( = 5.86 m/s)

Given, q = 1µc = 10 –6 C

& m = 2 × 10 –3 Kg and

λ = 0.8 m

Let u be the speed of the particle at its lowest point and v its speed at highest point.

At highest point, three forces are acting on the particle.

(i) Electrostatic repulsion

F e = . (outwards)

(ii) Weight W = mg (inwards), and

(iii) Tension T (inwards)

T = 0, if the particle has just to complete the circle and the necessary centripetal force is provided by

W – F e i.e., = W – F e

or v 2 =

v 2 = m 2 / s 2

or v 2 = 2.4 m 2 / s 2 .....(1)

Now the electrostatic potential energy at the lowest and highest points are equal. Hence from conservation of mechanical energy

Increase in gravitational potential energy = Decrease in kinetic energy

or mg(2l) = m (u 2 – v 2 )

or u 2 = v 2 + 4 gl

Substituting the values of v 2 from equation (1) we get

u 2 = 2.4 + 4 (10) (0.8) = 34.4 m 2 / s 2 .

∴ u = 5.86 m/s

Therefore, minimum horizontal velocity imparted to the lower ball, so that it can make compete revolution, is 5.86 m/s.