A 5 kg positively charged sphere hangs from a vertical spring (k=10⁴ N m⁻¹), initially unstretched and held by a thread. An equally charged negative sphere is fixed 50 cm directly below; each charge has magnitude 100 μC. After cutting the thread, maximum spring extension is P×5 cm. Find P (g=10 m s⁻²).
Physics · Mixed Chapter
JEE Mains

Question

A 5 kg positively charged sphere hangs from a vertical spring (k=10⁴ N m⁻¹), initially unstretched and held by a thread. An equally charged negative sphere is fixed 50 cm directly below; each charge has magnitude 100 μC. After cutting the thread, maximum spring extension is P×5 cm. Find P (g=10 m s⁻²).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Energy change from the initial state to maximum extension x: Initial and final speeds are zero. The moving sphere gains gravitational energy mgx and Coulomb potential becomes more negative. Step 2 — Mechanical-energy equation: (1/2)kx² = mgx + ke q²[1/(r₀−x)−1/r₀], with r₀=0.50 m. Step 3 — Substitute numerical constants: ke q² = (9×10⁹)(100×10⁻⁶)² = 90 J·m. Thus 5000x² = 50x + 90[1/(0.5−x)−2]. Step 4 — Find the nonzero turning point: For x=0.10 m, left side =50 J. Right side =5+90(2.5−…