Electric dipoles

A molecule with two ends

A water molecule has no net charge. Yet water behaves as if its two ends were different. The oxygen end is slightly negative and the hydrogen end slightly positive. Water molecules turn in an electric field, and this is why a stream of water bends towards a charged comb.

A model of this is the electric dipole: two equal and opposite charges, $+q$ and $-q$, a small distance $2a$ apart. The total charge is zero. What matters is how large the charges are and how far apart they are, so we put them together in one quantity, the dipole moment:

It is a vector. It points from the negative charge to the positive charge. Its unit is the coulomb metre (C m). A water molecule has a dipole moment of about $6.2 \times 10^{-30}$ C m, a tiny number, but with enormous effects on how water behaves.

A point dipole is a dipole so small that we only care about $p$, not about $q$ and $2a$ one by one.

The field of a dipole on its axis

The axis of a dipole is the line through both charges. Take a point $P$ on the axis, on the side of $+q$, at a distance $r$ from the centre of the dipole.

The charge $+q$ is $r - a$ from $P$, and $-q$ is $r + a$ from $P$. The field of $+q$ points away from the dipole, and the field of $-q$ points towards it. So they are opposite, and the nearer charge wins.

The direction is along $\vec{p}$. Far from the dipole, $r \gg a$, so $r^2 - a^2 \approx r^2$ and

The field on the equatorial line

The equatorial line is the line at right angles to the axis, through the middle of the dipole. Take a point $Q$ on it, at a distance $r$ from the centre. Each charge is at a distance $\sqrt{r^2 + a^2}$ from $Q$, so both fields have the same size, $\dfrac{kq}{r^2 + a^2}$.

The field of $+q$ points away from $+q$, and the field of $-q$ points towards $-q$. Split each one into a part along the equatorial line and a part parallel to the axis. The parts along the equatorial line are opposite and cancel. The parts parallel to the axis both point from the $+q$ side towards the $-q$ side, so they add. Each is $\dfrac{kq}{r^2 + a^2}$ times $\dfrac{a}{\sqrt{r^2 + a^2}}$. So

The direction is opposite to $\vec{p}$. Far from the dipole,

Compare the two. At the same large distance, the field on the axis is twice the field on the equatorial line, and the two point in opposite directions. Both fall as $1/r^3$. This is faster than the $1/r^2$ of a single charge, because the two charges nearly cancel, and the cancelling gets better as you go away.

A worked example. A dipole is made of $+3.0$ nC and $-3.0$ nC, $4.0$ mm apart. Find the dipole moment and the field at $0.50$ m from its centre, on the axis and on the equatorial line.

1. The moment. $p = q \times 2a = 3.0 \times 10^{-9}\ \text{C} \times 4.0 \times 10^{-3}\ \text{m} = 1.2 \times 10^{-11}\ \text{C m}$.
2. On the axis. $E = \dfrac{2kp}{r^3} = \dfrac{2 \times 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times 1.2 \times 10^{-11}\ \text{C m}}{(0.50\ \text{m})^3} = 1.73\ \text{N/C}$, along $\vec{p}$.
3. On the equatorial line. $E = \dfrac{kp}{r^3} = 0.864\ \text{N/C}$, against $\vec{p}$.

The exact formulas give $1.728$ N/C and $0.864$ N/C. The distance is $125$ times the half-gap, so the short formulas are very good.

Increase the gap between the charges while the distance stays at $10$ cm. Do the exact fields move away from the short formulas? Now move to $50$ cm. Does the ratio stay near $2$?

A dipole in a uniform field

Put a dipole in a uniform field $\vec{E}$. The force on $+q$ is $q\vec{E}$ along the field, and the force on $-q$ is $q\vec{E}$ against it. The two forces are equal and opposite, so the net force is zero. The dipole does not move off. But the forces do not act along the same line, so they make a turning effect, a torque.

Let $\theta$ be the angle between $\vec{p}$ and $\vec{E}$. The two forces are $qE$ each, and the distance between their lines is $2a\sin\theta$. The torque of the pair is

In vector form $\vec{\tau} = \vec{p} \times \vec{E}$. The torque turns $\vec{p}$ towards $\vec{E}$. It is greatest, $pE$, when $\theta = 90^\circ$. It is zero when $\vec{p}$ is along $\vec{E}$ ($\theta = 0^\circ$) and also when it is against $\vec{E}$ ($\theta = 180^\circ$).

