschirmfoto 2018-05-26 um 21.35.05.png
    Bildschirmfoto 2018-05-26 um 21.35.05.png
    30.5 KB · Views: 545
  • #13
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.
 
  • Like
Likes   Reactions: mk9898
  • #14
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
Dale said:
There isn’t any secret here.
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.
 
  • Like
Likes   Reactions: mk9898 and Dale
  • #16
mk9898 said:
Would the change in the signs change the system I am referring to?
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.

mk9898 said:
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.
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:
  • Like
Likes   Reactions: mk9898 and sophiecentaur
  • #17
mk9898 said:
Can someone explain the problem here?

You are using two different definitions for ##W##. A very common source of confusion.
 
  • Like
Likes   Reactions: mk9898 and sophiecentaur
  • #18
Mister T said:
You are using two different definitions for ##W##. A very common source of confusion.

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
mk9898 said:
And is there some resource that alleviates this confusion? Or do all textbooks just skip over this and leave students not really understanding it.

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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 6 ·
Replies
6
Views
17K
  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 11 ·
Replies
11
Views
5K