A particle A having a charge of `2.0xx 10^(-6)` C and a mass of 100 g is placed at the bottom of a smoth inclined plane of inclination `30@` . Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?

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 break down the solution step by step.

### Step 1: Understand the Forces Acting on Particle B
- Particle A is at the bottom of the incline with a charge \( q_1 = 2.0 \times 10^{-6} \, \text{C} \) and mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \).
- Particle B, which has the same charge and mass, is placed at a distance \( x \) up the incline.
- The forces acting on particle B include:
1. The gravitational force (\( F_g \)) acting down the incline: \( F_g = mg \sin \theta \)
2. The electrostatic force (\( F_e \)) due to particle A: \( F_e = k \frac{q_1 q_2}{r^2} \), where \( r = x \).
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