A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and at a distance 2L apart so that distance of this charge from each charge located at end is L. Assuming that the middle charge moves along the line joining the fixed charges. The frequency of oscillation when it is displaced slightly is

Options

A. `q/(pisqrt(mpiepsilon_0L^3))`
B. `q/(4pi(mpiepsilon_0L^3))`
C. `(2q)/(pisqrt(mpiepsilon_0L^3))`
D. `q/(2pisqrt(mpiepsilon_0L^3))`

Answer

To solve the problem of finding the frequency of oscillation of a particle of mass \( m \) and charge \( q \) located midway between two fixed charged particles each having charge \( q \) and at a distance \( 2L \) apart, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have three charges: two fixed charges \( q \) located at points \( A \) and \( B \) which are \( 2L \) apart, and a movable charge \( q \) located at the midpoint \( M \) (distance \( L \) from both \( A \) and \( B \)).

2. **Displacement of the Middle Charge**:
- When the middle charge \( q \) is displaced slightly by a distance \( x \), the new distances from \( A \) and \( B \) become:
- Distance from \( A \): \( L - x \)
- Distance from \( B \): \( L + x \)

3. **Calculating Forces**:
- The force \( F_1 \) due to charge \( A \) on the middle charge \( q \) is given by Coulomb's law:
\[ F_1 = \frac{kq^2}{(L - x)^2} \]
- The force \( F_2 \) due to charge \( B \) on the middle charge \( q \) is:
\[ F_2 = \frac{kq^2}{(L + x)^2} \]

4. **Net Force on the Middle Charge**:
- The net force \( F \) acting on the middle charge \( q \) when displaced is:
\[ F = F_2 - F_1 = \frac{kq^2}{(L + x)^2} - \frac{kq^2}{(L - x)^2} \]

5. **Simplifying the Force Expression**:
- To simplify, we can use the approximation for small \( x \) (i.e., \( x \) is much smaller than \( L \)):
\[ F \approx kq^2 \left( \frac{1}{(L - x)^2} - \frac{1}{(L + x)^2} \right) \]
- Using the identity \( \frac{1}{a^2} - \frac{1}{b^2} = \frac{b^2 - a^2}{a^2 b^2} \), we can rewrite this as:
\[ F \approx kq^2 \cdot \frac{(L + x)^2 - (L - x)^2}{(L - x)^2 (L + x)^2} \]
- The numerator simplifies to \( 4Lx \) (since \( (L + x)^2 - (L - x)^2 = 4Lx \)), leading to:
\[ F \approx \frac{4kq^2Lx}{(L^2 - x^2)^2} \]
- For small \( x \), we can approximate \( (L^2 - x^2) \approx L^2 \):
\[ F \approx \frac{4kq^2Lx}{L^4} = \frac{4kq^2}{L^3} x \]

6. **Relating Force to Oscillation**:
- The net force can be related to the acceleration of the mass \( m \):
\[ F = ma \quad \text{where } a = \frac{d^2x}{dt^2} \]
- Thus, we can write:
\[ ma = -\frac{4kq^2}{L^3} x \]
- This is in the form of simple harmonic motion \( F = -kx \), where \( k = \frac{4kq^2}{L^3} \).

7. **Finding the Frequency**:
- The angular frequency \( \omega \) is given by:
\[ \omega^2 = \frac{4kq^2}{mL^3} \]
- The frequency \( f \) is related to \( \omega \) by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{4kq^2}{mL^3}} \]

8. **Substituting for \( k \)**:
- Recall that \( k = \frac{1}{4\pi\epsilon_0} \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{4 \cdot \frac{1}{4\pi\epsilon_0} \cdot q^2}{mL^3}} = \frac{q}{2\pi\sqrt{m\epsilon_0 L^3}} \]

### Final Answer:
The frequency of oscillation when the charge is displaced slightly is:
\[ f = \frac{q}{2\pi\sqrt{m\epsilon_0 L^3}} \]