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.

Respuesta :

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 :

[tex]F=kx\\\\F=238\times 0.029 \\\\F=6.902\ N[/tex]

This force will act between charges as :

[tex]F=k\dfrac{q^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{6.902\times (0.44+0.029)^2}{9\times 10^9}} \\\\q=1.29\times 10^{-5}\ C[/tex]

So, the magnitude of charge is [tex]1.29\times 10^{-5}\ C[/tex] and it has positive sign.