Question

Consider the four charges in the diagram below, in which d = 3.7 cm and q2 = +4 nC. The net force on the 1 nC charge is zero. What is q_1? nC

Answer 1

force on 1 nC due to q2

F = k x q x q2 / r^2

F = (9 x 10^9) x (1 x 10^-9) x (4 x 10^-9) / r23^2

Here r23 = sq.root ( 0.02^2 + 0.03^2)

r23 = 0.036 m

Therefore

F = (9 x 10^9) x (1 x 10^-9) x (4 x 10^-9) / 0.036^2

F = 2.78 x 10^-5 N

net force due to the two q2 charges

Fnet = 2*Fcos theta

now finding theta : tan theta = 3/2

theta = tan ^-1 3/2 = 56.3

new downward force = 2 x (2.78 x 10^-5) x cos 56.3

                              = 3.08 x 10^-5 N

this firce is balance by force due to q1

     force due to q1 = k x q x q1 / r13^2

=> 3.08 x 10^-5 N = (9 x 10^9) x (1 x 10^-9) x q1 / 0.017^2        

q1 = 0.9898 nC