Question

A particle of mass m and charge q is attached to a light rod of length L. The rod can rotate freely in the plane of paper about the other end, which is hinged at P. The entire assembly lies in a uniform electric field E also acting in the plane of paper as shown. The rod is released from rest when it makes an angle $\theta$ with the electric field direction. Determine the speed of the particle when the rod is parallel to the electric field.

• A. ${\dfrac{{2qEL(1 - \cos \theta )}}{m}}^{1/2}$
• B. ${\dfrac{{2qEL(1 - \sin \theta )}}{m}}^{1/2}$
• C. ${\dfrac{{qEL(1 - \cos \theta )}}{{2m}}}^{1/2}$
• D. ${\dfrac{{2qEL\cos \theta }}{m}}^{1/2}$

Answer 1

Answer: (A)

${\left( {\frac{{2qEL(1 - \cos \theta )}}{m}} \right)^{1/2}}$

Answer 2

Apply the work-energy theorem. The work done by the electric field is $W_e = qEL(1 - \cos \theta)$. Initial kinetic energy is zero. Final kinetic energy is $\frac{1}{2}mv^2$. Equating work done to change in kinetic energy: $qEL(1 - \cos \theta) = \frac{1}{2}mv^2$. Solve for $v$. Gravitational work is neglected as per standard problem context for such options.