Question

Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. A charge $q$ is placed at the origin. If $q$ is a small displacement along the $y$ direction, the force acting on $q$ is proportional to

• A. y
• B. -y
• C. 1/y
• D. -1/y

Answer 1

The Correct Option is B

Given: Two charges, each equal to \( -q \), are placed at \( (-a, 0) \) and \( (a, 0) \). A charge \( q \) is placed at the origin. We are asked to find the force acting on the charge \( q \) when it undergoes a small displacement along the \( y \)-direction.

Approach: The force on a charge in the presence of other charges is given by Coulomb's Law: \[ F = k_e \frac{q_1 q_2}{r^2} \] where: - \( k_e \) is Coulomb's constant, - \( q_1 \) and \( q_2 \) are the magnitudes of the charges, - \( r \) is the distance between the charges. In this case, the two charges \( -q \) at \( (-a, 0) \) and \( (a, 0) \) exert forces on the charge \( q \) placed at the origin. When the charge \( q \) is displaced slightly along the \( y \)-direction, the distance between \( q \) and the charges at \( (-a, 0) \) and \( (a, 0) \) changes. The displacement introduces small asymmetry in the forces from the two charges, which can be calculated using Coulomb's Law and considering the vector nature of the force. 

Resulting Force: After analyzing the force components, we find that the force on \( q \) due to a small displacement along the \( y \)-direction is proportional to the displacement \( y \), with a negative sign indicating that the force is directed toward the origin (restoring force). 

Conclusion: The force acting on \( q \) when it is displaced along the \( y \)-direction is proportional to \( -y \). 

Final Answer: The force is proportional to \( -y \).