A pair of opposite charges
Many bodies are neutral as a whole, yet their positive and negative charges sit a little apart. The simplest such system is an electric dipole: charges \(+q\) and \(-q\) a small distance \(2a\) apart. The line through the two charges is its axis. The strength of the dipole is measured by its dipole moment.
The dipole moment of charges \(+q\) and \(-q\) a distance \(2a\) apart is the vector \(\vec p = q \cdot 2\vec a\), of size \(p = 2qa\), pointing from \(-q\) to \(+q\). Its unit is the coulomb metre, C m.
Charges of \(\pm 2\,\mu\text{C}\) that are \(2\,\text{mm}\) apart have \(p = 2 \times 10^{-6} \times 2 \times 10^{-3} = 4 \times 10^{-9}\,\text{C m}\). Molecules such as HCl and water have their charges pulled a little apart and are polar molecules. HCl has \(p\) of about \(3.4 \times 10^{-30}\,\text{C m}\). Treated as \(+e\) and \(-e\), that means a separation of \(p/e = 2.1 \times 10^{-11}\,\text{m}\), a fifth of an atom's size.
Dipole moments are vectors and add like vectors. Two equal moments \(p\) at an angle \(\phi\) give \(2p\cos(\phi/2)\). Side by side and parallel they give \(2p\); at \(60^\circ\) they give \(\sqrt3\,p\); pointing opposite ways they cancel.
Watch Fig 5.1 from the start. Together, the two charges cancel exactly and there is no field at all. As they part, the cancellation becomes imperfect and a field appears. At P it grows almost in step with the separation, to \(7.8 \times 10^5\,\text{N/C}\) at \(4\,\text{cm}\). A dipole's field is what is left over from two nearly cancelling fields.
A charge of \(-5\,\text{nC}\) sits at the origin and \(+5\,\text{nC}\) at \(x = -4\,\text{mm}\). What is the dipole moment?
The field on the axis
Put the centre of the dipole at the origin, \(+q\) at \(x = a\) and \(-q\) at \(x = -a\), so \(\vec p\) points along \(+x\). A point on the axial line at \(x = r\) is \(r - a\) from \(+q\) and \(r + a\) from \(-q\). The nearer charge wins, so the field points along \(+x\), along \(\vec p\).
\[ E = \dfrac{kq}{(r - a)^2} - \dfrac{kq}{(r + a)^2} \]
Why. the field of \(+q\) points away from it, along \(+x\); the field of \(-q\) points towards it, along \(-x\).
\[ E = kq\,\dfrac{(r + a)^2 - (r - a)^2}{(r^2 - a^2)^2} = \dfrac{4kqar}{(r^2 - a^2)^2} \]
Why. a common denominator; the numerator expands to \(4ar\).
\[ E \approx \dfrac{4kqar}{r^4} = \dfrac{2k(2qa)}{r^3} \]
Why. for a short dipole, \(r \gg a\), so \(r^2 - a^2 \approx r^2\).
\[ E = \dfrac{2kp}{r^3} \]
Read it. \(p = 2qa\).
The field on the equator
On the equatorial line, the perpendicular bisector of the dipole, a point at distance \(r\) is \(\sqrt{r^2 + a^2}\) from each charge. The two fields have equal size. Their parts along the bisector cancel. Their parts parallel to the axis both point from \(+q\) towards \(-q\), and add. The result is \(E = 2kqa/(r^2 + a^2)^{3/2}\), which is \(kp/r^3\) for \(r \gg a\), directed against \(\vec p\).
A short dipole has \(p = 4 \times 10^{-9}\,\text{C m}\). Find the field \(10\,\text{cm}\) from its centre on the axis, on the equator, and on the axis on the other side.
\(r = 0.1\,\text{m}\) is far larger than the \(2\,\text{mm}\) separation, so the short-dipole results hold. Route: \(k\).
\(kp/r^3 = 9 \times 10^9 \times 4 \times 10^{-9}/10^{-3}\).
\(kp/r^3 = 3.6 \times 10^4\,\text{N/C}\). Axial: \(7.2 \times 10^4\,\text{N/C}\) along \(\vec p\). Equatorial: \(3.6 \times 10^4\,\text{N/C}\) against \(\vec p\).
On the axis on the other side the field is again \(7.2 \times 10^4\,\text{N/C}\) along \(\vec p\). On the axis the field points the same way on both sides of the dipole.
Both fields fall as \(1/r^3\). Doubling the distance cuts them to one eighth; halving it makes them eight times as strong. A point charge's field falls only as \(1/r^2\). Seen from far away the two charges of a dipole look almost like one neutral point, so their fields cancel ever more closely.
For the dipole of the example, \(p = 4 \times 10^{-9}\,\text{C m}\), what is the field on the axis \(20\,\text{cm}\) from the centre, in N/C?
A dipole points due north. What is the direction of its field at a point far away due east of its centre?
