Question: 3 An electric dipole is placed at an angle of 30^{\circ} with an electric field of intensity 2 \times 10^{5} \mathrm{NC}^{-1}. It experiences a torque equal to 4 N m . Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm .
(1) 4 mC
(2) 2 mC
(3) 8 mC
(4) 6 mC
Answer: Option (2)
Explanation:
The torque acting on an electric dipole placed in a uniform electric field is given by
\tau = p E \sin \theta.
The dipole moment is given by p = q \ell, where q is the charge
and \ell is the dipole length.
Given torque \tau = 4 \ \mathrm{N m}, electric field
E = 2 \times 10^{5} \ \mathrm{NC}^{-1}, angle \theta = 30^{\circ}
and dipole length \ell = 2 \ \mathrm{cm} = 2 \times 10^{-2} \ \mathrm{m}.
Substituting in the torque equation,
4 = q \times 2 \times 10^{-2} \times 2 \times 10^{5} \times \sin 30^{\circ}.
Using \sin 30^{\circ} = \frac{1}{2}, we get 4 = q \times 2000.
Thus, q = \frac{4}{2000} = 2 \times 10^{-3} \ \mathrm{C}.
Therefore, the magnitude of charge on the dipole is 2 \ \mathrm{mC}.