Potential of a Dipole

Every computational drug-discovery pipeline evaluates dipole potentials by the billion. A protein pocket and a candidate molecule are modelled as thousands of partial charges; the electrostatic potential each projects — monopole terms, dipole terms, the whole Legendre ladder — is computed on a grid, and docking software scores how well the candidate's hills nest into the pocket's hollows. The millivolt-scale, nanometre-range landscape you computed for one water molecule in the worked example is precisely the raw material: multiply it across a solvation shell and you have the electrostatics of biochemistry. The same physics runs the gadget in your kitchen. A microwave oven's 2.45 GHz field grabs water molecules by their dipole moments and twists them back and forth billions of times a second; the "thermal jostling" that fought the dipole in worked example 2 is exactly what the oven amplifies — friction between rocking dipoles becomes heat. Neutral molecules, non-zero $p$: dinner.

Build the landscape by superposition. A lone $+q$ raises a circular hill, $+kq/r$. A lone $-q$ digs a matching funnel, $-kq/r$. Now slide them toward each other until they sit a small distance $2a$ apart: almost everywhere, the hill and the funnel nearly cancel, because almost everywhere you are nearly equidistant from the two charges. The surviving terrain is subtle — a low ridge on the side facing $+q$, a shallow trench on the side facing $-q$, and between them a perfectly flat line at sea level: the plane of points exactly equidistant from both charges, where the cancellation is total.

Two features of this map decide every exam question. First, the sign compass: the potential is positive in the half-space the moment $\vec{p}$ points into (the $+q$ side), negative in the other half, zero on the boundary plane. The angle $\theta$ between $\vec{p}$ and your line of sight is all that decides the sign — through $\cos\theta$, positive for $\theta < 90°$, negative past it. Second, the falloff: double your distance from a point charge and $V$ drops to a quarter... no — to a half, since $V \propto 1/r$; but double your distance from a dipole and its potential drops to a quarter, $V \propto 1/r^2$. The dipole's influence dies faster because from far away the two charges overlap and erase each other; only their failure to sit at the same point keeps $V$ alive at all. Keep the family portrait in view — it is a one-mark question in three disguises:

The field column belongs to Field of a Dipole: Axial & Equatorial — we quote it here for the contrast and never re-derive it. Within each row, the field always falls one power faster than the potential: the field is the slope of the potential, and slopes of dying hills die faster, the rule you sharpened in Equipotentials & the Field–Potential Link.

Set up the standard geometry once, cleanly. Put the dipole at the origin with its axis along a line: $-q$ at distance $a$ on one side, $+q$ at distance $a$ on the other, total separation $2a$, moment $p = q(2a)$ pointing from $-q$ to $+q$. The observation point $P$ sits at distance $r$ from the centre, at angle $\theta$ from the moment's direction. Call the distances from $P$ to the two charges $r_+$ and $r_-$. Because potential is a scalar — the doctrine of Potential of Point Charges & Charge Systems — the exact answer is one line:

where: $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$; $r_\pm$ — the distances from $P$ to $+q$ and $-q$. The second form, with the two fractions combined, is the one to remember: it shows that everything hangs on how different the two distances are ($r_- - r_+$ upstairs) and how far away you are overall ($r_+ r_-$ downstairs). If $P$ is equidistant from both charges, the numerator dies and $V = 0$ — the equatorial plane, exactly, with no approximation. Far from a short dipole ($r \gg a$), two clean approximations turn equation (1) into the formula the boards want. The derivation card below earns them honestly; the results are $r_- - r_+ \approx 2a\cos\theta$ (the path difference is the projection of the separation onto your line of sight) and $r_+ r_- \approx r^2$. Substituting both:

where: $p$ — the dipole moment (C·m), direction $-q \to +q$; $\theta$ — the angle between $\vec{p}$ and the line from the dipole's centre to $P$; valid for $r \gg a$ (the "short dipole" regime). Read the anatomy: the $\cos\theta$ is the sign compass, the $r^2$ is the accelerated falloff, and $p$ is the only thing about the dipole the far zone can still see — not $q$ and $a$ separately, only their product, exactly the lesson of The Electric Dipole & Dipole Moment. The two special directions every paper asks about drop straight out of equation (2):

where: the $+$ sign holds on the $+q$ side ($\theta = 0$, $\cos\theta = +1$), the $-$ sign on the $-q$ side ($\theta = 180°$, $\cos\theta = -1$). Every intermediate direction interpolates through $\cos\theta$ — at $60°$ from the axis you get exactly half the axial value, a favourite quick-check in MCQs.

