Particle acceleration in electrical field

AI Thread Summary
The discussion focuses on calculating the acceleration of a charge placed at the origin due to two other charges located on the x and y axes. The user correctly identifies the need to use Newton's second law (F=ma) and Coulomb's law to find the forces acting on the charge. They express confusion about vector addition and whether to use calculated forces or side lengths for their calculations. The user has computed the forces as 3.03 N and 1.05 N but is unsure how to combine them. They also calculate the angle for part B as 33.7 degrees, indicating an understanding of the geometric relationship between the forces.
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Homework Statement


A charge is placed on the x-axis (q=+7.00 \muC, x=0.600 m), and another charge is placed on the y-axis (q=+9.00 \muC, x=0.400 m). A third charge (q=-6.00 \muC, m= 5.00 x 10^{-8} kg) is placed at the coordinate origin. If the charge at the origin were free to move, what would be (a) the magnitude of its acceleration and (b) the direction of its acceleration? Specify your answer in apart (b) as an angle relative to the +x axis.

Homework Equations



Okay, so I know that \sumF = ma, so a= sum(F)/a

Maybe this is relevant: F=k\frac{q1q2}{r^2}

The Attempt at a Solution



First off, I just started the second semester of General Physics. I took the first semester of it 2 1/2 years ago.. So some of the real basic stuff is fuzzy. I know essentially what the question is asking and how to work it. My problem seems to be with the addition of the vectors.

I have set it up as a right triangle (see attachment), so wouldn't I just use the Pythagorean theorem? And if so, do I use the calculated forces or the lengths of the sides? I calculated the F12 to be 3.03 N and F13 to be 1.05 N. After this, I am kind of lost.

And just to be sure for part B, I did theta = arctan(.400/.600) = 33.7 degrees.

Thanks!
 

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these forces act at 90 degree to each other take the vector product n u'll get total force
 
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