physics!
posted by cha on .
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?

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.