Question

In the figure, two conducting balls of identical mass m = 10 g and identical charge q hang from nonconducting threads of length L = 140 cm. If x = 7.3 cm, what is q? Since x is much smaller than L approximate sin(?) by ?.

Answer 1

Is x between the two balls? or is it to the 1/2 way point. I'll use between the two balls which means we will have to find the 1/2way point. x1 = x/2 = 3.65 cm

The tension in L is made of two forces: A vertical force (Fg) and a Horizontal Force an electrostatic force Fe.

We know the most about Fg, so let's find it first.
Fg = m * g
m = 10 grams = 10 grams * 1 kg/1000 grams = 0.010 kg
a = 9.81

Fg = 0.010 * 9.81 = 0.981 N

Sin(theta) = 3.65/140
Theta = 1.493degrees.

The tension in the chord is
Cos(theta) = Vertical force / Tension
Tension = Vertical Force / Cos(theta)
Tension = 0.0981 / cos(1.493)
Tension = 0.0981 N

Next find the Electrostatic Force.
Sin(theta) = Fe/Tension
Fe = T*sin(theta)
Fe = 0.0981*sin(1.493)
Fe = .002555 N

Now solve for q
Fe = k*q1*q2/r^2
Fe = 0.002555
r = 3.65 cm = 0.0365 m
q1 = q2 = q
k = 9*10^9

q^2 = Fe*r^2/9*10^9
q^2 = 0.002555*(0.0365)^2 / 9*10^9
q^2 =3.7821 * 10^-16
q =1.944 * 10^-8 coulombs.