Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

In moving from A to B along an electric field line, the electric field does 6.4\times 10^{-19} J of work on an electron. If \phi_1, \phi_2 are equipotential surfaces, then the potential difference (V_C - V_A) is  

  • Option 1)

    -4V

  • Option 2)

    4V

  • Option 3)

    Zero

  • Option 4)

    64V

 

Answers (1)

As we have learnt,

 

Potential energy Per unit charge -

V =\frac{W}{Q}=\frac{U}{Q}

- wherein

S.I unit is \frac{J}{C}.

 

 

 

Work done by the field

W = q(-dV) =-e (V_A - V_B) = -e(V_B - V_A) = e(V_C - V_A) \\*\Rightarrow (V_C - V_A) = \frac{W}{e} = \frac{6.4\times 10^{-19}}{1.6\times 10^{-19}} = 4V


Option 1)

-4V

Option 2)

4V

Option 3)

Zero

Option 4)

64V

Posted by

Vakul

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 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question