JEE Main 2019 — Physics solution | Fundamenthol

JEE Main2019Physics

Q.

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 :

A
$\frac{2}{m} (\frac{1}{15} \frac{q ^{2}}{4 π ε _{0} a})^{\frac{1}{2}}$
B
$\frac{2}{m} (\frac{1}{5} \frac{q ^{2}}{4 π ε _{0} a})^{\frac{1}{2}}$
C
$\frac{2}{m} (\frac{4}{15} \frac{q ^{2}}{4 π ε _{0} a})^{\frac{1}{2}}$
D
$\frac{2}{m} (\frac{2}{15} \frac{q ^{2}}{4 π ε _{0} a})^{\frac{1}{2}}$

Solution

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