At $\theta = 0^\circ$ the equilibrium is stable. A small turn brings a torque that returns the dipole. At $\theta = 180^\circ$ it is unstable. A small turn makes it swing away.

The energy of a dipole in a field

To turn a dipole against the field we must do work. To turn it from $\theta_1$ to $\theta_2$, the work against the torque is

We take the potential energy to be zero when the dipole is at right angles to the field. Then at an angle $\theta$

The energy is lowest, $-pE$, at $\theta = 0^\circ$. It is highest, $+pE$, at $\theta = 180^\circ$.

A worked example. A dipole has $p = 5.0 \times 10^{-8}$ C m. It lies in a uniform field of $3.0 \times 10^{4}$ N/C at $30^\circ$ to the field.

1. The torque. $\tau = pE\sin\theta = 5.0 \times 10^{-8} \times 3.0 \times 10^{4} \times 0.50\ \text{N m} = 7.5 \times 10^{-4}\ \text{N m}$.
2. The energy. $U = -pE\cos\theta = -1.5 \times 10^{-3}\ \text{J} \times 0.866 = -1.30 \times 10^{-3}\ \text{J}$.
3. To turn it right round. From $0^\circ$ to $180^\circ$ the work is $pE(\cos 0^\circ - \cos 180^\circ) = 2pE = 3.0 \times 10^{-3}\ \text{J}$.

Set the angle to $90^\circ$. Where is the torque greatest? Now set it to $0^\circ$, then to $180^\circ$. What is the torque, and what is the energy, at each?

Small swings, and a field that changes

If a dipole is turned a little from its stable position, the torque brings it back. For a small angle, $\sin\theta \approx \theta$, so the torque is $-pE\theta$. Newton's second law for turning, $\tau = I\alpha$, gives $\alpha = -\dfrac{pE}{I}\theta$. This is simple harmonic motion with

where $I$ is the moment of inertia of the dipole about its centre.

In a non-uniform field the two forces are no longer equal, so there is a net force as well as a torque. A dipole is pulled towards the stronger field. This is how a charged comb lifts a scrap of neutral paper. The field of the comb pulls the charges in the paper a little apart, so the paper becomes a small dipole. The side nearer the comb feels the stronger field, and the paper is pulled towards the comb.

Practise with new numbers

Try it in Python

The program finds the exact field of a dipole on its axis and on its equatorial line at four distances. It compares each with the short formula that is good for a long distance.

Common mistakes

• Saying a dipole has no field because its charge is zero. The two fields do not cancel, except where they are equal and opposite. A dipole has a field that falls as $1/r^3$.
• Forgetting the direction of $\vec{p}$. It points from the negative charge to the positive charge.
• Mixing up the axis and the equatorial line. The field on the axis is twice as large and along $\vec{p}$. On the equatorial line it is half as large and against $\vec{p}$.
• Writing a net force on a dipole in a uniform field. The net force is zero. There is only a torque.
• Using the cosine for the torque. The torque is $pE\sin\theta$. The energy uses $-pE\cos\theta$.
• Using $2a$ in place of $a$. In $p = q \times 2a$, the gap is $2a$. In the exact field formulas, $a$ is half the gap.

Remember these

• Dipole moment: $p = q \times 2a$, in C m, from $-q$ towards $+q$.
• On the axis, far away: $E = \dfrac{2kp}{r^3}$, along $\vec{p}$. On the equatorial line: $E = \dfrac{kp}{r^3}$, against $\vec{p}$.
• In a uniform field the net force is zero and the torque is $\tau = pE\sin\theta$, or $\vec{\tau} = \vec{p} \times \vec{E}$.
• Potential energy: $U = -pE\cos\theta$. Stable at $\theta = 0^\circ$, unstable at $180^\circ$. The work to turn from $\theta_1$ to $\theta_2$ is $pE(\cos\theta_1 - \cos\theta_2)$.
• Small swings have period $T = 2\pi\sqrt{I/(pE)}$. In a non-uniform field a dipole is also pulled towards the stronger field.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Charges of $+4.0\ \mu$C and $-4.0\ \mu$C are $1.5$ cm apart. What is the dipole moment, and which way does it point?

2. A small dipole has a field $E_a$ at a distance $r$ on its axis. What is the field at the same distance on its equatorial line?

3. A dipole is placed in a uniform field, at right angles to it. Is the net force zero? Is the torque zero?

4. A dipole is released at rest at $\theta = 180^\circ$ to a uniform field, with a slight turn to one side. What does it do, and why?