Class 12 NEET Physics Electrostatics Questions Part 1

Concept: For a system of charges to be in equilibrium, the net electrostatic force on each individual charge must be zero. Explanation: Let the two charges $Q$ be separated by a distance $2r$. The charge $q$ is placed at the center, so its distance from each charge $Q$ is $r$. For the system to be in equilibrium, we can consider the net force on any one of the charges to be zero. Let's analyze the forces on one of the charges $Q$. 1. The repulsive force on this charge $Q$ due to the other charge $Q$ is: $F_{QQ} = \frac{1}{4\pi\epsilon_0} \frac{Q \cdot Q}{(2r)^2} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{4r^2}$ 2. The force on the same charge $Q$ due to the central charge $q$ is: $F_{qQ} = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r^2}$ 3. For equilibrium, the net force on charge $Q$ is zero: $F_{net} = F_{QQ} + F_{qQ} = 0$ $\frac{1}{4\pi\epsilon_0} \frac{Q^2}{4r^2} + \frac{1}{4\pi\epsilon_0} \frac{qQ}{r^2} = 0$ 4. Simplifying the equation: $\frac{Q^2}{4r^2} = -\frac{qQ}{r^2}$ $\frac{Q}{4} = -q$ $q = -\frac{Q}{4}$ Answer: The correct answer is C. $-\frac{Q}{4}$.