A particle of mass 1 , mg and charge q is pla | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the distance between the two stationary charges be $2d = 2 \, m$,so $d = 1 \, m$. The mass of the particle is $m = 1 \, mg = 10^{-6} \, kg$.Th

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let the distance between the two stationary charges be $2d = 2 \, m$,so $d = 1 \, m$. The mass of the particle is $m = 1 \, mg = 10^{-6} \, kg$.The net force on the free charged particle when displaced by $x$ is:$F = \frac{kq^{2}}{(d-x)^{2}} - \frac{kq^{2}}{(d+x)^{2}}$$F = kq^{2} \left[ \frac{(d+x)^{2} - (d-x)^{2}}{(d^{2}-x^{2})^{2}} ight] = kq^{2} \left[ \frac{4dx}{(d^{2}-x^{2})^{2}} ight]$Since $x \ll d$,we can approximate $(d^{2}-x^{2})^{2} \approx d^{4}$:$F \approx \frac{4kq^{2}dx}{d^{4}} = \frac{4kq^{2}}{d^{3}} x$Since the force is directed towards the equilibrium position,$F = -kx_{spring} = -m\omega^{2}x$. Thus:$m\omega^{2} = \frac{4kq^{2}}{d^{3}}$$\omega = \sqrt{\frac{4kq^{2}}{md^{3}}}$Substituting the values $k = 9 	imes 10^{9} \, N \cdot m^{2}/C^{2}$,$q^{2} = 10 \, C^{2}$,$m = 10^{-6} \, kg$,and $d = 1 \, m$:$\omega = \sqrt{\frac{4 	imes 9 	imes 10^{9} 	imes 10}{10^{-6} 	imes 1^{3}}} = \sqrt{36 	imes 10^{16}} = 6 	imes 10^{8} \, rad/s$.Thus,the angular frequency is $6 	imes 10^{8} \, rad/s$.

Similar Questions

$A$ charge $+Q$ is placed at each of the diagonally opposite corners of a square. $A$ charge $-q$ is placed at each of the other diagonally opposite corners as shown. If the net electrical force on $+Q$ is zero,then $\frac{+Q}{-q}$ is equal to

State Coulomb's law and explain its scalar form.

Two point charges of $+2 \, \mu \text{C}$ and $+6 \, \mu \text{C}$ repel each other with a force of $12 \, \text{N}$. If a charge of $-4 \, \mu \text{C}$ is added to each,the force will be:

Five charges, each $+Q$, are placed at five vertices of a regular hexagon of side length $L$. What is the magnitude of the force on a charge $-Q$ placed at the center of the hexagon?

$A$ proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially,the $\alpha$-particle is at rest,but it is free to move. If gravity is neglected,then the minimum separation along the straight line of their motion will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module