Question

A dipole is formed by two point charge -q and +q, each of mass m, and both are connected by a rod of length $l$ and mass m. This dipole is placed in uniform. If the dipole is disturbed by a small angle $\theta$ from stable equilibrium position motion will be almost SHM. Also find its time period. If the dipole is disturbed by $\theta$ angle, $\tau_{net}$ = -PE sin$\theta$ (Here -ve sign indicates torque is opposite to $\theta$)

If $\theta$ is very small, sin$\theta$ $\approx$ $\theta$ $\therefore$ $\tau_{net}$ = - (PE)$\theta$ $\tau_{net}$ $\propto$ (-$\theta$) so motion will be almost SHM & C = PE (where, P = q$l$)

$\therefore$ T=2$\pi$ $\sqrt{\frac{I}{C}}$

$\therefore$ T = 2m$\sqrt{\frac{\frac{ml}{12} + 2m(\frac{l}{2})^2}{PE}}$ = 2$\pi$ $\sqrt{\frac{\frac{ml}{12} + \frac{ml^2}{2}}{q lE}}$ = 2$\pi$ $\sqrt{\frac{7ml^2}{12qlE}}$ = 2$\pi$ $\sqrt{\frac{7ml}{12qE}}$

T = $\pi$ $\sqrt{\frac{7ml}{3qE}}$

Answer 1

T = $\pi$ $\sqrt{\frac{7ml}{3qE}}$

Answer 2

Explanation of the Solution:

1. Torque on the Dipole: When an electric dipole with dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$, it experiences a torque given by $\vec{\tau} = \vec{p} \times \vec{E}$. The magnitude of this torque is $\tau = pE \sin\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. For a dipole formed by charges +q and -q separated by length $l$, the dipole moment is $p = ql$. The stable equilibrium position is when the dipole moment is aligned with the electric field ($\theta = 0^\circ$). If the dipole is disturbed by a small angle $\theta$ from this position, the torque acts as a restoring torque, trying to bring it back to equilibrium. Therefore, we write $\tau_{net} = -pE \sin\theta$. The negative sign indicates that the torque is opposite to the angular displacement.

2. Small Angle Approximation for SHM: For small angular displacements, $\sin\theta \approx \theta$ (in radians). Substituting this into the torque equation, we get $\tau_{net} = -pE\theta$. This equation is in the form of angular Simple Harmonic Motion (SHM), $\tau_{net} = -C\theta$, where $C$ is the torsional constant. By comparison, $C = pE = qlE$. Since the restoring torque is directly proportional to the negative of the angular displacement, the motion will be approximately SHM for small angles.

3. Moment of Inertia (I): The system consists of a rod of mass $m$ and length $l$, and two point charges, each of mass $m$, attached at the ends of the rod. The rotation occurs about the center of mass of the system, which is the midpoint of the rod.

   • Moment of inertia of the rod about its center: $I_{rod} = \frac{1}{12}ml^2$.
   • Moment of inertia of each point charge about the center of the rod: $I_{charge} = m r^2$. Here, $r = l/2$ (distance from the center to each charge). So, for one charge, $I_{charge} = m(l/2)^2 = \frac{1}{4}ml^2$.
   • Total moment of inertia of the two charges: $I_{2charges} = 2 \times \frac{1}{4}ml^2 = \frac{1}{2}ml^2$.
   • Total moment of inertia of the dipole system: $I = I_{rod} + I_{2charges} = \frac{1}{12}ml^2 + \frac{1}{2}ml^2$ To add these, find a common denominator (12): $I = \frac{ml^2}{12} + \frac{6ml^2}{12} = \frac{7ml^2}{12}$.

4. Time Period of SHM: The time period for angular SHM is given by the formula $T = 2\pi \sqrt{\frac{I}{C}}$. Substitute the calculated values of $I$ and $C$: $T = 2\pi \sqrt{\frac{\frac{7ml^2}{12}}{qlE}}$ $T = 2\pi \sqrt{\frac{7ml^2}{12qlE}}$ Cancel one 'l' from the numerator and denominator: $T = 2\pi \sqrt{\frac{7ml}{12qE}}$ To simplify further, move the '2' inside the square root by squaring it ($2 = \sqrt{4}$): $T = \pi \sqrt{4 \times \frac{7ml}{12qE}}$ $T = \pi \sqrt{\frac{28ml}{12qE}}$ Divide both 28 and 12 by their greatest common divisor, which is 4: $T = \pi \sqrt{\frac{7ml}{3qE}}$

The final expression for the time period of oscillation is $T = \pi \sqrt{\frac{7ml}{3qE}}$.