Question

The charges and coordinates of two charged particles held fixed in the xy plane are: q1=1.40 μC, x1=-5.100 cm, y1=7.200 cm, and q2=9.00 μC, x2=-5.215 cm, y2=10.498 cm, Find the magnitude and direction of the electrostatic force on q2.

What are the x and y coordinates of a third charge q3=7.00 μC, if the net electrostatic force on q2 is zero?

Answer 1

The Coulomb's Law reads

F = q1 q2 /(4 pi epsilon0 r^2) u

where

F electrostatic force between the two particles
epsilon0 -> vacuum permittivity = 8.854 e-12 (approximatively)
q1, q2, charges of the two particles
u unit vector orientated along the vector M1M2 joining the two particles
r distance between the two particles

To know the sign of u, just remember that charges of different signs are attracted to each other.

a) We convert cm in meters, and microC in C
M1 = (-0.0510, 0.0720) M2 = (-0.0521, 0.0104)
M1M2 = (0.0026571, 0.0007488)
r = | M1M2| = sqrt(0.0026571^2 + 0.0007488^2) = 0.00276 m
q1 = 1.40 e-6 C
q2 = 9 e-6 C
|F| = 14866.3 Newtons

b) the two particles have opposite charges, so q2 will be attracted towards q1. The force will be orientated along vector M2 M1 = (0.0026571, 0.0007488)
If you want to get the angle teta between M2M1 and +x axis
cos(theta) = 0.0026571 / 0.00276
theta = 15.69 degrees counterclockwise