Question

Two particles are fixed on an x axis. Particle 1 of charge 57.8 μC is located at x = -12.7 cm; particle 2 of charge Q is located at x = 16.3 cm. Particle 3 of charge magnitude 40.6 μC is released from rest on the y axis at y = 12.7 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

Answer 1

here's the method.

Start with a clear diagram.
Angle formed between pt1 and pt 3, A = tan-1(12.7/-12.7) = -45 or 315
Angle formed between pt2 and pt 3, B = tan-1(12.7/16.3) = 37.923
Distance from pt1 to pt3, R = sqrt(12.7^2 + 12.7^2)= sqrt(322.58) (Note, you only need R^2 later)
Distance from pt2 to pt3, S = sqrt(12.7^2 + 16.3^2)= sqrt(426.98) (Note, you only need S^2 later)
Call the charges Q1, Q and Q3.

If the initial acceleration of particle 3 is along the positive x-direction, then the y components of F13 and F23 must cancel.
(Q1.Q3/R^2)sinA = (Q.Q3/S^2)sinB
Q3 cancels. You know all the values except Q. Simplify and calculate Q.
Q = Q1*S^2 *sin A /(sinB*R^2) = 88.02 x 10^-6 C

If the initial acceleration of particle 3 is along the positive y-direction, then the x components of F13 and F23 must cancel.
(Q1.Q3/R^2)cosA = (Q.Q3/S^2)cosB
Q3 cancels. You know all the values except Q. Simplify and calculate Q.
Q = Q1*S^2 *cos A /(cosB*R^2) = 68.579 x 10^-6 C

Note Q3 can have any value. The direction of the field at point 3 does not depend on it.