Two identical pith balls are charged by rubbing againt each other. The...
Problem Statement
Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through the strings of length 20cm each, the separation b/w the suspension points being 5cm. In equilibrium the separation b/w the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2×10^-8.
Calculating the mass of each ball
Let the mass of each ball be m.
From equilibrium, the tension in the strings is equal to the weight of each ball.
Therefore, T = mg, where T is the tension in the strings and g is the acceleration due to gravity.
Let the distance between the center of the balls be x.
Using Coulomb's law, the force of repulsion between the balls is given by F = (1/4πε) * ((2q)^2)/(x^2), where q is the charge on each ball and ε is the permittivity of free space.
At equilibrium, the forces acting on each ball are balanced. Therefore, the force of repulsion between the balls is balanced by the weight of each ball.
Therefore, (1/4πε) * ((2q)^2)/(x^2) = mg
Solving for m, we get m = ((1/4πε) * ((2q)^2)/(x^2))/g
Substituting the given values, we get m = 7.2 * 10^-14 kg.
Calculating the tension in the strings
Using the equation T = mg, we get T = 7.2 * 10^-14 kg * 9.8 m/s^2 = 7.1 * 10^-13 N.
Explanation
When two objects are rubbed together, they can transfer charge from one object to the other. In this case, the two pith balls become charged and repel each other due to the electrostatic force. The tension in the strings balances the weight of each ball and the force of repulsion between the balls. By analyzing the forces at equilibrium, we can calculate the mass of each ball and the tension in the strings.