Question

Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibrium, each string makes and angle θ =50 with the vertical. Find the size of the charge on each spere.

Answer 1

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:

$T*sin \theta = Fe$

And on the y-axis:

$T*cos \theta = m*g$   Solving for T and replacing its value on the X-axis equation we get:

$m*g*tan \theta = Fe$  Since $Fe = \frac{K*q^2}{D^2}$

$m*g*tan \theta = \frac{K*q^2}{D^2}$  Solving for q we get:

q = 7.88μC