an electric dipole of moment P is released in a uniform electric field E from the position of maximum torque angular speed of the dipole when P becomes parallel to e will be(I=moment of inertia)
a)√2PE/I
b)√PE/I
c)√4PE/I
d)√2 I/PE
Answers
Given:
An electric dipole of moment P is released in a uniform electric field E from the position of maximum torque
To find:
The angular speed of the dipole when P becomes parallel to e will be(I=moment of inertia)
Solution:
From given, we have,
An electric dipole of moment P is released in a uniform electric field E from the position of maximum torque
As the dipole experiences a torque pEsinθ tending to bring itself back in the direction of the field, so on being released the dipole oscillates about an axis through its centre of mass and perpendicular to the field.
The equation of motion considering the moment of inertia I, we have,
d²θ/dt² =−pEsinθ
For a smaller angular displacement/amplitude
sinθ ≈ θ
Thus d²θ/dt² = α = −(pE/I).θ=−ω²θ
where ω= (pE/I)
This is an S.H.M whose period of oscillation is T=2π √(I/PE)
Therefore the period of oscillation is T =2π √(I/PE)