Two charges of -4 μ C and +4 μ C are placed at the points A (1,0,4) m and B (2,-1,5) m located in an electric field vec E =0.20 hat i V / cm . The magnitude of the torque acting on the dipole is 8 √(α) × 10 -5 Nm , Where α= \ \ \ \ .

Physics · Electrostatics

JEE Main 2024 — 27 January, Shift 2 — Question 51

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

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

The displacement vector is

\vec{d} = (2-1)\hat{i} + (-1-0)\hat{j} + (5-4)\hat{k} = \hat{i} - \hat{j} + \hat{k}

Electric dipole moment:

\vec{p} = 4 \times 10^{-6} (\hat{i} - \hat{j} + \hat{k}) \ \text{C·m}

Electric field:

\vec{E} = 20\hat{i} \ \text{V/m}

Torque on dipole:

\vec{\tau} = \vec{p} \times \vec{E} = (80\hat{j} + 80\hat{k}) \times 10^{-6} \ \text{N·m}

Magnitude of torque:

|\vec{\tau}| = 80\sqrt{2} \times 10^{-6} = 8\sqrt{2} \times 10^{-5} \ \text{N·m}, \quad \alpha = 2

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Electrostatics
Topic: Electric Dipole

Answer: 2

Step-by-step solution

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