Question

Problem 7 A small conducting sphere is attached to a spring, as shown in figure (17.59). The charge on the sphere, @ 5.0uC and the mass of the conducting sphere, m- 300g The spring constant, k 150N/m. When the charge is at its equilibrium position it it 30cm above the floor. The sphere is pulled down 5cm from its equilibrium position and released. After the charge is released it will oscillate. Determine the electric field at point P, which is located directly below the charge on the floor, as a function of time.

Answer 1

Let when the elongation in spring is x, total force on the spherical ball is F = m g - k x - k l (l) along downward Here l is elongation in spring at the time of equilibrium so at equilibrium total force on the ball is zero as F_e = m g - k l = 0 implies m g = k l putting in (1) gives F = m g - k x - k l = k l - k x - k l F = -k x so F = m g = m d^2x/dt^2 gives m d^2x/dt^2 + k x = 0 or d^2x/dt^2 + k/m x = 0 let w^2 = k/m implies w = squareroot k/m so x^doubledot + w^2 x = 0 (2) Let x = A e^m t implies dx/dt = A m e^m t implies d^2x/dt^2 = A m^2 e^m t so putting in (2) gives A m^2 e^m t + w^2 A e^m t = 0