An electric dipole consists of 2.0 g spheres charged to 5.0 nC (positive and negative) at the ends of a 12 cm long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1400V, then released. What is the dipole’s angular velocity at the instant it is aligned with the electric field?

Answer 1

The torque acting on the dipole in electric field of strength E= 1400 V/m (!!!!- units) is M=F•(d/2), where d=12 cm and F=qE.
From mechanics (Newton’s second law for rotation)
M=I•ε,
where I=(md^2)/12 is the moment of inertia of the rod about the axis passing through the center, and ε is the angular acceleration of the rod.
Therefore,
qE •(d/2) = ε• (md^2)/12,
ε=6qE/md=0.175 rad/s.
Since the angular velocity is ω= εt and angular displacement is φ= ε t^2/2,
ω=sqroot(2 ε•φ)= sqroot(2 0.175•π/2)=0.74 rad/s.