# A particle of charge q and mass m is subjected to an electric field $E=E_{0}(1-ax^{2})$ in the x - direction, where a and $E_{0}$ are constants. Initially the particle was at rest at x=0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is: Option: 1 a Option: 2 Option: 3 Option: 4

$E=E_{0}\left(1-a x^{2}\right)$

As we know that force acting on a charge in an electric field is qE and the acceleration will be -

$a = \frac{qE}{m}$

$\begin{array}{l} \frac{v d v}{d x}=\frac{q E_{0}}{m}\left(1-a x^{2}\right) \\ \Rightarrow \quad \frac{v^{2}}{2}=\frac{q E_{0}}{m}\left[x-\frac{a x^{3}}{3}\right]=0 \\ \Rightarrow \quad x=\sqrt{\frac{3}{a}} \end{array}$

## Most Viewed Questions

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
##### Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-