Board level — axial, angled and equatorial in one sweep (CBSE/ISC pattern)

Two charges $\pm 1.0\ \mathrm{nC}$ are separated by $2.0\ \mathrm{cm}$. Find the electric potential (a) at a point $10\ \mathrm{cm}$ from the centre on the axis, on the $+q$ side; (b) at $10\ \mathrm{cm}$ from the centre at $60°$ from the axis; (c) on the equatorial plane. Take $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$.

Step 1. Assemble the moment first — it is the only dipole property the formula needs: $p = q(2a) = (1.0\times10^{-9})(2.0\times10^{-2}) = 2.0\times10^{-11}\ \mathrm{C\,m}$.

Step 2. Check the regime: $r = 0.10$ m against $a = 0.010$ m gives $r = 10a$ — comfortably "short dipole", so equation (2) applies (with ~1% honesty, worth one line in an answer).

Step 3. Axial, $+q$ side ($\theta = 0$): $V = \dfrac{kp}{r^2} = \dfrac{(9.0\times10^9)(2.0\times10^{-11})}{(0.10)^2} = \dfrac{0.18}{0.010} = \boxed{+18\ \mathrm{V}}$.

Step 4. At $60°$: multiply by $\cos 60° = 0.5$: $V = 18 \times 0.5 = \boxed{+9.0\ \mathrm{V}}$. No new computation — the axial answer is the amplitude, $\cos\theta$ is the dial.

Step 5. Equatorial ($\theta = 90°$): $\cos 90° = 0$, so $V = \boxed{0}$ — and state the reason for the mark: every equatorial point is equidistant from $+q$ and $-q$, so the two scalar contributions cancel exactly.

Competitive level — a water molecule versus thermal chaos (JEE-Main/NEET pattern)

A water molecule has dipole moment $p = 6.2\times10^{-30}\ \mathrm{C\,m}$. (a) Find the potential it produces at $r = 1.0\ \mathrm{nm}$ from its centre, at $\theta = 60°$ from its axis. (b) Find the potential energy of an electron placed at that point, and compare its magnitude with the thermal energy scale $k_BT$ at $300\ \mathrm{K}$ ($k_B = 1.38\times10^{-23}\ \mathrm{J/K}$). Use $k = 8.99\times10^9\ \mathrm{N\,m^2/C^2}$, $e = 1.6\times10^{-19}\ \mathrm{C}$.

Step 1. Potential from equation (2): $V = \dfrac{kp\cos\theta}{r^2} = \dfrac{(8.99\times10^9)(6.2\times10^{-30})(0.5)}{(1.0\times10^{-9})^2}$.

Step 2. Numerator: $8.99\times10^9 \times 6.2\times10^{-30} \times 0.5 = 2.79\times10^{-20}$. Denominator: $1.0\times10^{-18}$. So $V = 2.79\times10^{-2}\ \mathrm{V} \approx \boxed{28\ \mathrm{mV}}$ — millivolt terrain, a nanometre out.

Step 3. Potential energy of the electron there — the $U = qV$ bridge from Electric Potential & Potential Difference: $|U| = eV = (1.6\times10^{-19})(2.79\times10^{-2}) = \boxed{4.5\times10^{-21}\ \mathrm{J}}$. (Sign: the electron is negative, so at this positive-$V$ point its energy is $-4.5\times10^{-21}$ J — it is attracted toward the positive lobe.)

