Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle `'theta'` with the vertical ?

Options

A. `x=((q^(2)l)/(2pi epsilon_(0)mg))^(1//2)`
B. `x=((q^(2)l)/(2pi epsilon_(0)mg))^(1//3)`
C. `x=((q^(2)l^(2))/(2pi epsilon_(0)m^(2)g))^(1//3)`
D. `x=((q^(2)l^(2))/(2pi epsilon_(0)m^(2)g^(2)))^(1//3)`

Answer

To find the equilibrium separation \( R \) between two identical tennis balls, each having mass \( m \) and charge \( q \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Balls
Each ball experiences three forces:
1. Gravitational force \( mg \) acting downwards.
2. Tension \( T \) in the thread acting along the thread.
3. Electrostatic force \( F \) due to the repulsion between the two charges, given by Coulomb's law:
\[ F = \frac{k q^2}{R^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Analyze the Angles
Since the balls are making a small angle \( \theta \) with the vertical, we can resolve the tension \( T \) into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)

### Step 3: Set Up the Equations
From the vertical forces, we have:
\[ T \cos \theta = mg \quad \text{(1)} \]
From the horizontal forces, we have:
\[ T \sin \theta = \frac{k q^2}{R^2} \quad \text{(2)} \]

### Step 4: Divide the Equations
Dividing equation (2) by equation (1) to eliminate \( T \):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{k q^2}{R^2}}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{k q^2}{mg R^2} \quad \text{(3)} \]

### Step 5: Relate \( R \) to \( \theta \)
For small angles, we can use the approximation:
\[ \tan \theta \approx \theta \quad \text{(in radians)} \]
Thus, we can write:
\[ \theta \approx \frac{R/2}{L} \quad \text{(4)} \]
where \( R/2 \) is the vertical distance from the point of suspension to the ball.

### Step 6: Substitute \( \theta \) into Equation (3)
Substituting equation (4) into equation (3):
\[ \frac{R/2}{L} = \frac{k q^2}{mg R^2} \]
Cross-multiplying gives:
\[ R^3 = \frac{(k q^2) L}{2mg} \]

### Step 7: Solve for \( R \)
Now, substituting \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ R^3 = \frac{(1/4 \pi \epsilon_0) q^2 L}{2mg} \]
This simplifies to:
\[ R^3 = \frac{q^2 L}{8 \pi \epsilon_0 mg} \]
Taking the cube root:
\[ R = \left(\frac{q^2 L}{8 \pi \epsilon_0 mg}\right)^{1/3} \]

### Final Answer
Thus, the equilibrium separation \( R \) is given by:
\[ R = \left(\frac{q^2 L}{2 \pi \epsilon_0 mg}\right)^{1/3} \]