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Friday, 16 October

JEE Main 2023 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 11 April 2023, Shift 1, Physics Q55

Question 55 of 90 in this shift, Physics question 25 of 30, Section B.

As shown in the figure, a configuration of two equal point charges (q0=+2μq_0 = +2\mu C) 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×10−3h = x \times 10^{-3} m. The value of xx is ________. (Take 14πε0=9×109 N m2 C−2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \mathrm{N\,m^2\,C^{-2}}, g=10 m s−2g = 10\ \mathrm{m\,s^{-2}})

The figure, in words

Inclined plane making 30 degrees with the horizontal; one point charge q0 at the top of the incline at height h above the base, the other charge q0 at the bottom of the incline on the ground.

Official answer

300

NTA final key.

Chapter
Electric Charges and Fields
Topic
Coulomb's Law
Idea tested
Charge Equilibrium Analysis

Same topic in other shifts

All Coulomb's Law questions
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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.