Question

Consider the four charges in the diagram below, in which d = 4.4 cm and q2 = +4 nC. The net force on the 1 nC charge is zero. What is q1?

https://imgur.com/a/RQUwh

Answer 1

since net Force = 0

the force due to two q2 is upwards then the force due to q1 should be downwards

q3 = 1nC

distacne between q2 and q1

r = sqrt(3^2+2^2) = 3.6 cm = 0.036 m

tantheta = 2/3

theta = 33.7

force due to q2 on left

F2x = k*q2*q3*cos33.7/r^2 = 9*10^9*4*10^-9*1*10^-9*cos33.7/0.036^2 = 2.3*10^-5 N

F2y = k*q2*q3*sin33.7/r^2 = 9*10^9*4*10^-9*1*10^-9*sin33.7/0.036^2 = 1.54*10^-5 N

force due to q2 on right

F2x = -k*q2*q3*cos33.7/r^2 = -9*10^9*4*10^-9*1*10^-9*cos33.7/0.036^2 = -2.3*10^-5 N

F2y = k*q2*q3*sin33.7/r^2 = 9*10^9*4*10^-9*1*10^-9*sin33.7/0.036^2 = 1.54*10^-5 N

force due to q1

F1x = 0

F1y = -k*q1*q3/r13^2 = -9*10^9*q1*1*10^-9/(0.044-0.02)^2 = -15625*q1 N

F1y + F2y + F2y = 0

-15625*q1 + 1.54*10^-5 + 1.54*10^-5 = 0

q1 = 1.97 nC = +2 nC