A charged pith ball (mass 2.0g) is brought close to an identical,?

Question

equally charged pith ball that is suspended from a thread, as shown in the illustration. when the pith balls are 3.8cm apart, the thread sits n equilibrium at 14 degress from the vertical. what is the charge on each of teh pith balls?

Answer 1

I recommend starting with a free body diagram of the pith ball. From the FBD you should notice that the vertical component of tension in the thread opposes the force of gravity exerted on the pith ball, and the horizontal component of tension in the thread opposes the electric force exerted on the pith ball.

Let up and to the right be positive.

(1) Let G be the force of gravity exerted on the pith ball

=> G = mg = (0.0020 kg)(9.81 N/kg) = 0.01962 N [down]

(2) Let T be the tension in the thread

Since the pith ball is not moving vertically, the sum of all vertical forces equal zero, so:

Tcos(14°) - G = 0

=> Tcos(14°) = 0.01962N

=> T = 0.02022 N

(3) Let F be the electric force due to the other pith ball.

Since the pith ball is not moving horizontally, the sum of all horizontal force equal zero, so:

Tsin(14°) - F = 0

=> F = Tsin(14°)

=> F = 0.004892 N

According to Coulomb's Law,

F = k*q1*q2 / r²

where,

k = 9.0*10^9 Nm²/C²

r is the distance between the two charges (0.038 m)

q1 and q2 are the two charges; which, because they are identical in this situation, I'm going to rewrite the equation as:

F = 2kq / r²

=> 0.004892N = (2)(9.0*10^9 Nm²/C²)(q) / (0.038 m)

Solve for q to obtain:

q = 1.03 * 10^-14 C = 10.3 fC (femtocoulombs)

Therefore, the charge on each pith ball is:

1.03 * 10^-14 C or 10.3 fC.