1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

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

    Also another quick question...
    Figure 22-11 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: Sep 6, 2005
  2. jcsd
  3. Sep 6, 2005 #2


    User Avatar
    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

    User Avatar

    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

    User Avatar

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

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