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Potential Energy of a Charge

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data
    The electric field in a particular region of space is found to be uniform, with a magnitude of 700 N/C and parallel to the +y direction.

    (a) What is the change in electric potential energy of a charge [q = 6.5 mu or micro CC if it is moved from [(x, y) = (20 cm, 45 cm)] to [(5 cm, 30 cm)]? Answer in mJ.

    (b) What is the change in electric potential energy if the charge is moved the same distance along the x axis? Answer in mJ.

    Are my attempts correct?


    2. Relevant equations
    q*E*d



    3. The attempt at a solution
    (a) (6.5e-6)*700*(.015) = .06825 mJ
    (b) (6.5e-6)*700*.015*√(2) = .09652 mJ
     
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  3. Jan 30, 2014 #2

    SammyS

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    (a) looks right.

    (b) is not right. How much work must be done on the particle in part (b) ?
     
  4. Jan 30, 2014 #3

    collinsmark

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    15 cm is 0.15 m, not 0.015 m.

    By the way, I have to ask, where did you come up with the distance (I presume 15 cm)? I'm not necessarily saying your method was right or wrong, but there is one right way to get this.

    That doesn't look correct to me (which is the reason I asked the question above).

    The work done on the particle is [itex] W = q \vec E \cdot \vec d [/itex]. That dot isn't just a simple multiplication sign, it is the vector dot product.
     
  5. Jan 30, 2014 #4
    I got the distance from subtracting 45cm by 30cm
     
  6. Jan 30, 2014 #5

    collinsmark

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    Okay, that's the right approach! :approve: You used the difference in the y-component, which is correct, since the y-component is the one along the direction of electric field. (Well, actually it's opposite the direction of the electric field, but since we're talking about the particle's gain of potential energy rather than kinetic energy, the positive sign works out).

    [Edit: Oh, and by the way, don't forget to redo the calculation with 0.15 m instead of 0.015 m.]

    Now what about part (b)? What's the dot product in that case?
     
  7. Jan 30, 2014 #6
    Awesome! i was able to fix A properly. Thank you!

    b)The answer from A*.15?
     
  8. Jan 30, 2014 #7

    collinsmark

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    Sorry, that's not it either.

    Consider two vectors (and I'll just use 2-dimensions for this)
    [tex] \vec A = A_x \hat x + A_y \hat y [/tex]
    [tex] \vec B = B_x \hat x + B_y \hat y [/tex]
    (Some textbooks use [itex] \hat \imath [/itex] instead of [itex] \hat x [/itex], and [itex] \hat \jmath [/itex] instead of [itex] \hat y [/itex].)

    The dot product is
    [tex] \vec A \cdot \vec B = A_x B_x + A_y B_y [/tex]

    For this problem (part (b)), there are two vectors [itex] \vec E [/itex] and [itex] \vec d [/itex]. The vector [itex] \vec E [/itex] points along the y-axis and [itex] \vec d [/itex] points along the x-axis.

    So what are [itex] E_x, \ E_y, \ d_x [/itex] and [itex] d_y \ ? [/itex]
     
  9. Jan 31, 2014 #8

    collinsmark

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    If you're not comfortable with that approach, here is another. One can also say,
    [tex] \vec A \cdot \vec B = AB \cos \theta. [/tex]
    You know that [itex] \vec d [/itex] points along the x-axis (for part (b)). And you know [itex] \vec E [/itex] points along the y-axis. What is the angle θ between the x- and y- axes?

    So what does that make [itex] E d \cos \theta [/itex] ?
     
  10. Jan 31, 2014 #9
    Would theta be 0 degree or 6.41 degree? I got 6.41 by (45/20)^2+(30/5)^2=c^2 => sqrt(41.0625)=>6.41 degree. Can I do that?

    To plug into that equation it would be (.6825*10^-3)*.15*6.41=10.158 mJ
     
  11. Jan 31, 2014 #10

    collinsmark

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    Maybe we are not interpreting part (b) of the problem statement in the same way. Recall part (b):
    "(b) What is the change in electric potential energy if the charge is moved the same distance along the x axis?"​
    Personally, I interpret that as saying the charge is moved the same total distance as it was in part (a), but with a direction that is completely along the x axis. In other words, there is no displacement along the y axis or z axis. The only displacement is along the x axis.

    Is that not the way you interpret part (b) the problem statement? Maybe I'm interpreting it incorrectly. What do you think?
     
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