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 ?

27 cm

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