(v 0 = 3 m/s; K.E. at origin × × 10 -4 J = 2.5 × × 10 -4 J)

Sol.

In the figure q = 1 µC = 10 -6 C q 0 = 0.1 µC = 10 -7 C and m = 6 × 10 -4 Kg

and Q = 8µC = 8 × 10 -6 C

Let P be any point at a distance x from origin O. Then

AP = CP =

and BP = DP =

Electric potential at point P will be–

V P = – ; where, K = = 9 × 10 9 Nm 2 /C 2

∴ V P = 2 × 9 ×10 9

or V = 1.8 × 10 4 ........(1)

∴ Electric field at P is–

E P = – = 1.8 × 10 4 (2x)

E = 0 on axis where → x = 0

or = ⇒ =

∴ ( + x 2 ) = 4 ( + x 2 )

This equation gives, x = ± m

The least value of kinetic energy of the particle at infinity should be enough to take the particle upto x = + m because

at x = + m, E = 0 ⇒ Electrostatic force on charge q is zero or Fe = 0.

For x > m, E is repulsive (towards positive x–axis)

And For x < m, E is attractive (towards negative x–axis)

Now, from equation (1), potential at x = m

V P = 1.8 × 10 4

Applying energy conservation at x = ∞ and x = m

mv 0 2 = q 0 V.........(2)

∴ v 0 =

Substituting the values

v 0 =

or v 0 = 3 m/s

∴ Minimum value of v 0 is 3 m/s. Ans. (i)

From equation (1), potential at origin (x = 0) is

V 0 = 1.8 × 10 4

≈ 2.45 × 10 4 V

Let K be the kinetic energy of the particle at origin.

Applying energy conservation at x = 0 and at x = ∞

K + q 0 V 0 = mv 0 2 But, mv 0

2 = q 0 V from equation (2)

∴ K = q 0 (V – V 0 )

Or K = (10 –7 ) (2.7 × 10 4 – 2.45 × 10 4 ) ~ 2.5 × 10 –4 J Ans (ii)