Potential due to an Electric Dipole

The axial derivation CBSE sets for 3 marks

Let a dipole have $-q$ at $A$ and $+q$ at $B$, separated by $2a$, with centre $O$. Let $P$ lie on the axis at distance $r$ from $O$, on the $+q$ side, so $AP=r+a$ and $BP=r-a$. Potential is a scalar, so add algebraically:

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(-q)}{r+a}+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r-a}=\dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{r-a}-\dfrac{1}{r+a}\right]$$

$$V=\dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{2a}{r^{2}-a^{2}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^{2}-a^{2}}\qquad(p=q\times2a)$$

For a far-off point, $r\gg a$, so $a^{2}$ is dropped:

$$V_{\text{axial}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^{2}}$$

• On the $-q$ side the same working gives $V=-\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^{2}}$, so the axial result is written $\pm\dfrac{p}{4\pi\varepsilon_0r^{2}}$, positive for $\theta=0$ and negative for $\theta=\pi$.

The general point (NCERT's own derivation)

For $P$ at distance $r$ from the centre, making angle $\theta$ with $\vec p$, with $r_1$ from $+q$ and $r_2$ from $-q$:

$$V=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q}{r_1}-\dfrac{q}{r_2}\right)$$

Geometry gives $r_1^{2}=r^{2}+a^{2}-2ar\cos\theta$ and $r_2^{2}=r^{2}+a^{2}+2ar\cos\theta$. For $r\gg a$, keeping first order in $a/r$,

$$r_1\simeq r\left(1-\dfrac{a\cos\theta}{r}\right),\qquad r_2\simeq r\left(1+\dfrac{a\cos\theta}{r}\right)$$

and by the binomial theorem $\dfrac{1}{r_1}\simeq\dfrac1r\left(1+\dfrac{a\cos\theta}{r}\right)$, $\dfrac{1}{r_2}\simeq\dfrac1r\left(1-\dfrac{a\cos\theta}{r}\right)$. Substituting and using $p=2qa$,

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^{2}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec p\cdot\hat r}{r^{2}}\qquad(r\gg a)$$

• This is approximate for a physical dipole but exact for a point dipole at the origin.

The equatorial plane

Put $\theta=90^\circ$: $\cos\theta=0$, so $V=0$. Physically, every equatorial point is equidistant from $+q$ and $-q$, so the two potentials cancel exactly — at any distance, near or far. Hence the 1-mark result: if the axial potential at distance $r$ is $V$, the potential at the same distance on the equatorial line is zero.

Contrast with a single charge — the two points CBSE wants

• The dipole potential depends not only on $r$ but also on the angle $\theta$ between $\vec r$ and $\vec p$. It is, however, axially symmetric about $\vec p$: rotating $\vec r$ about $\vec p$ at fixed $\theta$ traces a cone of equal potential.
• It falls off as $1/r^{2}$ at large distance, faster than the $1/r$ of a single charge, because the net charge of a dipole is zero.

Diagram to describe in words

Draw $-q$ at $A$ and $+q$ at $B$ on a horizontal line, centre $O$, with $AB=2a$ and the dipole moment arrow $\vec p$ pointing from $-q$ to $+q$. For the axial case, mark $P$ on the extension beyond $B$ at distance $r$ from $O$, with $BP=r-a$ and $AP=r+a$ labelled. For the general case, draw $OP$ at angle $\theta$ above the axis, drop perpendiculars from $A$ and $B$ onto $OP$ to show the $a\cos\theta$ segments, and label $r_1=BP$ and $r_2=AP$.

Standard results