Question

Four charges each equal to Q are placed at the four Q.19 corners of a square and a charge q is placed at the centre of the square. If the system is in equilibrium then the value of q is

• A. $\frac{Q}{2}(1+2\sqrt{2})$
• B. $\frac{-Q}{4}(1+2\sqrt{2})$
• C. $\frac{Q}{4}(1+2\sqrt{2})$
• D. $\frac{-Q}{2}(1+2\sqrt{2})$

Answer 1

Answer: (B)

$\frac{-Q}{4}(1+2\sqrt{2})$

Answer 2

To achieve equilibrium, the net force on each charge in the system must be zero. Consider one of the corner charges (+Q). It experiences repulsive forces from the other three corner charges. The charge q at the center must exert an attractive force on this corner charge to balance the repulsive forces.

1. Forces from corner charges: Calculate the net repulsive force on a corner charge (+Q) due to the other three corner charges. This involves vector addition of the forces.

2. Force from the center charge: The force from the center charge q must counteract the net repulsive force calculated in step 1. Since the force must be attractive, q must be negative.

3. Equilibrium Condition: Set the attractive force equal to the repulsive force and solve for q. This gives $q = -\frac{Q}{4}(1+2\sqrt{2})$.