Question

A 5 cm spring is suspended with a mass of 1 g attached to it which extends the spring by 3.2 cm. The same spring is placed on a frictionless flat surface and charged beads are attached to each end of the spring. With the charged beads attached to the spring, the spring's extension is 0.01 m. What are the charges, in micro-Coulombs, of the beads?

Answer 1

First, we need to find the spring constant

Force of gravity = force of tension from spring

mg = kx

(1x10^-3)(9.8) = k (3.2x10^-2)

k = .30625

Now we can look at the other situation

Since the spring was moved 0.01 meters, we can find the force

F = kx = (.01)(.30625) = .0030625 N

Now we can set the electric force equal to the force of the beads

kq^2/r^2 = .0030625 N

q = 292 microCoulombs