A dipole consisting of `+10 nC and -10 nC` separated by a distance of 2cm oscillates in an electric field of strength 60,000 `Vm^(-1)`. The frequency of its oscillation is (M.I. about the axis of oscillations is `3 xx 10^(-10)kg m^(2)`)

Options

A. 20.2 Hz
B. 25.4 Hz
C. 31.38 Hz
D. 37.1 Hz

Answer

To find the frequency of oscillation of the dipole in the given electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dipole Moment (p)**:
The dipole moment \( p \) is given by the product of the charge and the separation distance.
\[ p = q \cdot d \]
where \( q = 10 \, \text{nC} = 10 \times 10^{-9} \, \text{C} \) and \( d = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} \).
\[ p = (10 \times 10^{-9} \, \text{C}) \cdot (2 \times 10^{-2} \, \text{m}) = 2 \times 10^{-10} \, \text{C m} \]

2. **Calculate the Angular Frequency (ω)**:
The angular frequency \( \omega \) of the oscillation can be calculated using the formula:
\[ \omega = \sqrt{\frac{p \cdot E}{I}} \]
where \( E = 60,000 \, \text{V/m} \) and \( I = 3 \times 10^{-10} \, \text{kg m}^2 \).
\[ \omega = \sqrt{\frac{(2 \times 10^{-10} \, \text{C m}) \cdot (60,000 \, \text{V/m})}{3 \times 10^{-10} \, \text{kg m}^2}} \]
\[ = \sqrt{\frac{(2 \times 10^{-10}) \cdot (60,000)}{3 \times 10^{-10}}} \]
\[ = \sqrt{\frac{1.2 \times 10^{-6}}{3 \times 10^{-10}}} = \sqrt{4 \times 10^{3}} = 2 \times 10^{2} \, \text{rad/s} \]

3. **Convert Angular Frequency to Frequency (f)**:
The frequency \( f \) is related to the angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
\[ f = \frac{2 \times 10^{2}}{2 \times 3.14} \approx \frac{200}{6.28} \approx 31.37 \, \text{Hz} \]

### Final Answer:
The frequency of oscillation of the dipole is approximately **31.37 Hz**.