Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Two charges, each equal to q, are kept at x=-a\; and\; x=a on the x- axis. A particle of mass m and charge q_{0}=\frac{q}{2} is placed at the origin. If charge q_{0} is given a small displacement (y<<a)\; along\; the\; y - axis,the net force acting on the particle is proportional to :

 

  • Option 1)

    -\frac{1}{y}

  • Option 2)

    y

  • Option 3)

    -y

  • Option 4)

    \frac{1}{y}

 

Answers (2)

best_answer

As we discussed in

Magnitude of the Resultant force -

F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }

- wherein

 

 F_{net} = 2Fcos\theta

F_{net} = \frac {2Kq(\frac{q}{2})Y}{(\sqrt{Y^2 + a^2})^2(\sqrt{Y^2 + a^2})}

F_{net} = \frac {2Kq(\frac{q}{2})Y}{{Y^2 + a^2})^\frac{3}{2}}=\frac{Kq^2Y}{a^3}

\therefore F \propto Y

 

 

 

 


Option 1)

-\frac{1}{y}

Option 2)

y

Option 3)

-y

Option 4)

\frac{1}{y}

Posted by

solutionqc

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

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

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question