Question

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.4 g mass from it. This stretches the spring to a length of 5.2 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.8 cm . What is the magnitude of the charge (in nCnC) on each bead?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Force constant of the spring = k

Unstretched length of the spring = L = 4 cm = 0.04 m

Mass attached to the spring = m = 1.4 g = 0.0014 kg

Length of the spring after the 1.4 g mass is attached to it = L₁ = 5.2 cm = 0.052 m

Amount the spring stretches due to this mass = d₁

d₁ = L₁ - L

d₁ = 0.052 - 0.04

d₁ = 0.012 m

The weight of the 1.4 g mass is supported by the spring therefore,

mg = kd₁

(0.0014)(9.81) = k(0.012)

k = 1.1445 N/m

Coulomb's constant = k_e = 8.99 x 10⁹ N.m²/C²

Magnitude of charge on each bead = q

Length of the spring after the plastic beads are attached to opposite ends of the spring = L₂ = 4.8 cm = 0.048 m

This is also the distance between the beads.

Amount the spring is stretched due to the beads = d₂

d₂ = L₂ - L

d₂ = 0.048 - 0.04

d₂ = 0.008 m

Electrical force between the beads = F_e

This electrical force on each bead is balanced by the spring force on them.

q = 48.4 x 10^-9 C

q = 48.4 nC

Magnitude of charge on each bead = 48.4 nC