Question

Consider the system shown below. If the charge is slightly displaced along the perpendicular to the wire from its equilibrium position then find out the time period of $\text{SHM}$ $(k=\frac{1}{4\pi {\epsilon }_o})$.

• A. $2\pi \sqrt{\frac{m{d}^2}{k\lambda q}}$
• B. $2\pi \sqrt{\frac{m{d}^2}{3k\lambda q}}$
  
• C. $\pi \sqrt{\frac{m{d}^2}{2k\lambda q}}$
  
• D. $2\pi \sqrt{\frac{m{d}^2}{2k\lambda q}}$
  

Answer 1

The correct option is D $2\pi \sqrt{\frac{m{d}^2}{2k\lambda q}}$


Given,
charge of the particle$=q$,
mass of the charge$=m$,
charge per unit length of the wire$=\lambda$,
perpendicular distance between the charge and wire$=d$.


At equilibrium position weight of the particle is balanced by the electric force, $$mg=qE...\left(1\right)$$ Electric field due to a infinite wire having charge density $\lambda$ is, $$E=\frac{2k\lambda }{d}...\left(2\right)$$From equation $\left(1\right)$ and $\left(2\right)$, we have

$mg=q\left(\frac{2k\lambda }{d}\right)...\left(3\right)$


Now if the particle is slightly displaced by a distance $x$ (where $x<<d$), the net force on the body,

$F_{net}=\frac{2k\lambda q}{d+x}-mg...\left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$, we have

$F_{net}=\frac{2k\lambda q}{d+x}-\frac{2k\lambda q}{d}$

$\Rightarrow F_{net}=\frac{-2k\lambda qx}{d(d+x)}$

As $x<<d$

$\Rightarrow F_{net}=\frac{-2k\lambda qx}{d^2}$

If $a$ is the acceleration of the charged particle due to force $F_{net}$, then

$\Rightarrow a=\frac{F_{net}}{m}-\frac{-2k\lambda qx}{m{d}^2}$

Comparing this with the equation of $\text{SHM}$, $a=-{\omega }^{2}x$ we get, angular velocity of the motion of particle,

${\omega }^2=\frac{2k\lambda q}{m{d}^2}$

$\Rightarrow \omega =\sqrt{\frac{2k\lambda q}{m{d}^2}}$

Therefore, the time period of the motion,

$T=\frac{2\pi }{\omega }$

$\therefore T=2\pi \sqrt{\frac{m{d}^2}{2k\lambda q}}$

Hence, option (d) is the correct answer.