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 1.2 g mass from it. This stretches the spring to a length of 4.8 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 4.5 cm .

What is the magnitude of the charge (in nC) on each bead?

Answer 1

In the first case the weight of the mass balanced by the force stored in the spring due to stretching.

m g = k x

here m =mass of the object = 1.2 g = 1.2 x 10^-3 kg.

g =acceleration due to gravity =9.8 m/s².

k =spring constant , x =stretch in the spring = 4.8 cm - 4.0 cm = 0.8 cm =0.8 x 10^-2 m

(1.2 x 10^-3 kg) (9.8 m/s²) = k (0.8 x 10^-2)

k =1.47 N /m.

In the second case the electric force between the beads is balanced by the force stored in the spring

K (q₁ q₂ ) / r²  = k x

q₁ = q₂ = q =chatge on each bed.

r =distance between the beads when the spring stretched = 4.5 cm = 4.5 x 10^-2 m

x =expansion in the spring = 4.5 cm - 4.0 cm = 0.5 cm = 0.5 x 10^-2 m

K =coulomb's constant = 9.0 x 10⁹ N. m² / C²

k =spring constant =1.47 N /m

on substituting all values we get

(9.0 x 10⁹ N. m² / C² ) (q² ) / (4.5 x 10^-2 m)²  = (1.47 N /m ) ( 0.5 x 10^-2 m)

q² = 1.6537 x 10^-15 C² = 16.537 x 10^-16 C²

q = 4.067 x 10^-8 C

q =40.67 nC

Therefore the charge on the each bed is 40.67 nC.