Question

A charged particle q of mass m is in equilibrium at a height h from a horizontal infinite line charge with uniform linear charge density $\lambda$. The charge lies in the vertical plane containing the line charge. If the particle is displaced slightly (vertically), prove that the motion of the charged particle will be simple harmonic. Also find its time period

Answer 1

At equilibrium $mg=qE$

$mg=\frac{q2k\lambda }{h}$ (i)

When particle is displaced to a distance x in upward direction, then net force in upward direction is

$F=q\left(\frac{2k\lambda }{h+x}\right)-mg$

$=\frac{2kq\lambda }{h}{(1+\frac{x}{h})}^{-1}-mg=\frac{2kq\lambda }{h}(1-\frac{x}{h})-mg$

$=-\left(\frac{2Kq\lambda }{h^2}\right)x=-K_{0}x$

$\therefore F\propto -x$ (Motion is SHM)

Time period of motion is

$T=2\pi \sqrt{\frac{m}{K_0}}=2\pi \sqrt{\frac{m{h}^2}{2kq\lambda }}$