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# 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)

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

Views

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

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

-

Potential energy Per unit charge -

$\dpi{100} V =\frac{W}{Q}=\frac{U}{Q}$

- wherein

S.I unit is $\dpi{100} \frac{J}{C}$.

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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