An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency `omega` is:

Options

A. `sqrt((q^E)/(2md))`
B. `sqrt((q^E)/(md))`
C. `sqrt((2q^E)/(md))`
D. `2sqrt((q^E)/(md))`

Explanation

To find the angular frequency \( \omega \) of an electric dipole in a uniform electric field when it is slightly rotated from its equilibrium position, we can follow these steps:

### Step 1: Understand the dipole moment
The electric dipole consists of two equal and opposite charges \( +q \) and \( -q \) separated by a distance \( d \). The dipole moment \( \mathbf{p} \) is given by:
\[ \mathbf{p} = q \cdot d \]

### Step 2: Torque on the dipole
When the dipole is placed in a uniform electric field \( \mathbf{E} \), the torque \( \tau \) acting on the dipole is given by:
\[ \tau = \mathbf{p} \times \mathbf{E} \]
For small angles \( \theta \), this can be approximated as:
\[ \tau \approx pE \sin \theta \approx pE \theta \]
since \( \sin \theta \approx \theta \) for small \( \theta \).

### Step 3: Relate torque to angular acceleration
The net torque is also related to the moment of inertia \( I \) and the angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
Thus, we can write:
\[ I \alpha = pE \theta \]

### Step 4: Express angular acceleration
From the above equation, we can express angular acceleration \( \alpha \) as:
\[ \alpha = \frac{pE}{I} \theta \]

### Step 5: Relate to simple harmonic motion
The equation \( \alpha = \frac{pE}{I} \theta \) resembles the form of simple harmonic motion, where:
\[ \alpha = -\omega^2 \theta \]
This implies:
\[ \omega^2 = \frac{pE}{I} \]

### Step 6: Substitute the dipole moment
Substituting \( p = qd \) into the equation gives:
\[ \omega^2 = \frac{qdE}{I} \]

### Step 7: Calculate the moment of inertia
For the dipole consisting of two masses \( m \) located at a distance \( d \) from the center, the moment of inertia \( I \) about the center of the dipole is:
\[ I = 2 \cdot \left( m \cdot \left( \frac{d}{2} \right)^2 \right) = 2m \cdot \frac{d^2}{4} = \frac{md^2}{2} \]

### Step 8: Substitute moment of inertia into the equation
Substituting \( I \) into the equation for \( \omega^2 \):
\[ \omega^2 = \frac{qdE}{\frac{md^2}{2}} = \frac{2qdE}{md^2} \]

### Step 9: Solve for angular frequency \( \omega \)
Taking the square root gives:
\[ \omega = \sqrt{\frac{2qdE}{md^2}} \]

### Final Answer
Thus, the angular frequency \( \omega \) of the dipole is:
\[ \omega = \sqrt{\frac{2qe}{md}} \]

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