In short

The potential energy of two point charges q_1 and q_2 separated by distance r is

U \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r} \;=\; \frac{kq_1 q_2}{r}.

It is the work done by an external agent to bring the charges from infinity to their current positions (and equivalently, the work the electric force would do if they flew apart back to infinity). For a system of N charges,

U \;=\; \sum_{\text{pairs } i<j} \frac{kq_i q_j}{r_{ij}} \;=\; \tfrac{1}{2}\sum_i q_i V_i,

where V_i is the potential at the location of q_i due to all the other charges. A charge q placed in an external potential V has energy U = qV. A dipole of moment \vec p in a uniform external field \vec E has energy

U_\text{dipole} \;=\; -\vec p\cdot\vec E \;=\; -pE\cos\theta,

minimum when \vec p aligns with \vec E — the statement that torque turns dipoles until they point along the field.

Strike a match in the dark and the whole room lights up. Where was that energy before you struck the match? Locked inside the atoms — specifically, in the electrical arrangement of the nuclei and electrons that make up the phosphorus tip and the paper. When those electrons rearrange during combustion, electrostatic energy pours out as light and heat. The match did not "create" energy; it merely let a configuration that had stored energy slide downhill into a configuration that stored less, with the difference emerging as photons and warm air. Every chemical reaction that has ever warmed a home in Siachen, every battery that has ever powered a Delhi Metro train, every kilogram of dal you have ever digested — all of it is the release of electrostatic potential energy locked up in specific arrangements of nuclei and electrons.

The bookkeeping is done by one quantity: the potential energy of the charge configuration, U. It is a single number that tells you how much work was needed to assemble the charges in their current places, starting from all of them infinitely far apart. Equivalently — and this is the payoff — it tells you, the instant you let the charges move, how much kinetic energy they can ultimately produce. The atoms-and-molecules version of "how much water you have stored behind the dam at Bhakra Nangal." Stored electrical energy is the same kind of quantity.

This article builds U from the ground up. First, U = kq_1q_2/r for two point charges — the molecule of all further formulas. Then the pair-sum rule for many charges, which turns out, by a beautiful accounting trick, to equal \tfrac{1}{2}\sum q_i V_i — the formula that connects configuration energy to the scalar potential you learned to work with in the last two chapters. Then U = qV, the energy of a charge in an external field, which is the version that matters for circuits and for a single ion in an accelerator. Finally, the dipole-in-a-field energy U = -\vec p\cdot \vec E, a formula so central that it governs the alignment of water molecules in every microwave oven in every Indian kitchen, the orientation of ionic dumbbells in the sun-baked salt flats of the Rann of Kutch, and the thermodynamics of ferroelectric materials being engineered at ISRO's semiconductor division for next-generation memory. Every step is derived; no step is punted.

Why potential energy exists for charges (one paragraph of conservative-force recap)

Recall from Electric Potential and Potential Difference that the electrostatic field is conservative: the work done in moving a charge from point B to point A depends only on the endpoints, not on the path. This is the necessary and sufficient condition for a potential energy to exist. Given any conservative force \vec F, the potential energy at a point is defined as the (negative of the) work done by the force from a chosen reference to that point:

U(\vec r) \;=\; -\int_\text{ref}^{\vec r} \vec F \cdot d\vec\ell.

For the electric force on a charge q in an external field \vec E, this gives (with infinity as the reference)

U(\vec r) \;=\; -\int_\infty^{\vec r} q\vec E \cdot d\vec\ell \;=\; q\,V(\vec r),

where V is the electric potential at \vec r. One line, but it packs the whole idea: the potential energy of a charge q in the field of other charges is simply q times the potential at the charge's location. This section makes that specific for systems of two charges, many charges, and dipoles.

Two point charges — the fundamental formula

Take charge q_1 fixed at position \vec r_1 and consider bringing charge q_2 from infinity to position \vec r_2. The potential due to q_1 at \vec r_2 is

V_1(\vec r_2) \;=\; \frac{kq_1}{|\vec r_2 - \vec r_1|} \;=\; \frac{kq_1}{r},

where r = |\vec r_2 - \vec r_1| is the distance between the two charges. The work done by an external agent to move q_2 from infinity (where the potential is zero) to \vec r_2 is, by equation (7) of the previous article,

W_\text{ext} \;=\; q_2 (V_1(\vec r_2) - 0) \;=\; \frac{kq_1 q_2}{r}.

This work is what is stored as potential energy of the two-charge configuration:

\boxed{\;U(r) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r} \;=\; \frac{kq_1 q_2}{r}\;} \tag{1}

A few things are worth noticing before the examples pile up.

Symmetry. Equation (1) is symmetric in q_1 and q_2 — we could just as easily have fixed q_2 and brought q_1 in from infinity. The same U comes out. Potential energy belongs to the pair, not to one of the two charges individually. (You will sometimes hear "the potential energy of charge q_2 in the field of q_1," but that is a shorthand; there is only one potential energy, the pair's.)

Sign. If q_1 and q_2 have the same sign (both positive or both negative), U > 0 — energy was put in by the external agent to push them together against their mutual repulsion. If they have opposite signs, U < 0 — the charges attracted each other; the external agent had to hold them back, doing negative work, while the electric field did positive work. The configuration ends up with less energy than its infinite-separation reference.

Distance dependence. U \propto 1/r, unlike the field's 1/r^2. As two charges are brought from infinity to a finite distance r, the stored energy grows in magnitude as 1/r. If you try to bring two same-sign point charges all the way together (r \to 0), U \to \infty — it takes infinite energy to squeeze two like point charges into the same place, which is what sets the effective "hard-core" size of ions and why ionic crystals do not collapse.

