Question

As shown in the figure, a configuration of two equal point charges (q₀ = + 2 \(\mu\)C) is placed on an inclined plane. Mass of each point charge is 20g. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height \( h = x \times 10^{-3} \, \text{m} \). The value of \( x \) is _____

Hint:

In problems involving point charges on an inclined plane, use equilibrium conditions and equate the electrostatic force to the component of the gravitational force along the plane.

Answer 1

Correct Answer: 300

The point charge is in equilibrium at rest. Hence, the forces on the point charge must balance out. The force due to the gravitational pull is counteracted by the electrostatic force.
The forces acting on the point charge are:
- The electrostatic force, \( F_e \), due to the other charge.
- The gravitational force, \( mg \), acting downward.

Since the system is in equilibrium, the electrostatic force is balanced by the component of gravitational force along the plane: \[ F_e = mg \sin \theta \] Now, we know the formula for the electrostatic force between two point charges: \[ F_e = \frac{k q_0^2}{r^2} \] Where:
- \( k = 9 \times 10^9 \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2} \),
- \( q_0 = 2 \times 10^{-6} \, \text{C} \) (the charge),
- \( r \) is the distance between the charges, which is related to \( h \) by the geometry of the inclined plane, where \( r = \frac{h}{\sin \theta} \).
Substituting the values into the equation: \[ \frac{k q_0^2}{r^2} = mg \sin 30^\circ \] Substituting the known values: \[ \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{\left( \frac{h}{\sin 30^\circ} \right)^2} = 20 \times 10^{-3} \times 10 \times \sin 30^\circ \] Solving this equation: \[ h^2 = 9 \times 10^{10} \Rightarrow h = 300 \times 10^{-3} \, \text{m} \] Thus, the value of \( x \) is 300.