The electrostatic potential inside a charged solid sphere is given by V(r) = 100 - 50 r^2 V, where r is the distance from the center in meters. A particle of mass 1 g and charge -40 C is released from rest at a distance of r = 2 m from the center. Assume the radius of the sphere is greater than 2 m. If the time taken by the particle to reach the center of the sphere is x seconds, the value of x is ______.

Step-by-step solution

The electric field inside the sphere is given by the negative gradient of the potential: E = - dV dr = - d dr (100 - 50 r^2) = 100 r V/m The force acting on the negatively charged particle is: F = qE = (-40 10⁻⁶) (100 r) = -4 10⁻³ r N The acceleration of the particle is: a = F m = -4 10⁻³ r 1 10⁻³ = -4 r m/s ^2 Since the acceleration is proportional to the negative of the displacement from the center ( a = - ^2 r ), the particle executes Simple Harmonic Motion (SHM). ^2 = 4 = 2 rad/s The time period of the SHM is: