Question

Two identically-charged particles are fastened to the two ends of a spring of spring constant 100 N m⁻¹ and natural length 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is 2.0 × 10⁻⁸ C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

Answer 1

Let the extension in the string be x.
Given:
Magnitude of the charges, q = 2.0 × 10⁻⁸ C
Separation between the charges, r = (0.1+ x) m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r^2}$
The spring force due to extension x,
$F=Kx$
For equilibrium,
Electrostatic force = Spring force
$$\begin{gathered} \frac{1}{4\mathrm{\pi }{\epsilon }_0}\times \frac{q^2}{{\left(x+0.1\right)}^2}=K\left(0.1+x\right) \\ \Rightarrow {\left(0.1+x\right)}^3=\frac{9\times {10}^9\times {\left(2.0\times {10}^{-8}\right)}^2}{{10}^{-2}} \\ =\frac{36\times {10}^9\times {10}^{-16}}{{10}^{-2}} \\ =36\times {10}^{-5} \\ \Rightarrow x=3.6\times {10}^{-6}\mathrm{m} \end{gathered}$$
Yes, the assumpton is justified. As two similar charges are present at the ends of the spring so they exert repulsive force on each other.Due to the repulsive force between the charges,an extension x is produced in the spring. Springs are made up of elastic material.When a spring is extended then a restoring force acts on it which is always proportioanal to the extension produced and directed opposite to the direction of applied force.The restoring force depends on the elasticity of the material. When the extension is small then only the restoring force is proportinal to the extension.If the extension s camparable to the natural length of the spring then the restoring force will depend on higher powers of the extension produced.