Answer:
q = 7.88μC
Explanation:
In the diagram attached, you can visualize the situation. Separation distance between the 2 charges is:
D = 2*L*sin θ = 0.1532m
From a sum of forces on the X-axis:
[tex]T*sin \theta = Fe[/tex]
And on the y-axis:
[tex]T*cos \theta = m*g[/tex] Solving for T and replacing its value on the X-axis equation we get:
[tex]m*g*tan \theta = Fe[/tex] Since [tex]Fe = \frac{K*q^2}{D^2}[/tex]
[tex]m*g*tan \theta = \frac{K*q^2}{D^2}[/tex] Solving for q we get:
q = 7.88μC
