JEE Main 2023PhysicsElectrostaticsHardNumerical

JEE Main 2023 — Electrostatics Question with Solution

JEE Main 2023 (11 Apr Shift 1)

Question

As shown in the figure, a configuration of two equal point charges q0=+2μC is placed on an inclined plane. Mass of each point charge is 20 g. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height h=x×10-3 m. The value of x is
 Take 14πε0=9×109 N m2 C-2, g=10 m s-2

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Show full solutionCorrect answer: 300
Correct answer
300

Step-by-step explanation

For the condition of equilibrium, the Coulomb force is equal to mgsinθ. Hence,

kq024h2=(mgsinθ)⇒kq024h2=20×10-3×10×12

⇒9×109×4×10−124h2=20×10-3×10×12⇒h=9100 m=0.3 m=300×10-3 m

h2=9100⇒h=310m=0.3m

=300×10−3m

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About this question

This is a previous-year question from JEE Main 2023, 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.