A thin copper ring of radius `a` is charged with `q` units of electricity.An electron is placed at the centre of the copper ring .If the electron is displaced a little, it will have frequency

Options

A. `(1)/(2pi)sqrt((eq)/(4piepsilon_(0)ma^(3)))`
B. `(1)/(2pi)sqrt((eq)/(4piepsilon_(0)ema^(3)))`
C. `sqrt((eq)/(4piepsilon_(0)ma))`
D. `sqrt((q)/(4piepsilon_(0)ema^(3)))`

Explanation

To find the frequency of oscillation of an electron placed at the center of a charged copper ring when it is slightly displaced, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a thin copper ring of radius \( a \) charged with \( q \) units of electricity.
- An electron is placed at the center of the ring.

2. **Electric Field at the Center**:
- The electric field \( E \) at the center of the ring due to the uniformly charged ring is zero. This is because the contributions to the electric field from all parts of the ring cancel each other out.

3. **Displacing the Electron**:
- When the electron is displaced a small distance \( x \) from the center, it experiences a force due to the electric field generated by the ring.

4. **Electric Field at a Distance \( x \)**:
- The electric field \( E \) at a distance \( x \) from the center along the axis of the ring can be given by:
\[ E = \frac{kq x}{(a^2 + x^2)^{3/2}} \]
- Here, \( k \) is Coulomb's constant.

5. **Force on the Electron**:
- The force \( F \) on the electron when displaced by \( x \) is:
\[ F = -eE = -e \cdot \frac{kq x}{(a^2 + x^2)^{3/2}} \]
- Where \( e \) is the charge of the electron.

6. **Simplifying for Small Displacements**:
- For small displacements where \( x \) is much smaller than \( a \) (i.e., \( x \ll a \)), we can approximate:
\[ E \approx \frac{kq x}{a^3} \]
- Thus, the force becomes:
\[ F \approx -\frac{kqe}{a^3} x \]

7. **Relating Force to Acceleration**:
- According to Newton's second law, \( F = ma \), where \( m \) is the mass of the electron. Therefore:
\[ ma = -\frac{kqe}{a^3} x \]
- This can be rewritten as:
\[ a = -\frac{kqe}{ma^3} x \]

8. **Identifying Simple Harmonic Motion**:
- The equation \( a = -\omega^2 x \) indicates that the motion is simple harmonic with:
\[ \omega^2 = \frac{kqe}{ma^3} \]

9. **Finding the Frequency**:
- The frequency \( f \) of the oscillation is given by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{kqe}{ma^3}} \]

### Final Answer:
The frequency of oscillation of the electron when displaced slightly from the center of the charged copper ring is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{kqe}{ma^3}} \]