Electric Dipole

Definition

An electric dipole is a pair of equal and opposite point charges $+q$ and $-q$ separated by a distance $2a$. The line joining them defines the dipole axis, and its midpoint is the centre of the dipole. By convention the direction of the dipole is from $-q$ to $+q$.

$$\vec{p} = q\,(2a)\,\hat{p}$$

• The dipole moment $\vec{p}$ is a vector of magnitude $2qa$ directed along the axis from $-q$ to $+q$. SI unit C m.
• The total charge of a dipole is zero, but its field is not zero: the fields of $+q$ and $-q$ do not cancel exactly because the charges sit at different places.
• At distances much greater than $2a$ they nearly cancel, so the dipole field falls off faster than $1/r^2$ — in fact as $1/r^3$.
• A short (or ideal, or point) dipole means $r\gg a$, and it is only then that the simple $1/r^3$ formulas hold.

Field on the axial line — the derivation CBSE sets

Let $P$ lie on the axis at distance $r$ from the centre, on the side of $+q$. Then $+q$ is at distance $r-a$ and $-q$ at distance $r+a$ from $P$. Let $\hat{p}$ be the unit vector from $-q$ to $+q$.

$$\vec{E}_{+q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2}\,\hat{p},\qquad \vec{E}_{-q} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2}\,\hat{p}$$

The first points away from $+q$, that is along $\hat{p}$; the second points towards $-q$, that is along $-\hat{p}$. Adding,

$$\vec{E}_{\text{axial}} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2}-\dfrac{1}{(r+a)^2}\right]\hat{p}$$

$$= \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2}\,\hat{p} = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ar}{(r^2-a^2)^2}\,\hat{p}$$

Using $p = q(2a)$,

$$\vec{E}_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2pr}{(r^2-a^2)^2}\,\hat{p}$$

For a short dipole, $r\gg a$, so $a^2$ may be neglected against $r^2$ and $(r^2-a^2)^2\to r^4$:

$$\vec{E}_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\vec{p}}{r^3}$$

directed along $\vec{p}$. Writing it with $x$ in place of $r$ changes nothing.

Field on the equatorial line — the derivation CBSE sets

Let $P$ lie on the perpendicular bisector at distance $r$ from the centre. Each charge is at distance $\sqrt{r^2+a^2}$ from $P$, so the two fields have equal magnitudes

$$E_{+q} = E_{-q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r^2+a^2)}$$

Draw them at $P$: the field of $+q$ points away from $+q$, the field of $-q$ points towards $-q$. Resolve each along and perpendicular to the dipole axis.

• The components perpendicular to the axis are equal and opposite, so they cancel.
• The components parallel to the axis are equal and both point from $+q$ towards $-q$, that is along $-\hat{p}$, so they add.

$$E_{\text{net}} = 2E_{+q}\cos\theta,\qquad \cos\theta = \dfrac{a}{(r^2+a^2)^{1/2}}$$

$$E_{\text{net}} = 2\cdot\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r^2+a^2)}\cdot\dfrac{a}{(r^2+a^2)^{1/2}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\,(2a)}{(r^2+a^2)^{3/2}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{(r^2+a^2)^{3/2}}$$

directed antiparallel to $\vec{p}$. For a short dipole, $r\gg a$:

$$\vec{E}_{\text{equatorial}} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}}{r^3}$$

The minus sign is worth a mark on its own: the equatorial field is opposite in direction to the dipole moment.

The diagram in words

• Axial case: draw the dipole horizontally, $-q$ on the left, $+q$ on the right, separation $2a$, centre $O$, and $\vec{p}$ as an arrow from $-q$ to $+q$. Mark $P$ on the axis to the right of $+q$ at distance $r$ from $O$. At $P$ draw a long arrow to the right for $\vec{E}_{+q}$ and a shorter one to the left for $\vec{E}_{-q}$.
• Equatorial case: same dipole, with $P$ on the perpendicular through $O$ at distance $r$. At $P$ draw $\vec{E}_{+q}$ pointing directly away from $+q$ and $\vec{E}_{-q}$ pointing directly towards $-q$, both making angle $\theta$ with the line $OP$ produced. Mark their resolved components and show the perpendicular pair cancelling.

Standard results

• The dipole field at large distances depends on $q$ and $a$ only through their product $qa$ — which is why the dipole moment is the natural quantity to define.
• The dipole field falls as $1/r^3$, in contrast to the $1/r^2$ of a point charge, and its magnitude and direction depend on the angle between $\vec{r}$ and $\vec{p}$ as well as on $r$.
• Torque and potential energy of a dipole, $\vec{\tau} = \vec{p}\times\vec{E}$ and $U = -\vec{p}\cdot\vec{E}$, are also examined under this heading: torque is maximum when $\vec{p}$ is perpendicular to $\vec{E}$, and the potential energy in stable equilibrium is $-pE = -qLE$ for a dipole of charge separation $L$.