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 smooth 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 particle A remains in equilibrium. Here’s a step-by-step solution:

### Step 1: Understand the forces acting on particle A
- Particle A has a charge \( q = 2.0 \times 10^{-6} \, \text{C} \) and mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \).
- The inclined plane has an angle of inclination \( \theta = 30^\circ \).
- The gravitational force acting on particle A can be resolved into two components: one perpendicular to the incline and the other parallel to the incline. The component acting down the incline is given by:
\[
F_{\text{gravity}} = mg \sin \theta
...