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Electric charge and potential energy

  1. May 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle with charge q experiences a force F when placed a distance d from a point charge. At a distance d, the electric field has a strength E and a potential V. Which of the following represents the potential energy of the charge?
    (A) Fd
    (B) Eqd
    (C) Vq
    (D) Ed

    2. Relevant equations
    U=Eqd; U=kq1q2/r; V = Ed; V = kq1/r.

    3. The attempt at a solution
    I chose B but the correct answer is C. We know that the potential energy of a charge in an electric field is the force times the displacement: U = Eqd. What is the difference between answer choice B and C? Thanks in advance and if I didn't provide enough explanation for my question, please let me know.
     
  2. jcsd
  3. May 20, 2015 #2
    The voltage has units of Volt or, what we are interested in is Joule per Coulomb so V.q has units of Joule, which can be energy, but also B) proposes a result with unit of Joules, so what's the difference ?
    Well, The ELECTRIC potential energy is a potential energy, it just depends on you distance to the source of electric field and by displacing the charge as B) proposes, you increase it's PE, or give it any form of energy, what I mean if d = r (the distance btw the charge) then B) would be true and in either case C) is the most general and the right one, Hope I've made it a little bit clearer and good luck .
     
    Last edited: May 20, 2015
  4. May 20, 2015 #3
    Right. When I look at the units for B and C, I notice they are both in Nm = J. However, can you clarify when to use U = Eqd and U = Vq? My book states that Eqd is the potential energy or work done, which makes sense but I cannot see how they get U =Vq. Thank you.
     
  5. May 20, 2015 #4

    TSny

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    The equation U = Vq is valid in general. The expression Eqd is used for the case of a charge q in a uniform electric field. Usually one writes ΔU = -Eqd where d is the distance the charge q moves parallel to the uniform field and ΔU is the change in potential energy of the charge.
     
  6. May 20, 2015 #5
    Ok. I also noticed that Vq = (J/C)(C), which is just Joules. I understand this by process of elimination. I wonder, did the problem imply a "uniform electric field" in this case? How would I know when the electric field is uniform or not?
     
  7. May 20, 2015 #6

    TSny

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    A point charges produce a field that gets weaker as you move farther from the charge. So, the electric field is not uniform.

    When you have two point charges, you can think of one of the charges as placed in the field of the other charge.

    A uniform field is a special situation. It can be thought of as produced by a large flat plate of uniform charge density. Or,it can be produced by two large, flat parallel plates of opposite charge.
     
  8. May 20, 2015 #7
    Eqd can be visualised as the change in potential, asas TSny said above ΔU =- Eqd, because d isn't necessery the distance btw the charges whereas the potential directly depend on that distance r and if you have noticed Vq = E*q*r that all about it !!
     
  9. May 21, 2015 #8
    Oh I see. I have a similar problem which states the same scenario but the charged particle is moved "from a great distance". In this situation, it would be appropriate to use Eqd, right?
     
  10. May 21, 2015 #9
    Edq is work, it's not allways a potential energy, doing work on a charge from a great distance give it enough kinetic energy to leave the field, E changes by the distance btw the charges, suppose that you're in a gravitation field, the work you do on a mass W = mgd isn't allways equal to the potential U = mgh,
    Imagine that you lift the mass from a 10 meter roof a 5 m upward, this case W = 5mg whereas U = 15mg, that's the difference btw Eqd and Vq, V = E*r it's like gh, just focuse on these, Vq is the potential and Eqd is work, it may not be potential !
     
    Last edited: May 21, 2015
  11. May 21, 2015 #10
    That is so very helpful! Thank you for your help!
     
  12. May 21, 2015 #11

    TSny

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    If will help if you state the problem exactly as given.
     
  13. May 21, 2015 #12
    well basically, voltage(potential difference) is equals to work done per unit charge. That means the amount of work done required to move a charge from one point to another point. That why it is called potential DIFFERENCE . To make things easier, theres an equation where V = W/Q.

    So, if you want to find the potential energy of the charge. Use this formula to get the answer ^_^
     
  14. May 21, 2015 #13
    Thank you. I will post this question in another thread.
     
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