To find the magnitude of the charge on the pith balls, we can use the concept of electrostatic force and equilibrium.
First, let's discuss the forces acting on each pith ball. There are two forces acting on each ball:
1. The force due to gravity, which is equal to the weight of the ball (mg), where m is the mass of the ball and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The electrostatic force between the two charged pith balls. Since the charges on the balls are equal, the electrostatic force will be repulsive.
We can use the fact that the system is in equilibrium, which means the net force on each ball is zero. This implies that the weight of the ball is balanced by the electrostatic force.
The electrostatic force between the two balls can be expressed as:
F = k * (q^2 / r^2)
where F is the force, k is the electrostatic constant (approximately 9 × 10^9 N·m^2/C^2), q is the magnitude of the charge on each ball, and r is the distance between the centers of the balls.
Since the balls are not moving and are in equilibrium, the net force on each ball is zero. This means that the weight of each ball is equal to the electrostatic force. We can equate these forces:
mg = k * (q^2 / r^2)
Now, let's solve for the charge q.
1. Convert the mass of the pith ball to kilograms:
m = 11.2 g = 0.0112 kg
2. Convert the angle to radians:
θ = 30.4° * π/180 = 0.531 rad
3. Find the distance between the centers of the balls using the given length and angle:
r = 1.7 m * sin(θ) = 1.7 m * sin(0.531) ≈ 1.19 m
4. Substitute the known values into the equation and solve for q:
q = √((mg * r^2) / k)
q = √((0.0112 kg * 9.8 m/s^2 * (1.19 m)^2) / (9 * 10^9 N·m^2/C^2))
q ≈ 0.000607 C ≈ 607 µC
Therefore, the magnitude of the charge on each pith ball is approximately 607 micro-Coulombs (µC).