Discussion
An electric dipole is placed at an angle of 30o with an electric field intensity 3×105NC-1 experiences a torque 3 Nm. If the length of dipole is 2 cm, the charge on the dipole is
(in milli coloumb)
Answer: D
The charge is 1 millicoulomb.
Torque on a dipole is \(\tau = pE\sin\theta\), with dipole moment \(p = q \times 2a\) (the length, here 2 cm).
\[ \begin{aligned} 3 &= q(0.02)(3 \times 10^{5})\sin 30^{\circ} \\ 3 &= q \times 3000 \\ q &= 10^{-3}\ \text{C} = 1\ \text{mC} \end{aligned} \]
Forgetting \(\sin 30^{\circ} = \frac{1}{2}\) gives 0.5 mC, and using 2 cm as half-length gives 0.5 mC as well; neither is offered.
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