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Physics · Electrostatics

JEE Main 2024 — 1 February, Shift 2 — Question 60

Suppose a uniformly charged wall provides a uniform electric field of 2×104 N/C2 \times 10^{4} \mathrm{~N} / \mathrm{C} normally. A charged particle of mass 2 g being suspended through a silk thread of length 20 cm and remain stayed at a distance of 10 cm from the wall. Then the charge on the particlewill be 1xμC\frac{1}{\sqrt{\mathrm{x}}}\mu\mathrm{C} where x=x= _______\_\_\_\_\_\_\_. [use g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} ]

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

sin⁡θ=1020=12\sin \theta=\frac{10}{20}=\frac{1}{2} θ=30∘\theta=30^{\circ} tan⁡θ=qEmg\tan \theta=\frac{q E}{m g}

tan⁡30∘=q×2×1041×10−3×10\tan 30^{\circ}=\frac{q \times 2 \times 10^{4}}{1 \times 10^{-3} \times 10}

13=q×106\frac{1}{\sqrt{3}}=\mathrm{q} \times 10^{6} q=13×10−6C\mathrm{q}=\frac{1}{\sqrt{3}} \times 10^{-6} \mathrm{C}

x=3\mathrm{x}=3

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Field & motion of charge

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