JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2025
- Easy
- MCQ
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 \(\varepsilon_0: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 1.12 Nm
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{E} & =\frac{\sigma}{2 \varepsilon_0} \\ \tau & =\mathrm{PE} \sin \theta\end{aligned}\)
\(\begin{aligned} & =\left[\left(2 \times 10^{-6}\right)\left(\frac{2}{10}\right)\right]\left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right]\left(\frac{1}{2}\right) \\ & =\frac{10}{8.85}=1.12 \mathrm{Nm}\end{aligned}\)
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