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  2. 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

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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)
Views
S solutionqc

 

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}}

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