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
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