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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 2 April · Morning Shift · Physics · Electrostatics

A small bob of mass 100 mg and charge +10 μ C is connected to an...

A small bob of mass 100 mg and charge +10μC+10 \mu \mathrm{C} is connected to an insulating string of length 1 m. It is brought near to an infinitely long nonconducting sheet of charge density ' σ\sigma ' as shown in figure. If string subtends an angle of 45∘45^{\circ} with the sheet at equilibrium the charge density of sheet will be :
(Given, ε0=8.85×10−12 F m\varepsilon_0=8.85 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}} and acceleration due to gravity, g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2)
A small bob of mass 100 mg and charge +10 μ C is connected to an insulating string of length 1 m. It is brought near to

Options

  1. A0.885nC/m20.885 \mathrm{nC} / \mathrm{m}^2
  2. B17.7nC/m217.7 \mathrm{nC} / \mathrm{m}^2
  3. C885nC/m2885 \mathrm{nC} / \mathrm{m}^2
  4. D1.77nC/m21.77 \mathrm{nC} / \mathrm{m}^2

Correct answer

D. 1.77nC/m21.77 \mathrm{nC} / \mathrm{m}^2

Step-by-step solution

A small bob of mass 100 mg and charge +10 μ C is connected to an insulating string of length 1 m. It is brought near to
qE=mgq[σ2ε0]=mgσ=2ε0mgqσ=2×8.85×10−12×100×10−6×1010×10−6σ=17.7×10−10C/m2σ=1.77nC/m2\begin{aligned} & \mathrm{qE}=\mathrm{mg} \\ & \mathrm{q}\left[\frac{\sigma}{2 \varepsilon_0}\right]=\mathrm{mg} \\ & \sigma=\frac{2 \varepsilon_0 \mathrm{mg}}{\mathrm{q}} \\ & \sigma=\frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}} \\ & \sigma=17.7 \times 10^{-10} \mathrm{C} / \mathrm{m}^2 \\ & \sigma=1.77 \mathrm{nC} / \mathrm{m}^2\end{aligned}
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