Question: (a) Explain what happens to the balls of Problem 42 if one of them is discharged (loses its charge qto, say, the ground). (b) Find the new equilibrium separation x, using the given values of Land mand the computed value of |q|.

1. One of the balls from the two conducting balls hanging to non-conducting stand loses all of its charges.
2. Given values to get the charge,$L=120cm,m=10g,andx=5.0cm$

Using the concept of Newton's laws of motion and the force due to Coulomb's law, we can get the formula of the separation equilibrium of the system. Using this formula and the given data, we can get the magnitude of the unknown charge.

Formulae:

The magnitude of the electrostatic force between any two particles,$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (i)

The separation equilibrium of this system according to previous problem, $x={\left(\frac{q^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{1/3}$ (ii)

$\theta =0^0$

If one of them is discharged, then using equation (i), we can say that there would no electrostatic repulsion between the two balls and they would both come to the position, making contact with each other.

A redistribution of the remaining charge would then occur, with each of the balls getting charge, q/2. Then they would again be separated due to electrostatic repulsion, which results in the new equilibrium separation, which is given using equation (ii) and the given data as:

$$\begin{gathered} x\text{'}={\left(\frac{{\left(q/2\right)}^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{1/3} \\ ={\left(\frac{1}{4}\right)}^{1/3}x \\ ={\left(\frac{1}{4}\right)}^{1/3}5.0cm \\ =3.1cm \\ \text{(}\because x={\left(\frac{q^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{1/3} \\ =5cm,\text{using the given data)} \end{gathered}$$

Hence, the new separation equilibrium is 3. 1 cm