To solve the problem step by step, we will analyze the situation of the three charged balls suspended from insulating threads.
### Part (a): One Ball is Discharged
1. **Initial Setup**:
- We have three identical conducting balls, each with charge \( Q \), suspended from a common point by insulating threads of length \( L \).
- When in equilibrium, the balls repel each other due to their identical charges, and the distance between any two balls is \( a \).
2. **Forces Acting on the Balls**:
- Each ball experiences two forces: the gravitational force \( mg \) acting downwards and the electrostatic force \( F_e \) acting horizontally due to the repulsion from the other charged balls.
- The electrostatic force between any two balls is given by:
\[
F_e = k \frac{Q^2}{a^2}
\]
- Here, \( k \) is Coulomb's constant.
3. **Equilibrium Condition**:
- In equilibrium, the vertical component of the tension in the thread balances the weight of the ball, and the horizontal component balances the electrostatic force.
- If we denote the angle between the thread and the vertical as \( \theta \), we can write:
\[
T \cos \theta = mg \quad \text{(1)}
\]
\[
T \sin \theta = F_e \quad \text{(2)}
\]
4. **Using Geometry**:
- The horizontal distance between the balls can be expressed as \( a = 2L \sin \theta \).
- From the geometry of the setup, we can also express \( \tan \theta \) as:
\[
\tan \theta = \frac{F_e}{mg}
\]
5. **Substituting Forces**:
- From equations (1) and (2):
\[
\tan \theta = \frac{k \frac{Q^2}{a^2}}{mg}
\]
- Substituting \( a = 2L \sin \theta \) into the equation gives us a relationship between \( a \), \( L \), \( Q \), and \( m \).
6. **Discharging One Ball**:
- When one ball is discharged, it loses its charge, and the remaining two balls will now have a charge \( Q \) each.
- The new electrostatic force between the two remaining charged balls is:
\[
F_e' = k \frac{Q^2}{x^2}
\]
- The distance between the two charged balls is now \( x \).
7. **New Equilibrium**:
- The new equilibrium condition can be set up as:
\[
T' \sin \theta' = 2F_e' = 2k \frac{Q^2}{x^2}
\]
- The vertical component remains the same:
\[
T' \cos \theta' = mg
\]
- Using the geometry again, we can relate \( x \) to \( L \) and \( \theta' \).
8. **Final Calculation**:
- By equating the forces and using the relationships derived, we can solve for \( x \):
\[
x^3 = \frac{2kQ^2L}{mg}
\]
- Thus, the new separation \( x \) can be calculated as:
\[
x = \left(\frac{2kQ^2L}{mg}\right)^{1/3}
\]
### Part (b): Two Balls are Discharged
1. **Discharging Two Balls**:
- When two balls are discharged, they will share their charge equally.
- Each ball will then have a charge of \( \frac{Q}{3} \).
2. **New Electrostatic Force**:
- The new electrostatic force between the two remaining charged balls is:
\[
F_e'' = k \frac{\left(\frac{Q}{3}\right)^2}{y^2} = k \frac{Q^2}{9y^2}
\]
- The distance between the two charged balls is now \( y \).
3. **New Equilibrium Condition**:
- The equilibrium condition will be similar to before:
\[
T'' \sin \theta'' = 2F_e'' = 2k \frac{Q^2}{9y^2}
\]
- The vertical component remains:
\[
T'' \cos \theta'' = mg
\]
4. **Final Calculation**:
- By setting up the equations and using the relationships derived, we can solve for \( y \):
\[
y^3 = \frac{3kQ^2L}{mg}
\]
- Thus, the new separation \( y \) can be calculated as:
\[
y = \left(\frac{3kQ^2L}{mg}\right)^{1/3}
\]
### Summary of Results:
- (a) The separation between the charged balls when one ball is discharged is:
\[
x = \left(\frac{2kQ^2L}{mg}\right)^{1/3}
\]
- (b) The separation between the balls when two balls are discharged is:
\[
y = \left(\frac{3kQ^2L}{mg}\right)^{1/3}
\]
