JEE MAIN - Physics (2025 - 4th April Morning Shift - No. 20)
Two small spherical balls of mass 10 g each with charges $-2 \mu \mathrm{C}$ and $2 \mu \mathrm{C}$, are attached to two ends of very light rigid rod of length 20 cm . The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of $100 \mu \mathrm{C} / \mathrm{m}^2$ such that length of rod makes an angle of $30^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:
(Take $\varepsilon_{\mathrm{o}}: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2$ )
1.12 Nm
2.24 Nm
11.2 Nm
112 Nm
Explanation
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$$\begin{aligned} & \mathrm{E}=\frac{\sigma}{2 \varepsilon_0} \\ & \tau=\mathrm{PE} \sin \theta \\ & =\left[\left(2 \times 10^{-6}\right)\left(\frac{2}{10}\right)\right]\left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right]\left(\frac{1}{2}\right) \\ & =\frac{10}{8.85}=1.12 \mathrm{Nm} \end{aligned}$$
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