Three charges q, q, and -2q are fixed on the vertices of an equilateral triangular plate of edge length a. This plate is in equilibrium between two very large plates having surfaces charge density `sigma_1 and sigma_2`, respectively. Find the time period of small anglular oscillations about an axis passing through its centroid and perpendicular to the plane. Moment of inertia of the system about this axis is l.

Options

A. `2pi sqrt((epsilon_0l)/(qa|sigma_1-sigma_2|)`
B. `2pi sqrt((epsilon_0l)/(2qa|sigma_1-sigma_2|)`
C. `2pi sqrt((2epsilon_0l)/(sqrt3qa|sigma_1-sigma_2|)`
D. `2pi sqrt((2epsilon_0l)/(qa|sigma_1-sigma_2|)`

Explanation

To solve the problem, we need to find the time period of small angular oscillations of a system of three charges arranged in an equilateral triangle, placed between two charged plates. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
We have three charges: \( q \), \( q \), and \( -2q \) located at the vertices of an equilateral triangle with side length \( a \). The triangle's centroid will be the axis about which we are considering the oscillations.

### Step 2: Calculate the Dipole Moment
The dipole moment \( \vec{P} \) can be calculated for the system. The dipole moment due to two charges \( q \) and \( -q \) can be calculated as:
\[ \vec{P} = q \cdot \vec{d} \]
where \( \vec{d} \) is the vector from the negative charge to the positive charge. For our configuration, we have two dipoles formed by the pairs of charges \( (q, -2q) \) and \( (q, -2q) \).

### Step 3: Calculate the Net Dipole Moment
Since the triangle is equilateral, the angle between the two dipoles is \( 60^\circ \). The net dipole moment \( P \) can be calculated using the formula:
\[ P_{\text{net}} = \sqrt{P_1^2 + P_2^2 + 2P_1P_2 \cos(60^\circ)} \]
Substituting \( \cos(60^\circ) = \frac{1}{2} \), we find:
\[ P_{\text{net}} = \sqrt{P^2 + P^2 + P^2} = \sqrt{3}P \]
where \( P = q \cdot a \).

### Step 4: Electric Field Between the Plates
The electric field \( E \) between two plates with surface charge densities \( \sigma_1 \) and \( \sigma_2 \) is given by:
\[ E = \frac{\sigma_1 - \sigma_2}{2\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 5: Calculate the Torque
The torque \( \tau \) on the dipole in an electric field is given by:
\[ \tau = \vec{P} \times \vec{E} = P E \sin(\theta) \]
For small angles, \( \sin(\theta) \approx \theta \).

### Step 6: Relate Torque to Angular Acceleration
Using Newton's second law for rotation, we have:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. Thus:
\[ P E \theta = I \alpha \]

### Step 7: Time Period of Oscillation
The time period \( T \) for small oscillations is given by:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} \]
Substituting \( \tau = P E \) gives us:
\[ T = 2\pi \sqrt{\frac{I}{P E}} \]

### Step 8: Substitute Values
Substituting \( P = \sqrt{3} q a \) and \( E = \frac{\sigma_1 - \sigma_2}{2\epsilon_0} \):
\[ T = 2\pi \sqrt{\frac{I}{\sqrt{3} q a \cdot \frac{\sigma_1 - \sigma_2}{2\epsilon_0}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{2\epsilon_0 I}{\sqrt{3} q a |\sigma_1 - \sigma_2|}} \]

### Final Answer
The time period of small angular oscillations about the centroid is:
\[ T = 2\pi \sqrt{\frac{2\epsilon_0 I}{\sqrt{3} q a |\sigma_1 - \sigma_2|}} \]