Question

At first the rod is in unstable equilibrium in a horizontal uniform electric 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:

• A. $\sqrt{\frac{2QEL}{m}}$
• B. $\sqrt{\frac{2QEL}{5m}}$
• C. $\sqrt{\frac{QEL}{5m}}$
• D. $\sqrt{\frac{4QEL}{5m}}$

Answer 1

The correct option is C $\sqrt{\frac{QEL}{5m}}$

Position (A) is of unstable equilibrium, because if displaced from this position, resultant torque will be in the direction of displacement and finally it will attain equilibrium position (B). At position (B), velocity will be maximum.
$W=\Delta KE$
or $QE\frac{3L}{2}-QE\frac{L}{2}=\frac{1}{2}m[v^2+(3v{)}^2]$
or $v=\sqrt{\frac{QEL}{5m}}$