JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2025
- Medium
- MCQ
A small bob of mass 100 mg and charge \(+10 \mu \mathrm{C}\) is connected to an insulating string of length 1 m. It is brought near to an infinitely long nonconducting sheet of charge density ' \(\sigma\) ' as shown in figure. If string subtends an angle of \(45^{\circ}\) with the sheet at equilibrium the charge density of sheet will be :
(Given, \(\varepsilon_0=8.85 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}\) and acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(1.77 \mathrm{nC} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{qE}=\mathrm{mg} \\ & \mathrm{q}\left[\frac{\sigma}{2 \varepsilon_0}\right]=\mathrm{mg} \\ & \sigma=\frac{2 \varepsilon_0 \mathrm{mg}}{\mathrm{q}} \\ & \sigma=\frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}} \\ & \sigma=17.7 \times 10^{-10} \mathrm{C} / \mathrm{m}^2 \\ & \sigma=1.77 \mathrm{nC} / \mathrm{m}^2\end{aligned}\)
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