# Coloumbs Law

1. Jul 8, 2008

### Laxman2974

1. The problem statement, all variables and given/known data
The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +3.5 µC, x1 = 1.5 cm, y1 = 0.50 cm, and q2 = -4.0 µC, x2 = -2.0 cm, y2 = 1.5 cm.
(a) Find the magnitude of the electrostatic force on q2. Done I got this part =
94.98868N
(b) Find the direction of this force. Got this also =
-15.9453959° (counterclockwise from the +x axis)
(c) 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?
Can't get this - saw problem solving methods for questions that are very similar in previous postings, but I can't get the math to work out

2. Relevant equations
Got as far as q1/x^2 = q2/(x+r)^2
r being the distance between q1 and q2 since q3 must lie to the right of q1

3. The attempt at a solution
many attempts at a solution - but its been 15 years since I have done significant math and I can't get anywhere now.

2. Jul 8, 2008

### cryptoguy

the equation you gave in part 2 is may not be right since q3 may not be located on the same Y value as q1. The equation would look more like:
$$\frac{q_1q_3}{x^2+y^2} = \frac{q_2q_3}{(r+\sqrt{x^2+y^2})^2}$$ (where x is distance from q1 x-coordinate to q3 x-coordinate and y is dist from q1 y-coord to q3 x-coord)

Now I'm not sure if this is the only way to solve this but I would start with equating the x-components of the forces aka $$F_{1-3x} = F_{2-3x}$$ and then do the same for y.

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