# Homework Help: Point Charge Problem

1. Jan 15, 2008

### azila

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. Jan 15, 2008

### silver-rose

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...

3. Jan 15, 2008

### azila

oh so it doesn't matter that they have the same charges... is that right?

4. Jan 15, 2008

### azila

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..