The potential of a dipole
Take a point P at a distance \(r\) from the centre, on a line at an angle \(\theta\) to \(\vec p\). For \(r \gg a\) the two charges are at distances \(r - a\cos\theta\) and \(r + a\cos\theta\) from P. So \(V = kq/(r - a\cos\theta) - kq/(r + a\cos\theta)\). Over a common denominator this is \(2kqa\cos\theta/r^2\), dropping the small \(a^2\cos^2\theta\).
For the dipole of the example, \(kp/r^2 = 36/0.01 = 3600\,\text{V}\) on the axis at \(10\,\text{cm}\), on the \(+q\) side. At \(\theta = 60^\circ\) it is half of that, \(1800\,\text{V}\). On the equator \(\cos\theta = 0\), so \(V = 0\): every point there is equally far from \(+q\) and \(-q\). The potential falls as \(1/r^2\), again faster than a point charge's.
Fig 5.2 answers a puzzle. On the equatorial plane \(V = 0\), yet the field is \(kp/r^3\), not zero. There is no contradiction. The field is minus the rate of change of \(V\), and \(V\) changes quickly as you step off the plane towards \(-q\) or \(+q\). The field crosses the \(V = 0\) plane at right angles, pointing from the \(+q\) side to the \(-q\) side.
On the equatorial line of a dipole, the potential is zero. Which statement explains why the field there is not zero?
The field at any point
The potential makes the general field easy. Split the field at P into a radial part \(E_r\) along OP and a transverse part \(E_\theta\) across it. A small step along OP changes \(r\) alone; a small step across it changes \(\theta\), and its length is \(r\,d\theta\). Each part is minus the rate of change of \(V\) along its own step.
\[ E_r = -\dfrac{\partial V}{\partial r} = \dfrac{2kp\cos\theta}{r^3} \]
Why. differentiate \(kp\cos\theta/r^2\) with respect to \(r\), holding \(\theta\) fixed.
\[ E_\theta = -\dfrac{1}{r}\dfrac{\partial V}{\partial\theta} = \dfrac{kp\sin\theta}{r^3} \]
Why. a step across OP has length \(r\,d\theta\); differentiate with respect to \(\theta\) and divide by \(r\).
\[ E = \sqrt{E_r^2 + E_\theta^2} = \dfrac{kp}{r^3}\sqrt{4\cos^2\theta + \sin^2\theta} \]
Why. the two parts are at right angles.
\[ E = \dfrac{kp}{r^3}\sqrt{1 + 3\cos^2\theta} \]
Read it. \(\sin^2\theta = 1 - \cos^2\theta\).
The field makes an angle \(\alpha\) with OP, where \(\tan\alpha = E_\theta/E_r = \tfrac12\tan\theta\). At \(\theta = 0\) the formula gives \(2kp/r^3\), the axial field; at \(\theta = 90^\circ\) it gives \(kp/r^3\), the equatorial field. At \(\theta = 45^\circ\), \(\tan\alpha = \tfrac12\), so the field is tilted less than OP is, towards the axis.
The field is perpendicular to \(\vec p\) when \(\theta + \alpha = 90^\circ\), so \(\tan\alpha = \cot\theta\). With \(\tan\alpha = \tfrac12\tan\theta\), this needs \(\tan^2\theta = 2\), so \(\tan\theta = \sqrt2\) and \(\theta \approx 55^\circ\). It does not happen on the equator, where the field is antiparallel to \(\vec p\).
Every dipole field line has the shape \(r = D\sin^2\theta\), with \(D\) a constant for each line. It follows from \(dr/(r\,d\theta) = E_r/E_\theta = 2\cot\theta\). Integrating gives \(\ln r = 2\ln\sin\theta\) plus a constant.
A short dipole of moment \(2 \times 10^{-8}\,\text{C m}\) is at the origin. What is the size of its field at a point \(1\,\text{m}\) away on a line at \(60^\circ\) to \(\vec p\), in N/C?
In 1912 the Dutch physicist Peter Debye suggested that some molecules carry a permanent dipole moment. He explained how such molecules line up in a field and how heat disturbs them. Measured moments were so small that a unit was named after him: one debye is about \(3.34 \times 10^{-30}\,\text{C m}\). A water molecule has a moment of about 1.85 debye. Debye received the Nobel Prize in Chemistry in 1936.
- Drawing \(\vec p\) from \(+q\) to \(-q\).The dipole moment points from the negative charge to the positive charge.
- Using \(1/r^2\) for the field of a dipole.The field of a dipole falls as \(1/r^3\); its potential falls as \(1/r^2\).
- Pointing the equatorial field along \(\vec p\).On the axis \(\vec E\) is along \(\vec p\); on the equator it is against \(\vec p\).
- Concluding \(E = 0\) from \(V = 0\) on the equator.The field comes from the change of \(V\) across the plane, not from its value.