Electrostatics

JEE Main level, single correct, medium

163 of 311

Question 163 of 311

Two small spherical balls of mass 10 g each with charges −2μC-2 \mu \mathrm{C} and 2μC2 \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 100μC/m2100 \mu \mathrm{C} / \mathrm{m}^2 such that length of rod makes an angle of 30∘30^{\circ} with electric field generated by charge sheet. Net torque acting on the rod is:
(Take
ε0:8.85×10−12C2/Nm2\varepsilon_0: 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2)

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Answer: (B) 1.12 Nm


E=σ2ε0τ=PEsin⁡θ\begin{aligned} \mathrm{E} & =\frac{\sigma}{2 \varepsilon_0} \\ \tau & =\mathrm{PE} \sin \theta\end{aligned}
=[(2×10−6)(210)][100×10−62×8.85×10−12](12)=108.85=1.12Nm\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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