A small bob carrying charge +Q and mass 5 kg is suspended from a spring (k = 10⁴ N m⁻¹) and held at rest 50 cm above a fixed point charge −Q. The spring is initially unstretched. Upon release its maximum elongation is 10 cm. If Q = 25n μC, find n. (g = 10 m s⁻².)
Physics · Mixed Chapter
JEE Mains

Question

A small bob carrying charge +Q and mass 5 kg is suspended from a spring (k = 10⁴ N m⁻¹) and held at rest 50 cm above a fixed point charge −Q. The spring is initially unstretched. Upon release its maximum elongation is 10 cm. If Q = 25n μC, find n. (g = 10 m s⁻².)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Explanation: At the maximum downward displacement x=0.1 m, speed is zero. The initial-to-final loss of gravitational-plus-Coulomb potential equals elastic energy: mgx+kₑQ²(1/0.4−1/0.5)=(1/2)kx². Substitute m=5, g=10, k=10⁴, kₑ=9×10⁹ to get 5+4.5×10⁹Q²=50, or Q=100 μC. Since Q=25n μC, n=4.