Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is a(the length of thread `Lgtgta`). Then one of the balls is discharged. what will be the distance `b(bltltL)` between the balls when equilibrium is restored?

Explanation

To solve the problem step-by-step, we will analyze the forces acting on the charged balls and apply the principles of equilibrium and electrostatics.

### Step 1: Understand the Initial Setup
We have two identical small conducting balls, each with charge \( q \), suspended from a common point by threads of length \( L \). Initially, they are in equilibrium at a distance \( a \) apart. The forces acting on each ball are:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the thread acting along the thread.
- The electrostatic force \( F_e \) acting horizontally due to the repulsion between the two charges.

### Step 2: Analyze the Forces
In equilibrium, the vertical and horizontal components of the forces must balance.

1. **Vertical Forces**:
\[ T \cos \theta = mg \quad \text{(1)} \]

2. **Horizontal Forces**:
\[ T \sin \theta = F_e = \frac{k q^2}{a^2} \quad \text{(2)} \]
where \( k \) is Coulomb's constant.

### Step 3: Relate the Angles to the Distances
Using the geometry of the situation, we can express \( \sin \theta \) and \( \cos \theta \) in terms of \( a \) and \( L \):
- From the triangle formed, we have:
\[ \sin \theta \approx \frac{a/2}{L} \quad \text{(for small angles)} \]
\[ \cos \theta \approx 1 \quad \text{(for small angles)} \]

### Step 4: Substitute and Simplify
Substituting \( \sin \theta \) and \( \cos \theta \) into the equations (1) and (2):
1. From (1):
\[ T = \frac{mg}{\cos \theta} \approx mg \]

2. From (2):
\[ T \sin \theta = \frac{mg}{L} \cdot \frac{a/2}{L} = \frac{k q^2}{a^2} \]

### Step 5: Find the Relationship Between \( a \) and \( L \)
Using the two equations, we can derive a relationship:
\[ \frac{mg \cdot \frac{a/2}{L}}{mg} = \frac{k q^2}{a^2} \]
This simplifies to:
\[ \frac{a^3}{2L} = \frac{k q^2}{mg} \]
Let’s denote this as Equation (1).

### Step 6: Discharge One Ball
When one ball is discharged, it loses its charge, and the other ball retains a charge \( q \). The charge on the discharged ball becomes \( 0 \), and the remaining ball will have a charge of \( \frac{q}{2} \) when they touch and share charge.

### Step 7: New Equilibrium Condition
After the discharge, we need to find the new distance \( b \) between the balls when they come to equilibrium again. The new electrostatic force is:
\[ F_e' = \frac{k \left(\frac{q}{2}\right)^2}{b^2} = \frac{k q^2}{4b^2} \]

Using similar force balance as before:
1. **Vertical Forces**:
\[ T' \cos \theta' = mg \quad \text{(3)} \]

2. **Horizontal Forces**:
\[ T' \sin \theta' = \frac{k q^2}{4b^2} \quad \text{(4)} \]

### Step 8: Substitute and Relate \( b \) and \( a \)
Using the same approach as before, we can write:
\[ \frac{mg \cdot \frac{b/2}{L}}{mg} = \frac{k q^2}{4b^2} \]
This simplifies to:
\[ \frac{b^3}{2L} = \frac{k q^2}{4mg} \]
Let’s denote this as Equation (2).

### Step 9: Divide Equations (1) and (2)
Dividing Equation (2) by Equation (1):
\[ \frac{b^3 / 2L}{a^3 / 2L} = \frac{k q^2 / 4mg}{k q^2 / mg} \]
This simplifies to:
\[ \frac{b^3}{a^3} = \frac{1}{4} \]
Thus:
\[ b^3 = \frac{a^3}{4} \]
Taking the cube root:
\[ b = \frac{a}{\sqrt[3]{4}} = \frac{a}{2^{2/3}} \]

### Final Answer
The distance \( b \) between the balls when equilibrium is restored is:
\[ b = \frac{a}{2^{2/3}} \]