Question

0.22
The diagram shows a small bead of mass m carrying charge q. The bead can
freely move on the smooth fixedring placed on a smooth horizontal plane In the
same plane a charge has also been fixed as shown. The potential atthe point
P due to + is V. The velocity with which the bead should projected from the
point P so that it can complete a circle should be greater than
​

Answer 1

It is such an interesting question.

at first, we have to find a situation in which the bead completes a circle after that we can easily solve using law of conservation of energy.

see diagram, at point S, electrostatic force opposes beam to go circular motion, similarly at point T, electrostatic force opposes the beam to go circular motion and this happens upto nearest point to R. after that electrostatic force supports the beam to go circular. so, To complete a circle, The beam has to move from P to R.

given, potential at P due to +Q = V

so, V = kQ/4a

or, 4V = KQ/a

potential at point R, V' = KQ/a = 4V

from law of conservation of energy,

energy at P = energy at R

or, $K.E_i+U_i=K.E_f+U_f$

Let initial velocity of bead is v and velocity at R = 0

so, K.Ei = 1/2mv² , K.Ef = 1/2 m × 0 = 0

Ui = qV [ electrostatic potential energy] and Uf = 4qV

so, 1/2mv² + qV = 0 + 4qV

or, 1/2mv² = 3qV

or, v = √{6qV/m}

hence, answer is $\sqrt{\frac{6qV}{m}}$

Answer 2

Solution in attachment

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