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:

• A. \( 0.5 \times 10^{-6} \ \text{Nm} \)
• B. \( 1.0 \times 10^{-8} \ \text{Nm} \)
• C. \( 2.0 \times 10^{-8} \ \text{Nm} \)
• D. \( 4.0 \times 10^{-8} \ \text{Nm} \)

Hint:

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.

Answer 1

The Correct Option is B

The torque \( \vec{\tau} \) on a dipole in an electric field is \( \vec{\tau} = \vec{p} \times \vec{E} \).Given magnitudes:\(|\vec{p}| = 1.0 \times 10^{-12} \ \text{Cm}\) and \(|\vec{E}| = 2.0 \times 10^4 \ \text{NC}^{-1}\).Given directions:\( \hat{p} = \hat{i} \) and \( \hat{E} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \).The cross product is calculated as \( \vec{\tau} = pE \sin\theta \).The angle \( \theta \) between \( \hat{p} \) and \( \hat{E} \) yields \( \sin\theta = \left| \hat{i} \times \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) \right| = \frac{1}{2} \).Therefore, the magnitude of the torque is \( \tau = pE \sin\theta = (1.0 \times 10^{-12})(2.0 \times 10^4) \left( \frac{1}{2} \right) = 1.0 \times 10^{-8} \ \text{Nm} \).