A particle ‘A’ having a charge of 2 × 10-6 C and a mass of 100 g is fixed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, having same charge and mass be placed on the incline so that it may remain in equilibrium ?
Answer & explanation
Correct answer: option 3
First of all draw the F.B.D. of the masses.
For equilibrium ∑F = 0
N = mg cos30° . . . (1)
Fe = mg sin30° . . . (2)
From (2)
From (2) $\frac{kq^2}{x^2}=\frac{mg}{2}$
$\Rightarrow x=\sqrt{\frac{2 kq^2}{mg}}$ = 27 cm