Question

Four point charges with magnitude 5.0 microcoulumbs are placed at the corners of a square that is 30.0 cm on a side. Two charges, diagonally opposite each other, are positive, and the other two are negative. What are the magnitude and the direction of the force on one of the charges?

Answer 1

given

magnitude of the charge q = 5*10^-6 C

side of the square d = 0.3 m

let q1 = q4 = -5*10^-6 C

and

q2 = q3 = 5*10^-6 C

force between two point charges is F = kq1q2/r^2

forces on q1 due to q2 and q3 are attractive.

therefore magnitude of F2 = magnitude of F3

F3 = F2 = 9*10^9*5*10^-6*5*10^-6/(0.3)^2

F3 = F2 = 2.5 N

resultant force of F2 and F3 = sqrt(2.5^2+2.5^2) = 3.54 N

force on q1 due to q4 is repulsive

F' = 9*10^9*5*10^-6*5*10^-6 C/(0.3*sqrt(2))^2 = 1.25 N

net force F net = 3.54-1.25 = 2.29 N

direction : it acts from q1 to q4.