Question

An unstrained horizontal spring has a length of 0.44 m and a spring constant of 238 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 0.029 m relative to its unstrained length. Determine the possible algebraic signs and the magnitude of the charges.

Answer 1

Step-by-step explanation:

Given that,

Length of the spring, l = 0.44 m

Spring constant of the spring, k = 238 N/m

Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes.

Displacement in the spring, x = 0.029 m

The force acting on the spring is given by :

`F=kx\\\\F=238* 0.029 \\\\F=6.902\ N`

This force will act between charges as :

`F=k(q^2)/(r^2)\\\\q=\sqrt{(Fr^2)/(k)} \\\\q=\sqrt{(6.902* (0.44+0.029)^2)/(9* 10^9)} \\\\q=1.29* 10^(-5)\ C`

So, the magnitude of charge is
`1.29* 10^(-5)\ C` and it has positive sign.