# A small bead of mass m carrying charge q. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge +Q has also been fixed as shown. The potential atthe point P due to +Q 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

Akshita Jasoria
26 Points
6 years ago
this is an easy but little tricky ques BUT WE CAN SOLVE IT EASILY.....
potential at B is 3V.....To complete the circle....
therefore
1/2mv2= 3Vq
v2=(6Vq)/m
daksh
13 Points
5 years ago
We only have to provide the bead enough energy to travel the semicircle i.e. to the diametrically opposite point to the point P.
Thus, PD accross the two points= 4V-V = 3V
Change in KE= q*PD
3 years ago
Hello student
This question can be solved by applying conservation of mechanical energy as only electrostatic force is acting which is a conservatve force
For the bead to complete the circle and bound to the system it must have a kinetic energy just greater than 0 at the point of max potential energy
Hence,
Ki + Ui = Kf + Uf
½ mv2 + qV = 0 + 4qV
or,
v = (6qV/m)1/2

Hope it helps
Regards,
Kushagra