There is an insulator rod of length L and of negligible mass with two small balls of mass m and electric charge Q attached to its ends. The rod can rotate in the horizontal plane around a vertical axis crossing it at a distance `L//4` from one of its ends.
At first the rod is in unstable equilibrium in a horizontal uniform electric field of field strength E. Then we gently displace it from this position. Determine the maximum velocity attained by the ball that is closer to the axis in the subsequent motion

Options

A. `sqrt((2QEL)/m)`
B. `sqrt((2QEL)/(5m)`
C. `sqrt((QEL)/(5m)`
D. `sqrt((4QEL)/(5m)`

Answer

C

Explanation

Position of maximum velocity is 1' and 2' for the balls as it is minimum potential energy position
Potential difference between 1 and `1'=V_(1)-V_(1)=Exx((3L)/4xx2)`
Potential difference between 2 and `2'=V_(2)-V_(2)=-Exx(L/4xx2)`
Change in electrostatic potential energy `=Qxx(V_(1)-V_(1))+Qxx(V_(2)-V_(2))=-QEL (loss in PE)`
=Gain in KE`=1/2mV^(2)+1/2m(2v)^(2)`
(where velocity of the ball closer to the axis is V & farther from axis is 2V)
`Rightarrow v=sqrt((QEL)/(5m))`