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Grade 12Electrostatics

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

VNDT

9 Years agoGrade 12

3 Answers

Akshita Jasoria

Approved Tutor Answer9 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

v^2=(6Vq)/m

daksh

8 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

Now, get the answer by,

Change in KE= q*PD

Kushagra Madhukar

6 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

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