An electric dipole of length 2 cm is placed at an angle of 30° with an electric field $3 × 10^5 N C^{-1}$. If the dipole experiences a torque of $9 × 10^{-3} N m$, the magnitude of either charge of the dipole is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3 μC
Given:
Dipole length: $l = 2\ \text{cm} = 0.02\ \text{m}$
Electric field: $E = 3 \times 10^5\ \text{N/C}$
Torque: $\tau = 9 \times 10^{-3}\ \text{Nm}$
Angle with field: $\theta = 30^\circ$
Torque on dipole: $\tau = p E \sin\theta = q l E \sin\theta$
Solving for charge $q$:
$q = \frac{\tau}{l E \sin\theta} = \frac{9 \times 10^{-3}}{0.02 \cdot 3 \times 10^5 \cdot \frac{1}{2}}$
$q = \frac{9 \times 10^{-3}}{0.02 \cdot 3 \times 10^5 \cdot 0.5} = \frac{9 \times 10^{-3}}{3000} = 3 \times 10^{-6}\ \text{C}$
∴ Magnitude of either charge of the dipole = 3 μC