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Charge problem, used a quadtratic equation, can you see if i did it right?

  1. Sep 6, 2005 #1
    Hello everyone. I need to see if I did this right. the problem is this:
    Particle 1 of charge 1.0E-6 C and particle 2 of charge -2.0E-6 C are held at seperation L = 13.0cm on an x axis. If particle 3 of unkonwn charge q3 is to be located such that the net electrostatifc force on it from particles 1 and 2 is zero what must be the coordinates of particle 3?

    Here is my drawing and work:
    http://img231.imageshack.us/img231/2886/quad5nx.jpg [Broken]

    So would the answer be X = 31.385 cm Y = 0 cm?

    Also another quick question...
    http://img76.imageshack.us/img76/510/22115ph.gif [Broken] shows four situations in which charged particles are fixed in place on an axis.

    In which situations is there a point to the left of the particles where an electron will be in equilibrium? (Select all that apply.)
    situation b and a were the only ones that made senes to me. Does that seem right?

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 6, 2005 #2


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    Homework Helper

    with the smaller charge q at x=0, and the -2q at x=13m,
    the E-field contributions cancel farther from the -2q.
    How much farther? sqrt(2) farther than from the small q.
    (so ratio 2q/(sqrt(2)*d)^2 = 1)
    (d+130mm) = 1.414 (d) , so .414d = 130mm
    no quadratic needed
  4. Sep 7, 2005 #3
    My professor told me to use the quadratic, so my solution is incorrect? If i convert your 130mm to cm's it doesn't come even close to my answer. If you would use my method, what did i screw up? thanks for the reply
  5. Sep 7, 2005 #4

    Doc Al

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    Staff: Mentor

    Your solution is correct.
  6. Sep 7, 2005 #5
    Thanks Doc, quick question, Is it suppose to be negative answer or postive answer? becuase I placed the charge on the negative x-axis. and it is wanting the Y coordinates and x coordinates. Y = 0 cm, but would X = -31 or X = 31?
  7. Sep 7, 2005 #6
    I believe you are perfectly correct with your answer however your method was over-laborious. If you had cancelled the 10^6 and expressed d in terms of a separation you could get the quadratic terms out much more quickly

    the roots i got in the form 13(1 + sqrt(2)) and 13(1 - sqrt(2)) confirm what you got.

    note however that it the co-ordinate system you sketch the charge is at (-31.385,0)
  8. Sep 7, 2005 #7

    Doc Al

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    Staff: Mentor

    As gnpatterson already explained, if you put the positive charge at X = 0 and the negative charge at X = 13 cm, then the coordinates of your answer will be X = -31 cm.

    In your equation, d stands for the distance from the charge at X = 0. Once you found the solution of the equation, it is up to you to translate the answer into the format required: a coordinate along the x-axis.
  9. Sep 7, 2005 #8
    thanks guys!
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