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)/(4 pi in_0 ma^3))`
B. `(1)/( 2pi ) sqrt(( q)/(4 pi in_0 e ma^3))`
C. `(1)/( 2pi ) sqrt(( eq)/(4 pi in_0 ma))`
D. `(1)/( 2pi ) sqrt(( q)/(4 pi in_0 e ma^3))`

Answer

To solve the problem, we need to determine the frequency of an electron placed at the center of a charged copper ring when it is slightly displaced. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a thin copper ring of radius \( a \) charged with \( q \) units of electricity. An electron is placed at the center of this ring.

### Step 2: Electric Field at the Center of the Ring
The electric field \( E \) at the center of the ring due to the charge on the ring is zero. However, when the electron is displaced slightly from the center, it experiences a force due to the electric field created by the charge \( q \).

### Step 3: Displacement of the Electron
Let’s denote the displacement of the electron from the center as \( x \). The force \( F \) acting on the electron when it is displaced can be derived from the electric field \( E \) at that point.

### Step 4: Calculate the Electric Field
The electric field \( E \) at a distance \( r \) from the center of the ring can be given by:
\[ E = \frac{kq}{(a^2 + x^2)^{3/2}} \cdot x \]
where \( k \) is Coulomb's constant.

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

### Step 6: Approximation for Small Displacements
For small displacements \( x \) (where \( x \ll a \)), we can approximate \( (a^2 + x^2)^{3/2} \approx a^3 \). Therefore, the force simplifies to:
\[ F \approx -\frac{ekq}{a^3} x \]

### Step 7: Relate Force to Acceleration
Using Newton's second law, \( F = ma \), where \( m \) is the mass of the electron, we have:
\[ ma = -\frac{ekq}{a^3} x \]
This can be rewritten as:
\[ a = -\frac{ekq}{ma^3} x \]

### Step 8: Identify the Frequency
The equation \( a = -\omega^2 x \) suggests a simple harmonic motion where:
\[ \omega^2 = \frac{ekq}{ma^3} \]
Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{ekq}{ma^3}} \]

### Step 9: Calculate the Frequency
The frequency \( f \) is related to the angular frequency by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{ekq}{ma^3}} \]

### Step 10: Final Expression
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{q}{4\pi \epsilon_0 ma^3}} \]

### Conclusion
The frequency of the electron when displaced slightly from the center of the charged copper ring is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{q}{4\pi \epsilon_0 ma^3}} \]