Question

An electric dipole with dipole moment $4\times10^{-9}\,\text{C m}$ is aligned at $30°$ with the direction of a uniform electric field of magnitude $5\times10^{4}\,\text{N C}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

Hint:

Use \(\tau = pE\sin\theta\) with \(\theta = 30^\circ\) so \(\sin\theta = 0.5\).

Answer 1

Given: dipole moment \(p = 4\times10^{-9}\,\text{C m}\), angle with field \(\theta = 30^\circ\), uniform field \(E = 5\times10^{4}\,\text{N C}^{-1}\).

Step 1: Torque on a dipole. A dipole of moment \(\vec p\) in a uniform field \(\vec E\) experiences a torque of magnitude

\[ \tau = pE\sin\theta. \]

Step 2: Substitute the values.

\[ \tau = (4\times10^{-9})(5\times10^{4})\sin 30^\circ. \]

Step 3: Use \(\sin 30^\circ = 0.5\) and compute.

\[ \tau = (4\times10^{-9})(5\times10^{4})(0.5). \]

\[ \tau = (20\times10^{-5})(0.5). \]

\[ \tau = 10\times10^{-5}. \]

\[ \tau = 1\times10^{-4}\,\text{N m}. \]

\[\boxed{\tau = 1\times10^{-4}\,\text{N m}}\]

Answer 2

Expert method: torque as the cross product \(\vec\tau = \vec p \times \vec E\), with a dimensional check.

Step 1: The fundamental relation is the vector form \(\vec\tau = \vec p \times \vec E\). Its magnitude is \(|\vec\tau| = pE\sin\theta\), where \(\theta\) is the angle between \(\vec p\) and \(\vec E\). The cross product makes explicit that only the component of the field perpendicular to the dipole produces turning effect.

Step 2 (perpendicular-field viewpoint): Resolve \(\vec E\) into a part along \(\vec p\) and a part perpendicular to it. The perpendicular part is \(E\sin\theta = (5\times10^{4})\sin 30^\circ = 2.5\times10^{4}\,\text{N C}^{-1}\). Only this part twists the dipole, so

\[ \tau = p\,(E\sin\theta) = (4\times10^{-9})(2.5\times10^{4}) = 1\times10^{-4}\,\text{N m}. \]

Step 3 (dimensional check): \([p] = \text{C m}\), \([E] = \text{N C}^{-1}\). Then \([pE] = \text{C m}\times\text{N C}^{-1} = \text{N m}\), which is the correct unit of torque. The sine factor is dimensionless, so the dimensions are consistent.

Step 4: Both routes give the same magnitude. The torque tends to rotate \(\vec p\) into alignment with \(\vec E\) (restoring toward \(\theta = 0\)).

\[\boxed{\tau = 1\times10^{-4}\,\text{N m}}\]