Help with Coulombs law Physics?

Question

The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +2.5 µC, x1 = 4.0 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.
i got that to be 24.2 N
(b) Find the direction of this force.
° (counterclockwise from the +x axis)
(c) At what coordinates should a third charge q3 = +4.0 µC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?

Answer 1

a) Be careful with rounding; I got 24.3 N.

b) Since the charges are unlike, the force is attractive; the force on q2 acts towards q1. If you plot the points, you'll see that q1 is to the right and below q2, so we already know the angle will be between 270 and 360 degrees. Using basic trig, we find that the reference angle is arctan(|Δy| / |Δx|) = arctan(1 / 6) = 9.46 degrees. We need to subtract it from 360 to get the right orientation, so the direction is 360 - 9.46 = 350.54 degrees.

c) From your answer to a), it's clear that you understand Coulomb's law. You need to assume q3 is at an unknown point (x3, y3) and set up Coulomb's law to calculate the magnitude of the force due to each charge. Then, use the method from b) to also determine the direction of each force; the positively charged q3 will be repelled from q1 and attracted to q2. You'll need to use the angle to split the forces into components, and then solve for forces in the x-direction summing to zero and forces in the y-direction summing to zero. With two equations, you can solve for the two unknowns x3 and y3.

Answer 2

This problem could have been harder with arbitrary vectors. As it is it is just one dimensional along the x axis. We will use symmetry and save some vector headaches. After all the form of the equation you stated is the scalor form. From symmetry we know that the y and z components of force is zero. We will define a scalor force that is positive if pointing toward the positive direction of the x axis. Q1 will be the one that is at the origin. The rest of the charges will combine to form Q2. When we sum we will flip the sing of the component charges if they are on the negative side of the x axis. Also since opposite charges attract I will make the whole equation negative. _> F = -k Q1 (Qa/da^2 + Qb/db^2 ...) _> F = -k 8e-6 (7e-6/.05^2 - 4e-6/.04^2 - 8e-6/.05^2) Notice that the 7uC charge is positive in the equation because it is on the negative side of x axis.