Two similar metal spheres are suspended by silk threads from the same point. When the spheres are given equal charges of `2 mu C` the distance between them becomes 6cm. If length of each thread is 5 cm, the mass of each sphere is (g=10 m/s2)

Options

A. 4 kg
B. 3kg
C. `4/3 kg `
D. `1/3` kg

Answer

To solve the problem, we need to analyze the forces acting on the two charged spheres suspended by silk threads. Here's a step-by-step solution:

### Step 1: Understand the Setup
We have two identical metal spheres, each with a charge of \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \). The distance between the spheres when charged is \( r = 6 \, cm = 0.06 \, m \). The length of each thread is \( L = 5 \, cm = 0.05 \, m \). The acceleration due to gravity is \( g = 10 \, m/s^2 \).

### Step 2: Draw a Diagram
Draw a diagram showing the two spheres, the threads, and the angles involved. Label the angles as \( \theta \). The distance between the spheres is \( 6 \, cm \), and each thread makes an angle \( \theta \) with the vertical.

### Step 3: Identify Forces
Each sphere experiences:
- Gravitational force \( F_g = mg \) acting downward.
- Tension \( T \) in the thread acting along the thread.
- Electrostatic force \( F_e \) acting horizontally due to the charges on the spheres.

### Step 4: Resolve Forces
From the diagram:
- The vertical component of tension: \( T \cos \theta = mg \)
- The horizontal component of tension: \( T \sin \theta = F_e \)

### Step 5: Express Electrostatic Force
The electrostatic force between the two charges is given by Coulomb's law:
\[ F_e = k \frac{q^2}{r^2} \]
where \( k = 9 \times 10^9 \, N m^2/C^2 \).

### Step 6: Calculate Electrostatic Force
Substituting the values:
\[ F_e = 9 \times 10^9 \frac{(2 \times 10^{-6})^2}{(0.06)^2} \]
Calculating \( F_e \):
\[ F_e = 9 \times 10^9 \frac{4 \times 10^{-12}}{0.0036} = 9 \times 10^9 \times \frac{4}{3.6} \times 10^{-12} \]
\[ F_e = 10 \times 10^{-3} = 0.01 \, N \]

### Step 7: Use Trigonometric Relationships
From the geometry of the setup, we can relate \( \tan \theta \):
\[ \tan \theta = \frac{3}{4} \quad \text{(since the horizontal distance is half of 6 cm, which is 3 cm, and the vertical distance is 4 cm)} \]

### Step 8: Substitute into Force Equations
Using \( \tan \theta = \frac{mg}{F_e} \):
\[ \frac{3}{4} = \frac{mg}{0.01} \]
Substituting \( g = 10 \, m/s^2 \):
\[ \frac{3}{4} = \frac{m \cdot 10}{0.01} \]
\[ m = \frac{3 \times 0.01}{4 \times 10} = \frac{0.03}{40} = 0.00075 \, kg = 0.75 \, g \]

### Final Answer
The mass of each sphere is \( 0.75 \, g \).