JEE Main 2019PhysicsElectrostaticsMediumMCQ

JEE Main 2019 — Electrostatics Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

The bob of a simple pendulum has mass 2 g and a charge of 5.0 μC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m At equilibrium, the angle that the pendulum makes with the vertical is:
take g=10 m/s2

Choose an option

Show full solutionCorrect option: C
Correct answer
C tan-10.5

Step-by-step explanation


tan⁡θ=QEmg=5×10-6×20002×10-3×10=12
⇒θ=tan-1⁡0.5

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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.