An electric dipole of dipole moment vecP is released in a uniform electric filed vecE from the position of maximum torque. The angular speed of the dipole when vecP becomes parallel to vecE will be [I is moment of inertia of dipole]

Question

An electric dipole of dipole moment vecP is released in a uniform electric filed vecE from the position of maximum torque. The angular speed of the dipole when vecP becomes parallel to vecE will be [I is moment of inertia of dipole] (A) √(P E/I) (B) √(2 P E/I) (C) √(4 P E/3 I) (D) √(2 I/3 P E)

Tardigrade Admin · Accepted Answer

Δ K=Δ U (1/2) I ω2=P E cos 0°-P E cos 90° (1/2) I ω2=P E ω=√(2 P E/I)

Q. An electric dipole of dipole moment $P$ is released in a uniform electric filed $E$ from the position of maximum torque. The angular speed of the dipole when $P$ becomes parallel to $E$ will be $[I$ is moment of inertia of dipole]

$\frac{PE}{I}$

$\frac{2 PE}{I}$

$\frac{4 PE}{3 I}$

$\frac{2 I}{3 PE}$

$Δ K = Δ U$
$\frac{1}{2} I ω^{2} = PE cos 0^{\circ} - PE cos 9 0^{\circ}$
$\frac{1}{2} I ω^{2} = PE$
$ω = \frac{2 PE}{I}$

Q. An electric dipole of dipole moment P is released in a uniform electric filed E from the position of maximum torque. The angular speed of the dipole when P becomes parallel to E will be [I is moment of inertia of dipole]

IPE​​

I2PE​​

3I4PE​​

3PE2I​​