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Homework Help: Work Needed to Move a Charge

  1. Feb 6, 2008 #1
    [SOLVED] Work Needed to Move a Charge

    1. The problem statement, all variables and given/known data
    Three identical point charges Q1, Q2, and Q3 all having charge 3 microCoulombs are located at the vertices of an equilateral triangle with sides s=3 m. How much work would I have to do to move Q1 to a point on the line connecting Q2 and !3 (which are fixed) if this point is a distance .4 s from Q2 and a distance .6 s from Q3?

    http://img129.imageshack.us/img129/1091/trianglejp1.th.gif [Broken]
    (Picture obviously not drawn to scale)

    2. Relevant equations
    W = Fd = qEd
    E = kQ/d

    3. The attempt at a solution
    I am using q = 3*10^-6 Coulombs in all of my attempts. The thing that is throwing me off is what to use as the distance and electric field.

    I did two different approaches:

    Attempt 1
    First I tried to find the distance between Q1 and its destination point. I just made a right triangle:
    http://img98.imageshack.us/img98/1240/88380904mi7.gif [Broken]
    And the hypotenuse was sqrt(2.34).


    E = kq/d
    E = (9*10^9)(3*10^-6)/sqrt(2.34)
    E = 17650.45216 N/C

    So I plugged in E to the formula for Work

    W = qEd
    W = (3*10^-6)(17650.45216)(sqrt(2.34))
    W = .081 J

    Attempt 2
    I tried to calculate the electric field at the point Q1 between Q2 and Q3. For distance i used 1.2 (.4*3m) and 1.8 (.6*3m).


    E_tot = E1 + E2
    E=kq/d^2 (electric field at a point between two charges)

    E2=-8333.34 (rounded up, because it was .3 repeating, negative because the direction of the field that acting on Q1 is to the left)

    E_tot = E1+E2
    E_tot = 18750 - 8333.34 = 10416.67


    W = qEd
    d will remain the same as in attempt 1.
    W = (3*10^-6)(10416.67)(sqrt(2.34))
    W = .0478 J


    There you have it. Neither of them were the right answer. Please help. I know I am close but like I said, I am confused as what to use for E and d. Thank you.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 7, 2008 #2

    Shooting Star

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    Homework Helper

    Work done = difference in PE between final position and initial position.

    You know the Potential at the initial and final posn due to Q2 and Q3.
  4. Feb 7, 2008 #3
    UE at initial point:

    UE = k(3*10^-6)^2/3 = .027
    Since that is the same for both charges, the answer is .054.

    UE at bottom:

    UE = k(3*10^-6)^2/1.8 = .045
    UE = k((3*10^-6)^2/1.2 = .0675
    Tot = .1125

    UE_ini - UE_final = .1125-.054 = .0585 J Which was the correct answer.

    Thanks a bunch! I really appreciate the quick reply. That helped a lot. :)
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