Coulomb's Law: Practice Problem & Solution
Two charges each of magnitude Q are fixed at 2a distance apart. A third charge (–q of mass 'm') is placed at the mid point of the two charges; now –q charge is slightly displaced perpendicular to the line joining the charges then find its time period:
Solution Explained:
To solve this problem, we apply the core principles of Coulomb's Law. Understanding the underlying formula is key to arriving at the correct answer below:
To find the time period of oscillation of the third charge \(-q\) when displaced perpendicular to the line joining the two charges of magnitude \(+Q\), we can follow these steps:
1. Restoring Force: When the charge \(-q\) is displaced by a small distance \(y\) perpendicular to the line joining the two charges, the net force on it due to the two fixed charges \(+Q\) has a restoring nature and acts toward the equilibrium position.
2. Approximation: For small \(y\), the restoring force \(F\) is proportional to \(y\), making it a simple harmonic motion (SHM) problem.
3. Electric Field: The electric field at the midpoint due to each charge \(+Q\) is \(E = \frac{Q}{4\pi\epsilon_0 (a^2 + y^2)}\). Using \(y \ll a\), we approximate the force on \(-q\) by expanding for small \(y\), resulting in a force \(F = -k y\), where \(k = \frac{2Qq}{4\pi \epsilon_0 a^3}\).
4. Angular Frequency: For SHM, the angular frequency \(\omega\) is \(\sqrt{\frac{k}{m}} = \sqrt{\frac{2Qq}{4\pi \epsilon_0 a^3 m}}\).
5. Time Period: The time period \(T\) is given by:
\[
T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2ma^3 \pi \epsilon_0}{Qq}}
\]
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