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Electric potential, Can Someone Check This?

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data

    A particle charge of 7.6μC is in a uniform electric field directred to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest it moves to the right. After it has moved 8cm, the additional force has done 6.5x10^-5J of work and the particle has 4.35x10^-5 J of kinetic energy.

    1. Calculate work done by Electric Field
    2. Find the potential of the starting point with respect to the end point
    3. Find the Magnitude of Electric Field
    • W(F) = 6.5x10^-5J
    • W(E) = unknown
    • Kb = 4.35x10^-5J
    • Ka = 0
    • Ua = unknown
    • Ub = unknown
    • q = 7.6μC
    • r = 0.08m
    2. Relevant equations
    W = Ua - Ub = Kb - Ka
    K = 0.5MV^2
    U = V = 1/4πε * q/r
    E = 1/4πε * q/r^3 * r

    3. The attempt at a solution
    1. Using the work-energy theorem, I can say that the work done by the electric field is 4.35x10^-5J - 6.5x10^-5J.
    2. The potential between 2 points equals the amount of work that would be required to move a unit charge between those points: So I Believe I should just plug in variables for V.
    3. Same as 2. I'll plug in for E, using the -i unit vector to represent the direction.
    Am I over simplifying the equations? This seems not right, but I can't be sure. Thank you!
     
  2. jcsd
  3. May 4, 2015 #2

    BvU

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    1. Agree
    2. plug in where ? What do you get ?
    3. plug in where ? What do you get ?
    No, but I don't see what the last two equations have to do with the problem at hand: you are given an E field and it isn't coming from a point charge.
    Let me "reassure" you: it isn't right
    You're welcome :wink:

    Sorry, I couldn't resist. On a serious note: 2. for 7.6μC you know the energy difference. All you have to do (as you say) is convert to 1 C.
    For 3.: I leave you to it, in the hope that with the insight won in 2. you can figure out 3. by yourself.
     
  4. May 4, 2015 #3
    I'm sorry, maybe I'm naive, but I don't see how converting units will help me in this endeavor... It's 7.6x10^-6 C, but I don't see how that helps me. Especially if my equations were wrong.
     
  5. May 5, 2015 #4

    BvU

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    I suppose I created some comfusion, sorry. In 1. you found how much work was needed and gave an answer. I get the same result.

    In 2. You only say: plug in variables for V. But you don't say where you plug them in, or what else you "plug in", or what you calculate.

    Your notation doesn't help: you have a V in equation 2 and in equation 3. OK, it's not the one in 2. But equation 3 isn't applicable -- aside from it being utterly wrong (check the dimensions). The second = is for a point charge. A point charge doesn't give the E-field you were given in this exercise.

    In the part 'on a serious note' I make a little jump: I figured you had by now realized that it isn't U = V in the second equation, but U = qV.

    One way for you to avoid this kind of thing in the future is to make a list of given/unknown variables with their dimensions and check that the equations you bring in have the correct dimensions. (That's why the template is so extremely useful). It sounds corny, I know. On the other hand you are already doing the right thing most of the way, so it's not much of an extra effort. And after a while it's a mental routine and you don't have to spell it all out. Up to you.
    From U = qV ( with a known q and U ) it's really only a small step to V. And it is not a unit conversion.

    Part 3 is equally easy. I leave it to you. (but please do post... :wink:)

    --
     
  6. May 5, 2015 #5
    Equations 2,3 and 4 are not relevant. You need delta potential energy = qEr in the right units.
    And WF+WE=Kb-Ka or something similar.
    In words: some work is put in by F but the final kinetic energy is less than that amount. The difference is what it took to move the charge against the electric field.
    You can write down all the equations you like but if you don't have a grasp of what is going on your answer will be a gamble.
    This problem is the electromechanical equivalent of the following gravitational problem. You are riding a bike putting in work (ignore the drag). You're going faster so your kinetic energy increases, but your riding up a slope. So you are not accelerating as fast as you would on a flat road. On top of the slope your kinetic energy is disappointing. (But the ride down will be fun). Invent some numbers and ask the student what the slope of the road is.
     
    Last edited: May 5, 2015
  7. May 5, 2015 #6

    BvU

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    Think before you post, please. And give the thread starter a chance to digest the reply.
     
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