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Electrostatics question

2025 · 2 Apr · Shift 1 · Q70
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Electrostatics question

2025 · 2 Apr · Shift 1 · Q70

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A small bob of mass 100 mg and charge +10μC+10 \mu \mathrm{C}+10μC is connected to an insulating string of length 1 m . It is brought near to an infinitely long non-conducting sheet of charge density 'σ\sigmaσ' as shown in figure. If string subtends an angle of 45∘45^{\circ}45∘ with the sheet at equilibrium the charge density of sheet will be. (Given, ϵ0=8.85×10−12 F m\epsilon_0=8.85 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}ϵ0​=8.85×10−12 m F​ and acceleration due to gravity, g=10 m s2\mathrm{g}=10 \frac{\mathrm{~m}}{\mathrm{~s}^2}g=10 s2 m​ ) JEE Main 2025 (Online) 2nd April Morning Shift Physics - Electrostatics Question 13 English
  1. A
    1.77 nC/m21.77 \mathrm{~nC} / \mathrm{m}^21.77 nC/m2
  2. B
    0.885 nC/m20.885 \mathrm{~nC} / \mathrm{m}^20.885 nC/m2
  3. C
    885 nC/m2885 \mathrm{~nC} / \mathrm{m}^2885 nC/m2
  4. D
    17.7 nC/m217.7 \mathrm{~nC} / \mathrm{m}^217.7 nC/m2
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