Step 4. Thermal comparison: $k_BT = (1.38\times10^{-23})(300) = 4.1\times10^{-21}\ \mathrm{J}$. The ratio is $4.5/4.1 \approx 1.1$ — the dipole's electrical grip at a nanometre is only just as strong as ambient thermal jostling. That single comparison is why water's dipoles do not lock rigidly to every passing ion at room temperature, and why polar interactions in liquids are constantly made and broken — the microscopic backdrop for Dielectrics & Polarisation, where thermal randomisation fights the field for control of the molecules. Timing note: 3 minutes. The mark-getters are the clean powers of ten in Step 2 (practise $10^9 \times 10^{-30}/10^{-18}$-type collections until they take five seconds) and remembering that (b) is one multiplication by $e$ — not a new formula. If the paper asks "at what angle is V half the axial value", answer $60°$ from $\cos\theta = \tfrac12$ without touching the calculator.

Mixing the dipole's $V$ and $E$ formulas. $V \propto 1/r^2$; the field magnitudes go as $1/r^3$ (axial $2kp/r^3$, equatorial $kp/r^3$ — from Field of a Dipole). Writing $kp\cos\theta/r^3$ for the potential, or $kp/r^2$ for a field, is the most common single error. Anchor: potential is always one power slower than field.

Measuring $r$ from a charge instead of the centre. In the board example above, taking $r$ from the $+q$ charge (0.09 m... or worse, 0.10 m to $+q$ and 0.12 m to $-q$) gives 15 V instead of 18 V — a wrong answer that looks plausible. The formula's $r$ starts at the dipole's midpoint. Always.

Forgetting the sign of the axial potential. $+kp/r^2$ only on the $+q$ side; the $-q$ side is $-kp/r^2$. Read the question's geometry before writing the sign, and remember $\vec{p}$ points from $-q$ to $+q$ — the direction convention itself is a one-mark question.

θ measured from the wrong line. $\theta$ is the angle from the axis (the direction of $\vec{p}$), not from the equatorial plane. A point "30° above the equatorial plane" sits at $\theta = 60°$ — translate before substituting.

Using the far-field formula too close. Equation (2) needs $r \gg a$: at $r = 5a$ it is already 4% off, and JEE-Advanced setters exploit this by asking for the exact two-term sum (1) at moderate range. If $r$ and $a$ are within a factor of ~5, compute both distances and use (1).

"V = 0 means no work ever". Bringing a charge from infinity to an equatorial point costs zero net work — true, and a favourite assertion-reason item. But moving it from an equatorial point to an axial point costs $q\,\Delta V \ne 0$. Zero is a value of the landscape at some places, not a promise about all journeys.

Constants. $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ is accepted by CBSE/ISC/NEET; carry $8.99\times10^9$ in JEE-Advanced when options sit close. Dipole moments of molecules are often quoted in debye: $1\ \mathrm{D} = 3.34\times10^{-30}\ \mathrm{C\,m}$ — water's $6.2\times10^{-30}\ \mathrm{C\,m}$ is 1.85 D; convert to SI before any formula.

Texts. NCERT Physics Part 1, Ch. 2 §2.4 — the board-canonical derivation this module dresses in full · HC Verma, Concepts of Physics 2, Ch. 29 · Halliday, Resnick & Walker Ch. 24. For the multipole ladder the $\cos\theta$ came from — and the dipole field in coordinate-free form — see Electric Dipole Moment (UG deep dive).

Provenance. Standard results per NCERT §2.4 and HRW Ch. 24. A verifier should recompute: $p = (1.0\times10^{-9})(0.020) = 2.0\times10^{-11}$ C·m; axial $V = kp/r^2 = 18.0$ V at $r = 0.10$ m vs exact $kq(1/0.09 - 1/0.11) = 18.2$ V (1.0% low); $V(60°) = 9.0$ V; water molecule $V = (8.99\times10^9)(6.2\times10^{-30})(0.5)/10^{-18} = 27.9$ mV; $|U| = eV = 4.5\times10^{-21}$ J vs $k_BT(300\,\mathrm{K}) = 4.1\times10^{-21}$ J, ratio 1.08; far-field error 4% at $r = 5a$.

Prerequisites. The Electric Dipole & Dipole Moment · Potential of Point Charges & Charge Systems. Leads to. Dielectrics & Polarisation · Electric Dipole Moment (UG deep dive).