Question:

An electric dipole of dipole moment \( 1.0 \times 10^{-12} \ \text{Cm} \) lies along the x-axis. An electric field of magnitude \( 2.0 \times 10^4 \ \text{NC}^{-1} \) is switched on at an instant in the region. The unit vector along the electric field is \( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \). The magnitude of the torque acting on the dipole at that instant is:

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To find torque on a dipole in a non-parallel field, compute the cross product \( \vec{p} \times \vec{E} \) or use \( \tau = pE \sin\theta \) where \( \theta \) is the angle between dipole and field directions.
Updated On: Jun 24, 2025
  • \( 0.5 \times 10^{-6} \ \text{Nm} \)
  • \( 1.0 \times 10^{-8} \ \text{Nm} \)
  • \( 2.0 \times 10^{-8} \ \text{Nm} \)
  • \( 4.0 \times 10^{-8} \ \text{Nm} \)
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The Correct Option is B

Solution and Explanation

The torque \( \vec{\tau} \) on a dipole in an electric field is given by: \[ \vec{\tau} = \vec{p} \times \vec{E} \] Given: \[ |\vec{p}| = 1.0 \times 10^{-12} \ \text{Cm}, \quad |\vec{E}| = 2.0 \times 10^4 \ \text{NC}^{-1} \] \[ \hat{p} = \hat{i}, \quad \hat{E} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \] Now compute the cross product: \[ \vec{\tau} = pE \sin\theta \] Angle \( \theta \) between \( \hat{i} \) and the vector \( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \) can be found using: \[ \sin\theta = \left| \hat{i} \times \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) \right| = \left| \frac{\sqrt{3}}{2} (\hat{i} \times \hat{i}) + \frac{1}{2} (\hat{i} \times \hat{j}) \right| = \left| 0 + \frac{1}{2} \hat{k} \right| = \frac{1}{2} \] Thus: \[ \tau = pE \sin\theta = (1.0 \times 10^{-12})(2.0 \times 10^4) \left( \frac{1}{2} \right) = 1.0 \times 10^{-8} \ \text{Nm} \]
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