Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve it, A body of mass M and charge q isconnected to a spring of spring constantk. It is oscillating along x-direction aboutits equilibrium position, taken to be at x=0,with an amplitude A. An electric field Eis applied along the x-direction. Which

A body of mass M and charge q is
connected to a spring of spring constant
k. It is oscillating along x-direction about
its equilibrium position, taken to be at x=0,
with an amplitude A. An electric field E
is applied along the x-direction. Which of
the following statements is correct?

  • Option 1)

    The new equilibrium position is at a
    distance \frac{qE}{2k}  from x=0.

     

     


     

     

  • Option 2)

    The total energy of the system is \frac{1}{2}mw^{2} A^{2}+\frac{1}{2}\frac{q^{2}E^{2}}{k}.

  • Option 3)

    The total energy of the system is\frac{1}{2}mw^{2} A^{2}-\frac{1}{2}\frac{q^{2}E^{2}}{k}.

  • Option 4)

    The new equilibrium position is at a
    distance \frac{2 q E}{k} from x=0.

 

Answers (1)

best_answer

As we learned 

at new equilibrium position 

f_{net}=0

\Rightarrow qE - Kx =0

x=\frac{qE}{k}  new equilibrium position 

Energy = \frac{1}{2} \cdot mw^{2}\left [ A^{2} +\left ( \frac{qE}{k}\right )^{2}\right ]

=\frac{1}{2} \cdot mw^{2} A^{2} +\frac{1}{2}\frac{q^{2}E^{2}}{k}


Option 1)

The new equilibrium position is at a
distance \frac{qE}{2k}  from x=0.

 

 


 

 

Option 2)

The total energy of the system is \frac{1}{2}mw^{2} A^{2}+\frac{1}{2}\frac{q^{2}E^{2}}{k}.

Option 3)

The total energy of the system is\frac{1}{2}mw^{2} A^{2}-\frac{1}{2}\frac{q^{2}E^{2}}{k}.

Option 4)

The new equilibrium position is at a
distance \frac{2 q E}{k} from x=0.

Posted by

Avinash

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