A particle of charge q and mass m is subjecte | vedclass

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

Question

(D) The work done $W$ by the electric field on the charge $q$ as it moves from $x = 0$ to $x = x_{0}$ is given by the integral of the force $F = qE$ w

Vedclass · Accepted Answer

$\sqrt{\frac{3}{a}}$

Vedclass · Answer

(D) The work done $W$ by the electric field on the charge $q$ as it moves from $x = 0$ to $x = x_{0}$ is given by the integral of the force $F = qE$ with respect to displacement $dx$.$W = \int_{0}^{x_{0}} qE \, dx = qE_{0} \int_{0}^{x_{0}} (1 - ax^{2}) \, dx$Integrating the expression,we get:$W = qE_{0} \left[ x - \frac{ax^{3}}{3} ight]_{0}^{x_{0}} = qE_{0} \left( x_{0} - \frac{ax_{0}^{3}}{3} ight)$According to the work-energy theorem,the change in kinetic energy $\Delta KE$ is equal to the work done $W$. Since the particle starts from rest and we want to find the position where the kinetic energy is zero again,$\Delta KE = 0$,which implies $W = 0$.Setting $W = 0$:$qE_{0} \left( x_{0} - \frac{ax_{0}^{3}}{3} ight) = 0$Since $q 
eq 0$ and $E_{0} 
eq 0$,we have:$x_{0} - \frac{ax_{0}^{3}}{3} = 0$$x_{0} (1 - \frac{ax_{0}^{2}}{3}) = 0$Ignoring the initial position $x_{0} = 0$,we solve for $x_{0}$:$1 - \frac{ax_{0}^{2}}{3} = 0$$ax_{0}^{2} = 3$$x_{0} = \sqrt{\frac{3}{a}}$

Similar Questions

$A$ particle of mass $2 \ g$ and charge $6 \ \mu C$ is accelerated from rest through a potential difference of $60 \ V$. The speed acquired by the particle is (in $ms^{-1}$)

An oil drop of radius $2 \, mm$ with a density $3 \, g \, cm^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5} \, V \, m^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Consider $g = 9.81 \, m \, s^{-2}$)

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

$A$ positively charged particle moving along the $x$-axis with a certain velocity enters a uniform electric field directed along the positive $y$-axis. Its

Vedclass Test Series

Exam Paper Generator

Online Exam Module