The Dipole in an Electric Field

A couple, and no net force

Place a dipole in a uniform field \(\vec E\), with \(\vec p\) at an angle \(\theta\) to \(\vec E\). The field pushes \(+q\) with \(q\vec E\) and \(-q\) with \(-q\vec E\). The two forces are equal and opposite, so the net force is zero and the centre of the dipole does not move off. But the forces act along different lines. In the chapter on Torque and Equilibrium, such a pair was a couple.

The torque of a couple is one force times the perpendicular distance between the lines of action. That distance is \(2a\sin\theta\). So \(\tau = qE \times 2a\sin\theta = pE\sin\theta\). As a vector, \(\vec\tau = \vec p \times \vec E\). The torque turns \(\vec p\) towards \(\vec E\).

The torque is largest, \(pE = 8 \times 10^{-4}\,\text{N m}\), when \(\vec p\) is perpendicular to \(\vec E\). It is zero at \(0^\circ\) and at \(180^\circ\), when the two forces act along one line.

The energy of a turned dipole

Turn the dipole slowly from \(\theta_1\) to \(\theta_2\). The outside agent must supply a torque equal and opposite to \(pE\sin\theta\) at every angle. Its work is \(\int_{\theta_1}^{\theta_2} pE\sin\theta\,d\theta = pE(\cos\theta_1 - \cos\theta_2)\). That work is stored as potential energy of the dipole in the field.

Choose the zero of energy at \(\theta = 90^\circ\). Then turning from \(90^\circ\) to \(\theta\) gives \(U = pE(\cos 90^\circ - \cos\theta) = -pE\cos\theta\). In vector form, \(U = -\vec p\cdot\vec E\).

For \(pE = 8 \times 10^{-4}\,\text{J}\): at \(0^\circ\), \(U = -8 \times 10^{-4}\,\text{J}\); at \(90^\circ\), \(U = 0\); at \(180^\circ\), \(U = +8 \times 10^{-4}\,\text{J}\). Turning from \(0^\circ\) to \(90^\circ\) takes \(8 \times 10^{-4}\,\text{J}\). Turning all the way to \(180^\circ\) takes twice that, \(1.6 \times 10^{-3}\,\text{J}\).

Two zeros of torque, one stable

The torque is zero at \(0^\circ\) and at \(180^\circ\), so both are equilibria. The energy graph in the sim tells them apart. At \(0^\circ\) the energy is lowest: a small turn raises \(U\), and the torque brings the dipole back. At \(180^\circ\) the energy is highest: the slightest tilt lowers \(U\), and the dipole topples over.

The torque and the energy are linked as force and energy were. The torque is minus the rate of change of energy with angle, \(\tau = -dU/d\theta\). With \(U = -pE\cos\theta\), \(dU/d\theta = pE\sin\theta\). The minus sign says the torque turns the dipole towards lower energy, that is, towards smaller \(\theta\).

The dipole as a pendulum

Press play in the sim at a small angle. The dipole swings through the field direction, overshoots and swings back, like a pendulum. For small angles \(\sin\theta \approx \theta\), so the torque is \(-pE\theta\), proportional to the angle and opposite to it. In the chapter on the Dynamics of SHM, a restoring torque of this form gave simple harmonic motion.

Newton's second law for rotation reads \(I\,d^2\theta/dt^2 = -pE\theta\), where \(I\) is the moment of inertia of the dipole about its centre. This is SHM with \(\omega^2 = pE/I\).

For \(p = 4 \times 10^{-9}\,\text{C m}\) in \(2 \times 10^5\,\text{N/C}\), \(pE = 8 \times 10^{-4}\,\text{N m}\). With \(I = 8 \times 10^{-9}\,\text{kg m}^2\), \(T = 2\pi\sqrt{10^{-5}}\) \(= 2\pi \times 3.16 \times 10^{-3}\) \(\approx 0.02\,\text{s}\). A stronger field gives a quicker swing: four times the field halves the period.

The sim also shows large swings. Released from \(150^\circ\), the dipole swings to \(-150^\circ\) and back, and each swing takes longer than a small one. Released from exactly \(180^\circ\), it does not move at all.

When the field is not uniform

Now let the field change from place to place. Put a dipole along the \(x\) axis, in a field \(E(x)\) along \(x\), with \(-q\) at \(x\) and \(+q\) at \(x + 2a\). The net force is \(q[E(x + 2a) - E(x)]\). For a short dipole this is \(q \cdot 2a \cdot dE/dx\), so \(F = p\,dE/dx\). A dipole along the field is pulled towards stronger field.

For \(p = 4 \times 10^{-9}\,\text{C m}\) along a field \(E_x = 5 \times 10^6\,x\), the field grows by \(5 \times 10^6\,\text{N/C}\) per metre. So \(F = 4 \times 10^{-9} \times 5 \times 10^6 = 0.02\,\text{N}\), along \(+x\), towards the stronger field.

Now the stream of water. A water molecule is a dipole. Near the charged comb the field is strong and far from uniform. First the torque turns the molecules to line up with the field. Then, lined up, each is pulled towards the stronger field, towards the comb. The whole stream bends, though it carries no net charge.