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QJEEM24Ph2e30: JEE Main 2024 Physics - Class 12 JEEM24Ph2e30 Solution: Two charges of and are placed at the points and located in an…

Two charges of −4μC-4 \mu \mathrm{C} and +4μC+4 \mu \mathrm{C} are placed at the points A(1,0,4)m\mathrm{A}(1,0,4) \mathrm{m} and B(2,−1,5)m\mathrm{B}(2,-1,5) \mathrm{m} located in an electric field E→=0.20i^ V/cm\overrightarrow{\mathrm{E}}=0.20 \hat{i} \mathrm{~V} / \mathrm{cm} . The magnitude of the torque acting on the dipole is 8α×10−5Nm8 \sqrt{\alpha} \times 10^{-5} \mathrm{Nm} , where α=\alpha= _________.

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JEE Main 2024 (Online) 27th January Evening Shift Physics - Electrostatics Question 43 English Explanation

amp;τ⃗=p⃗×E⃗amp;p⃗=qℓ⃗amp;E→=0.2Vcm=20Vmamp;p→=4×(i^−j^+k^)amp;=(4i^−4j^+4k^)μC−mamp;τ⃗=(4i^−4j^+4k^)×(20i^)×10−6Nmamp;=(8k^+8j^)×10−5=82×10−5amp;α=2\begin{aligned}& \vec{\tau}=\vec{p} \times \vec{E} \\& \vec{p}=q \vec{\ell} \\& \overrightarrow{\mathrm{E}}=0.2 \frac{\mathrm{V}}{\mathrm{cm}}=20 \frac{\mathrm{V}}{\mathrm{m}} \\& \overrightarrow{\mathrm{p}}=4 \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\& =(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mu \mathrm{C}-\mathrm{m} \\& \vec{\tau}=(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(20 \hat{\mathrm{i}}) \times 10^{-6} \mathrm{Nm} \\& =(8 \hat{\mathrm{k}}+8 \hat{\mathrm{j}}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \\& \alpha=2\end{aligned}

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