Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A particle of charge q and mass m is subjected to an electric field  in the x - direction, where a and 

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 \sqrt{\frac{2}{a}}
Option: 3 \sqrt{\frac{3}{a}}
Option: 4 \sqrt{\frac{1}{a}}

Answers (1)

best_answer

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}

 

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question