Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, - Electrostatics - JEE Main-2

A parallel plate capacitor with area 200 cm^{2} and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25 ×10^{-6} N, the value of V is approximately :

\left [ \epsilon _{0}= 8.85*10^{-12}\frac{c^{2}}{Nm^{2}} \right ]

  • Option 1)

    250 V

  • Option 2)

    100 V

  • Option 3)

    300 V

  • Option 4)

    150 V

 

Answers (2)

best_answer

As we learnt

Force between Parallel Plates Capacitor -

\dpi{100} F=\frac{\sigma ^{2}A}{2\epsilon _{0}}=\frac{Q^{2}}{2\epsilon _{0}A}=\frac{CV^{2}}{2d}

 

- wherein

\sigma -Surface\: charge\: density.

 

 

Force=\frac{\sigma ^{2}A}{2\epsilon _{0}}=\frac{1}{2}\epsilon _{0}E ^{2}A

F=\frac{1}{2}\epsilon _{0}\left ( \frac{v}{d} \right ) ^{2}A

v=d.\sqrt{\frac{2F}{\epsilon _{0}A}}

v=1.5*10^{-2}*\sqrt{\frac{2*25*10^{-6}}{8.85*10^{-12}*200*10^{-4}}}

v=1.5*10^{-2}*\sqrt{\frac{25}{8.85}*10^{8}}

     =\frac{1.5*10^{-2}*5*10^{4}}{3}=2.5*10^{2}v=250v

 

 


Option 1)

250 V

Option 2)

100 V

Option 3)

300 V

Option 4)

150 V

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question