Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An electron moving with the speed of  m/s is shot parallel to the electric field of

An electron moving with the speed of 5\times 10^6 m/s is shot parallel to the electric field of intensity 7\times 10^3 N/C. The field is responsible for the retardation of the motion of electrons. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of e = 9\times 10^{-31}kg charge =1.6\times 10^{-19}C)

Option: 1

1 cm


Option: 2

1 mm


Option: 3

1 m


Option: 4

1km


Answers (1)

best_answer

Electric force qE = ma \Rightarrow a = \frac{QE}{m} \therefore a = \frac{1.6\times 10^{-19}\times 7\times 10^3}{9\times 10^{-31}} = \frac{7\times 1.6}{9}\times 10^{15}

u = 5\times 10^6 and v = 0 

\therefore from v^2 = u^2 -2as\Rightarrow s = \frac{u^2}{2a}

\therefore Distance\;= s = \frac{(5\times 10^6)^2\times 9}{2\times 7\times 1.6\times 10^{15}} = 1 cm (approx)

 

Posted by

Gautam harsolia

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

NTA Exam Changes Live: Reforms in NEET UG, JEE Main, CUET; test centres, question papers, biometric data
anam-khan

NTA to not hold recruitment exams; will be restructured in 2025: Dharmendra Pradhan
anam-khan

JEE Main 2025: BTech ECE remains among top engineering choices; NIT cut-offs

Latest Question