Question

A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2m' 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 ______ × 10⁵ rad/s if q² = 10 C².

Answer 1

6000

Answer 2

The two stationary charges are located at $x = -a$ and $x = a$, where $2a = 2$ m, so $a = 1$ m. The free particle of mass $m = 1$ mg $= 10^{-6}$ kg and charge $q$ is initially at $x=0$. When displaced by a small distance $x$ ($x << a$), the net force on the particle is given by $F = \frac{kq^2}{(a+x)^2} - \frac{kq^2}{(a-x)^2}$. For small $x$, this simplifies to $F \approx -\frac{4kq^2}{a^3}x$. This is a restoring force of the form $F = -Kx$, where $K = \frac{4kq^2}{a^3}$. The angular frequency of SHM is $\omega = \sqrt{\frac{K}{m}}$. Substituting the given values $m = 10^{-6}$ kg, $a = 1$ m, $q^2 = 10$ C², and $k = 9 \times 10^9$ Nm²/C², we get $\omega = \sqrt{\frac{4 \times (9 \times 10^9) \times 10}{(10^{-6}) \times (1)^3}} = 6 \times 10^8$ rad/s. The question asks for the value in the format ______ × 10⁵ rad/s, which is $\frac{6 \times 10^8}{10^5} = 6000$.