Question

A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same charge 'q'. If the free charged particle is displaced from its equilibrium position through distance 'x' ($x << 1$ m). The particle executes SHM. Its angular frequency of oscillation will be ________ $\times 10^5$ rad/s if $q^2 = 10 \text{ C}^2$.

Hint:

For a central charge \(q\) between two fixed charges \(Q\) with separation \(2d\), the longitudinal frequency is \(\omega = \sqrt{\frac{4kQq}{md^3}}\). Always convert mass to kg to avoid power-of-ten errors!

Answer 1

Correct Answer: 6000

Step 1: Understanding the Concept:
When a charge is placed at the midpoint between two identical charges, the net force is zero.
Displacing the middle charge along the line joining the fixed charges creates a restoring force.
For a small displacement \(x\), this force is proportional to \(x\), which is the condition for Simple Harmonic Motion (SHM).
Step 2: Key Formula or Approach:
The restoring force constant for a longitudinal displacement \(x\) is:
\[ k_{SHM} = \frac{4 k q^2}{d^3} \]
Where \(d\) is the distance from the midpoint to one of the fixed charges (\(d = 1 \text{ m}\)).
The angular frequency is:
\[ \omega = \sqrt{\frac{k_{SHM}}{m}} \]
Step 3: Detailed Explanation:
Given:
Total distance between fixed charges = \(2 \text{ m}\) $\implies$ \(d = 1 \text{ m}\).
Mass \(m = 1 \text{ mg} = 10^{-6} \text{ kg}\).
Charge parameter \(q^2 = 10 \text{ C}^2\).
Electrostatic constant \(k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\).

Calculate the restoring force constant \(k_{SHM}\):
\[ k_{SHM} = \frac{4 \times 9 \times 10^9 \times 10}{1^3} = 3.6 \times 10^{11} \text{ N/m} \]

Calculate the angular frequency \(\omega\):
\[ \omega = \sqrt{\frac{3.6 \times 10^{11}}{10^{-6}}} = \sqrt{36 \times 10^{16}} = 6 \times 10^8 \text{ rad/s} \]

Expressing in the form \(....... \times 10^5\):
\[ \omega = 6000 \times 10^5 \text{ rad/s} \]
Step 4: Final Answer:
The angular frequency is 6000 $\times 10^5$ rad/s.