An electric dipole is placed at an angle of 30° with an electric field of intensity $3 × 10^5\, N/C$. It experiences a torque equal to 6 N m. If the dipole length is 2 cm, the value of charge will be

$2 × 10^{-3} C$

The correct answer is Option (4) → $2 × 10^{-3} C$

$\tau = pE \sin\theta$

$6 = p \cdot (3 \times 10^{5}) \cdot \sin 30^\circ$

$6 = p \cdot (3 \times 10^{5}) \cdot \frac{1}{2}$

$6 = p \cdot 1.5 \times 10^{5}$

$p = \frac{6}{1.5 \times 10^{5}} = 4 \times 10^{-5} \ \text{C·m}$

$p = q \cdot d$

$d = 2 \ \text{cm} = 2 \times 10^{-2} \ \text{m}$

$q = \frac{4 \times 10^{-5}}{2 \times 10^{-2}}$

$q = 2 \times 10^{-3} \ \text{C}$

The charge is $2 \times 10^{-3} \ \text{C}$.