(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Coulomb's Law

The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +3.5 µC, x1 = 3.0 cm, y1 = 0.50 cm, and q2 = -4.0 µC, x2 = -2.0 cm, y2 = 1.5 cm.

At what coordinates should a third charge q3 = +5.5 µC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?

I set both Forces together, F13= F23. When i simplified i i get the equation (q1/x^2)=(q2/(x-l)^2). I used the distance formula to find l and i came up with the answer .026.

When i solve for x i get .0126m or 1.26 cm. I understand i have to use sin and cosine to get the answers. I used the formula x= x2-x1cos -11.3 and the same for y using different numbers. X1 is the value i just solved for. by doing this i get an answer of -3.57 for x and -1.686 which are incorrect. I was able to find the angle by using the equation theta=Tan-1((y2-y1)/(x2-x1). i know that this value is correct because the system i used called webassign checks the answers immediately. Can someone please tell me what i did wrong.

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# Homework Help: Coulomb's Law and forces

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