Question

A particle A with a charge of 2.0 × 10⁻⁶ C and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, with the same charge and mass, be placed on the incline so that it may remain in equilibrium?

Answer 1

Given:
Magnitude of charge on particles A and B, q = 2.0 × 10⁻⁶ C
Mass of particles A and B, m = 100 g = 0.1 kg
Let the separation between the charges be r along the plane.
By Coulomb's Law, force (F) on B due to A,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r^2}$
For equilibrium:

For equilibrium along the plane,
$$\begin{gathered} \mathrm{F}=mg\sin \theta \\ \Rightarrow \frac{9\times {10}^9\times {\left(2\times {10}^{-6}\right)}^2}{r^2}=0.1\times 9.8\times \frac{1}{2} \\ \Rightarrow r^{{}_2}=7.34\times {10}^{-2}\mathrm{m} \\ \Rightarrow r=0.271\mathrm{m} \end{gathered}$$
So, the charge should be placed at a distance of 27 cm from the bottom.