Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve this problem A uniformly charged ring of radius 3a and total charge q is placedin xy-plane centred at origin. A point charge q is moving towardsthe ring along the z-axis and has speed v at z=4a. The minimumvalue of v such that it crosses the or

A uniformly charged ring of radius 3a and total charge q is placed

in xy-plane centred at origin. A point charge q is moving towards

the ring along the z-axis and has speed v at z=4a. The minimum

value of v such that it crosses the origin is :

  • Option 1)

    \sqrt{(\frac{2}{m})}(\frac{4}{15}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

  • Option 2)

    \sqrt{(\frac{2}{m})}(\frac{1}{5}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

  • Option 3)

    \sqrt{(\frac{2}{m})}(\frac{2}{15}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

  • Option 4)

    \sqrt{(\frac{2}{m})}(\frac{1}{15}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

 

Answers (1)

best_answer

 

E and V at a point P that lies on the axis of ring -

\dpi{100} E_{x}=\frac{kQx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}  ,   V=\frac{kQ}{\left ( x^{2}+R^{2} \right )^{\frac{1}{2}}}

-

 

Potential energy Per unit charge -

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

- wherein

S.I unit is \frac{J}{C}.

 

 

Use energy conservation

\Delta K.E+\Delta U=0

0-\frac{1}{2}mv^{2}=q(\frac{kq}{5a}-\frac{kq}{3a})

\frac{1}{2}mv^{2}=\frac{2kq^{2}}{15a}

v=(\frac{4}{15})(\frac{kq^{2}}{am})

v=\sqrt{\frac{2}{m}}\times (\frac{2}{15}\times (\frac{q^{2}}{4\pi\varepsilon _o})\times \frac{1}{a})^{\frac{1}{2}}


Option 1)

\sqrt{(\frac{2}{m})}(\frac{4}{15}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

Option 2)

\sqrt{(\frac{2}{m})}(\frac{1}{5}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

Option 3)

\sqrt{(\frac{2}{m})}(\frac{2}{15}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

Option 4)

\sqrt{(\frac{2}{m})}(\frac{1}{15}\frac{q^{2}}{4\pi\varepsilon _oa})^{\frac{1}{2}}

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