JEE Main 2019PhysicsElectrostaticsHardMCQ

JEE Main 2019Electrostatics Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

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:

Choose an option

Show full solutionCorrect option: D
Correct answer
D2m215q24πϵ0a1/2

Step-by-step explanation


From the conservation of energy,
14πε0q23a2+4a2+12mv2=14πε0q23a
v=q24πε0a×215×2m
=2m215q24πε0a1/2

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About this question

This is a previous-year question from JEE Main 2019, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.