Net force exerted by two charges on a 3rd charge

AI Thread Summary
The discussion revolves around calculating the net force exerted by two point charges on a third charge in a coordinate system. The charges are q1 = +9.00 µC, q2 = -5.00 µC, and q3 = +5.00 µC, positioned at specific angles. The initial approach incorrectly assumed that the forces from q2 and q3 canceled each other out, leading to a net force calculation of 0.48 N along the x-axis. Clarification is needed on how to incorporate both horizontal and vertical components of the forces, emphasizing the importance of using trigonometric functions to resolve the forces correctly. Understanding these components is crucial for accurately determining the net force and subsequent acceleration of the third charge.
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Homework Statement



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)


Homework Equations



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

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 :(
 
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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)
 
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