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,      

\[F = \frac{1}{4\pi \epsilon_0}\frac{q^2}{r^2}\] For equilibrium:

For equilibrium along the plane,

\[F = \text{ mg}\sin\theta\] 

\[ \Rightarrow   \frac{9 \times {10}^9 \times \left( 2 \times {10}^{- 6} \right)^2}{r^2} = 0 . 1 \times 9 . 8 \times \frac{1}{2}\] 

\[ \Rightarrow  r^{{}_2}  = 7 . 34 \times  {10}^{- 2} \]m

\[ \Rightarrow   r = 0 . 271 \]m

So, the charge should be placed at a distance of 27 cm from the bottom.