1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Point Charge Problem

  1. Jan 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Two Positive charges q1 = q2= 2.0 microC are located at x=0, y = .30 m and x =0, y =-.30m. Third point charge Q =4.0 microC is located at x=.40 m, y = 0. What is the net force magnitude and the direction on charge q1 exerted by the other two charges.

    2. Relevant equations
    coulomb's law

    3. The attempt at a solution

    Ok, so this is what I did. I drew the charges and then for the point charge Q acting on q1, i drew out the components and everything. So, first I just solved for the F(Q on q1)=k(Qq1)/(.50^2). I found the .50 through the pythagorean theorem. So I get .29 N. Then for the components. I solved that by having the x component be negative since cos is negative in the second quadrant and y component is positive since sin is positive in the second quadrant. I got the respective values to be -.17 and .23. Then I did F(q2 on q1). There is no x component and there is only a y component. The y component through coulomb's law, comes out to -.1 since they both are positive and so they repel. So, the Fx is -.17 and Fy is .23-.1 which is .13 N. I found the resultant to be .21 and theta to be 37 degrees. However, this is not the right answer. Any thing that I am doing wrong??Or am i not even following the steps correctly..I would appreciate any help. Thanks..
  2. jcsd
  3. Jan 15, 2008 #2
    Force is a vector. If you take it that a force in the positive y direction is positive (as you have done so when finding F(Q on q1), then the F(q2 on q1) is also positive, as it exists in the direction of the positive y axis...
  4. Jan 15, 2008 #3
    oh so it doesn't matter that they have the same charges... is that right?
  5. Jan 15, 2008 #4
    ok, i got the magnitude but the angle is not working. I get 63 and then i subtracted from 90 to get Counterclockwise, to be 27 but it does not work..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook