JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 20)
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 non-conducting 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, $\epsilon_0=8.85 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}$ and acceleration due to gravity, $\mathrm{g}=10 \frac{\mathrm{~m}}{\mathrm{~s}^2}$ )
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Explanation
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$$\begin{aligned} & \mathrm{qE}=\mathrm{mg} \\ & \mathrm{q}\left[\frac{\sigma}{2 \varepsilon_0}\right]=\mathrm{mg} \\ & \sigma=\frac{2 \varepsilon_0 \mathrm{mg}}{\mathrm{q}} \\ & \sigma=\frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}} \\ & \sigma=17.7 \times 10^{-10} \mathrm{C} / \mathrm{m}^2 \\ & \sigma=1.77 \mathrm{nC} / \mathrm{m}^2 \end{aligned}$$
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