Three small equally charged identical conducting balls are suspended from identical insulating threads secured at one point. Length (L) of the threads is large compared to the equilibrium separation (a) between any two balls. (a) One of the balls is suddenly discharged. Find the separation between the charged balls when equilibrium is restored. Assume that the threads do not interfere and balls do not collide. (b) If two of the balls are suddenly discharged, how will the balls behave after this? Find the separation between the balls when equilibrium is restored. The threads do not interfere

Answer

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:
\[ y = \left(\frac{3kQ^2L}{mg}\right)^{1/3} \]