A particle A having charge of `2.0xx10^-6C` and a mass of 100 g is fixed at the bottom of a smooth inclined plane of inclination `30^@`. Where should another particle B having same charge and mass, be placed on the inclined plane so that B may remain in equilibrium?

Options

A. 8 cm from the bottom
B. 13 cm from the bottomk
C. 21 cm from the bottom
D. 27 cm from the bottom

Explanation

To solve the problem, we need to find the position of particle B on the inclined plane such that it remains in equilibrium under the influence of gravitational and electrostatic forces. Let's go through the solution step by step.

### Step 1: Identify the forces acting on particle B
- The gravitational force acting on particle B can be resolved into two components: one perpendicular to the incline and one parallel to the incline. The component acting down the incline is given by:
\[ F_g = mg \sin \theta \]
where \( m \) is the mass of particle B, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination.

### Step 2: Calculate the gravitational force
- Given:
- Mass of particle B, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Angle of inclination, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g \approx 9.8 \, \text{m/s}^2 \)

Using the formula:
\[ F_g = mg \sin \theta = 0.1 \times 9.8 \times \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ F_g = 0.1 \times 9.8 \times \frac{1}{2} = 0.49 \, \text{N} \]

### Step 3: Calculate the electrostatic force between the two charges
- The electrostatic force \( F_e \) between two charges is given by Coulomb's law:
\[ F_e = k \frac{q_1 q_2}{r^2} \]
where:
- \( k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 = q_2 = 2.0 \times 10^{-6} \, \text{C} \)
- \( r \) is the distance between the two charges.

Substituting the values:
\[ F_e = 9 \times 10^9 \frac{(2.0 \times 10^{-6})^2}{r^2} = 9 \times 10^9 \frac{4 \times 10^{-12}}{r^2} = \frac{36 \times 10^{-3}}{r^2} \, \text{N} \]

### Step 4: Set the forces equal for equilibrium
For particle B to be in equilibrium, the gravitational force must equal the electrostatic force:
\[ F_g = F_e \]
Thus:
\[ 0.49 = \frac{36 \times 10^{-3}}{r^2} \]

### Step 5: Solve for \( r^2 \)
Rearranging the equation gives:
\[ r^2 = \frac{36 \times 10^{-3}}{0.49} \]
Calculating \( r^2 \):
\[ r^2 \approx \frac{36 \times 10^{-3}}{0.49} \approx 0.073469 \]

### Step 6: Calculate \( r \)
Taking the square root:
\[ r \approx \sqrt{0.073469} \approx 0.271 \, \text{m} \]

### Conclusion
The distance \( r \) from the bottom of the incline where particle B should be placed to remain in equilibrium is approximately **0.271 meters**.