Question

An unstrained horizontal spring has a length of 0.34 m and a spring constant of 320 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.039 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Answer 1

1.41131 × 10⁻⁸ = q2

Step-by-step explanation:

Given:

Length = 0.34m

Spring constant =320 N/m

Computaion:

Using coulomb law

Force applied on spring = (Spring constant)(spring stretch)

Force applied on spring = (320) (0.039) = 12.48N

Force applied on spring = q1q2² / r²

12.48 = (9 × 10⁹)q2² / (0.34+0.039)²

1.79263968 = (9 × 10⁹)q2²

1.99182187 × 10⁻⁸ = q2²

1.41131 × 10⁻⁸ = q2