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B Negative work and electric potential energy

  1. Nov 20, 2017 #1

    Abu

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    I am confused how a charge could have negative work done.

    To clarify, I was doing a problem earlier in which a positive charge and negative charge are moving towards each other. I used the equation
    work = Δv * q

    And when I was doing this, the change in electric potential, Δv, was negative, and the charge i was moving, q, was positive. That means the work was negative. Let's say that it was -5 Joules.

    I realize that this work, -5 Joules, is the change in electric potential energy. I assume it is negative because when the two unlike charges move together, no work is required because the two are moving together naturally, which means the electric potential energy is transferring to kinetic energy, thus making a negative change in electric potential energy because it is decreasing.

    However, lets say I wanted to use this work value, -5 Joules, in the formula w = f * d to find the force for the distance d that the charge traveled. Do I choose to use -5 Joules or positive 5 Joules? I am conflicted because:
    1. The charge is moving in the direction of the attractive force between it and the negative charge, thus making me want to choose the positive 5 Joules

    2. No work was required to even move the positive charge to the negative charge, so it could be wrong to use w = f*d

    3. The charge had a w value of -5 Joules, so if I just used that directly for the w = f * d equation, my force would be negative which does not make any sense at all.

    I guess what I'm really asking is:
    Is work done when a positive charge and a negative charge move together, and if so, is that work positive or negative?

    I realize that my question might be extremely confusing, if you want me to explain something that I said I will be more than happy to.

    Thank you
     
    Last edited: Nov 20, 2017
  2. jcsd
  3. Nov 21, 2017 #2

    Doc Al

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    Staff: Mentor

    The work done is positive when the object moves in the same direction as the applied force. That's certainly the case for charges attracted to each other and moving toward each other.

    And that should make sense, since the kinetic energy of each charge increases.
     
  4. Nov 21, 2017 #3

    Abu

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    Okay thanks! So if I get -5 joules for the
    change in electric potential energy, I should use positive 5 joules in the work formula, w = f*d ?
     
  5. Nov 21, 2017 #4

    Doc Al

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    Staff: Mentor

    For this sort of situation I would not use that work formula. (For one thing, the force varies with distance.) Instead I would treat the two charges as a system. And since the forces are internal, mechanical energy is conserved: So ΔPE + ΔKE = 0. If the change in PE is -5 Joules, the total KE increases by 5 Joules.
     
  6. Nov 21, 2017 #5

    Abu

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    So if the question was how much work is required for a negative charge and positive charge to move together to a distance d apart, the work that you would have to do would be zero, correct? Because the natural movement for those two charges is to come together. But the attracting force does work though, so imagining that the change in electric potential energy was -5 joules again, does that mean that the attractive force does positive 5 joules of work to move the two opposite charges to a distance d apart, while you yourself do 0 work?

    Thank you for your patience
     
  7. Nov 21, 2017 #6

    Doc Al

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    Staff: Mentor

    Sure, if you just separate the charges and let them go, then you are not even involved and certainly do no work on them. But if you slowly bring the charges together, you will have to do negative work to keep the charges from speeding up. (Example: What work do you do when lowering a book some distance at uniform speed?)

    Yes. The forces between the charges do positive work, adding KE. And you have nothing to do with it, since you let them go.
     
  8. Nov 21, 2017 #7

    Mister T

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    Science Advisor

    ##W=-\Delta U##. Assuming that ##W## is the work done on the particle by the electric field, as is the usual convention.

    But, as others noted, the force would have to be constant to be able to say that ##W=Fd##.
     
  9. Nov 21, 2017 #8

    Abu

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    Oh. I had no idea that the formula had a negative sign. My class excluded that from the formula. Is there a reason why some choose to use the formula without the negative sign?
     
  10. Nov 21, 2017 #9
    Mister T is correct. W=−ΔU. That is the definition of potential energy
     
  11. Nov 21, 2017 #10

    Mister T

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    Oh, yeah, sorry about that. If ##W=q\Delta V## then ##W=\Delta U##. Therefore you are using the convention where ##W## is the work done by the particle. A rather strange, but valid, way of doing things.

    Pose the following scenario to your instructor. A ball falls from ceiling to floor. Is ##W## positive or negative? Because the gravitational potential energy decreases, making ##\Delta U## negative. So if your instructor says ##W## is positive then he's defining ##W## to be the work done on the ball, and ##W=-\Delta U##. Note that as the ball falls the force and displacement are positive, and that makes ##W## positive!
     
  12. Nov 21, 2017 #11
    Basically, there is nothing wrong with negative work. Confusion arises if you don't use the correct definition for work. W = Fd is not correct. It is valid under only very specific cases. If you don't know vectors, then the definition of work, FOR A CONSTANT FORCE, is W = Fd cosθ, where θ is the angle between the force and the displacement. For example, if a shopping cart is rolling down an incline, and you want to stop it, by exerting a force in the direction opposite to its motion, then the angle between the force and the displacement is 1800, and the work done is negative.
     
  13. Nov 21, 2017 #12

    Abu

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    Thank you everyone for your replies and I apologize for the late response. I really appreciate the effort you guys put in to helping others understand these physics topics. I am currently in the process of digesting all of this useful information and will follow up with my instructor about what convention he prefers. I'll make another comment on this same thread if I have any other questions about it.

    But really I appreciate everyone's help, it means a lot.
     
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