# Net force exerted by two charges on a 3rd charge

1. Sep 3, 2007

### itsagulati

1. The problem statement, all variables and given/known data

The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +9.00 µC; the other two have identical magnitudes, but opposite signs: q2 = -5.00 µC and q3 = +5.00 µC.

For the problem imagine the x-y coordinate grid with a + and - 23 degree line coming off of it. and 1.3m out you get a -5 micro C (this is the +23 degree charge...q2) and a +5 micro C charge for the -23 degree (q3)

2. Relevant equations

i went with F = K q1 q2 / r^2
and then i doubled it since they are identical charges

3. The attempt at a solution

So, i assume the left and the right canceled...obviously wrong, but i am not sure what else to do.

I set up the problem as: 2 * [ 9e9 * 9e-6 * 5e-6/ 1.3^2)] = .48 N and since i assumed it was in the middle (due to the right and left sides canceling out due to the opposite and equal charges) it went straight along the x-axis (0 degrees).

so my answer was .48 N at 0 degrees

and then for the second part it asks for the acceleration of the particle with a 1.3 g mass...

so i took: .48 N / .0013 kg = 369 m/s^2

help please i am lost in physics :(

2. Sep 3, 2007

### learningphysics

If the horizontal components cancel... then you add the vertical components... you didn't use the vertical component... you used the net force... you should have 2q1q2/r^2 * cos(theta) (or maybe it's sin theta... I'm a little confused as to the orientation of the problem)