U(r) for two charges — positive branch (like charges) and negative branch (unlike charges) Graph of U versus separation r. Upper red curve: like charges, U = +kq1q2/r, positive and decreasing toward zero at large r. Lower dark curve: unlike charges, U = -k|q1q2|/r, negative and increasing toward zero at large r. separation r U(r) 0 U = +kq₁q₂/r (like charges — repulsive) U = −k|q₁q₂|/r (unlike charges — attractive)
The potential energy of a pair of point charges. For like charges, $U > 0$: energy stored in the repulsion, available to drive them apart. For unlike charges, $U < 0$: the pair is in a bound state relative to infinite separation, and you must supply energy to pull them apart.

More than two charges — sum over all pairs

The potential energy of a system of three or more point charges is built up by bringing them in one at a time from infinity and summing the work at each step.

Step 1. Start with all charges infinitely far apart. U = 0 by definition (the reference).

Step 2. Bring in q_1 to its position. No work done, because there are no other charges to interact with yet. Running total: U = 0.

Step 3. Bring in q_2 to its position, while q_1 is fixed. Work done = kq_1q_2/r_{12}. Running total: U = kq_1q_2/r_{12}.

Step 4. Bring in q_3 while q_1 and q_2 are fixed. The potential at q_3's destination is V = kq_1/r_{13} + kq_2/r_{23}, so the work done is q_3 V = kq_1q_3/r_{13} + kq_2q_3/r_{23}. Running total:

U \;=\; \frac{kq_1q_2}{r_{12}} + \frac{kq_1q_3}{r_{13}} + \frac{kq_2q_3}{r_{23}}.

Why: at each step, a new charge is brought in through the field of all charges already present, and the work done is the new charge times the sum of potentials from the already-placed charges. Summing these stepwise contributions produces exactly the three distinct pair terms.

Continuing this procedure for N charges gives one term for every pair:

\boxed{\;U \;=\; \sum_{\text{pairs } i < j} \frac{kq_i q_j}{r_{ij}}\;} \tag{2}

There are \binom{N}{2} = N(N-1)/2 such pairs. The i < j constraint is there to count each pair once — pair (1,2) is the same pair as (2,1), so we fix a convention (say, always writing the smaller index first) to avoid double counting.

The equivalent "half the sum qV" form

There is a gorgeous re-expression of (2) that appears constantly in capacitor problems and in more advanced electrostatics. Start from (2) and symmetrise the sum over all ordered pairs (then divide by 2):

U \;=\; \sum_{i < j} \frac{kq_iq_j}{r_{ij}} \;=\; \tfrac{1}{2}\sum_{i \neq j} \frac{kq_iq_j}{r_{ij}}.

Why: summing over all ordered pairs with i \neq j counts each unordered pair twice, so dividing by 2 recovers the original count.

Now factor out q_i from each term and recognise the inner sum as the potential at \vec r_i due to all the other charges:

U \;=\; \tfrac{1}{2}\sum_i q_i \underbrace{\sum_{j \neq i} \frac{kq_j}{r_{ij}}}_{V_i\,=\,\text{potential at }\vec r_i\text{ from all other charges}} \;=\; \tfrac{1}{2}\sum_i q_i V_i.

So:

\boxed{\;U \;=\; \tfrac{1}{2}\sum_i q_i V_i\;} \tag{3}

Why: the factor of \tfrac{1}{2} is not a convention — it is the correction for not double-counting pair energies. Each pair (i,j) contributes once to q_iV_i (with V_i including the contribution of j) and once to q_jV_j (with V_j including the contribution of i). The two contributions together equal 2 \times kq_iq_j/r_{ij}, so multiplying by \tfrac{1}{2} recovers the single pair term. The rule "half the sum of q_iV_i over all charges" is exact and general.

Formulas (2) and (3) are equivalent. Formula (2) is better when you can easily write down the pair distances (three charges on a triangle, four at the corners of a square). Formula (3) is better when the potential at each charge's location is already known from a higher-level calculation (as in a capacitor, where each conductor is at a known potential V_i and the total charge on each is q_i).

A charge in an external field — U = qV

Step back from "system of charges" to the simpler picture: a single charge q placed at a point where some other, external charges have already established a potential V. The potential energy of that single charge is

\boxed{\;U \;=\; qV\;} \tag{4}

Why: this is just equation (4) of the previous article, restated in energy language. The work needed to bring a charge q from infinity (where V = 0) to a point of potential V is qV, and that work is stored as potential energy.

Equation (4) is for the interaction of the charge q with the external source distribution. It does not include the energy of the external source distribution itself — that is a separate, already-stored quantity we are treating as a fixed background. This is often the picture you want in practice: a test ion in a mass spectrometer, an electron in a CRT, a charged ball hanging in the field of a cloud. The external field is given, and you only track the energy associated with placing your test charge somewhere in it.

The distinction between (3) and (4) is worth spelling out. In (3), every charge is both a source and a feeler — the sum \sum q_iV_i counts each charge's interaction with every other charge, and the factor \tfrac{1}{2} removes the double count. In (4), the charge q is only a feeler; the sources producing V are not included in the sum because their internal energy is not being recomputed. Equation (4) is U_\text{interaction}; equation (3) is U_\text{total} of the charge system minus the self-energy of each point charge (which is infinite and usually ignored).

The dipole in an external field

A particularly important application of U = qV. A dipole is a pair of equal and opposite charges +q and -q separated by a small distance \vec d (the displacement pointing from -q to +q). The dipole moment is

\vec p \;=\; q\vec d.

Place this dipole in an external, approximately uniform electric field \vec E. The two charges are at different points in space, at different potentials. Let V(\vec r_+) be the potential at the location of +q and V(\vec r_-) at the location of -q. The total potential energy of the dipole is

U \;=\; (+q)V(\vec r_+) + (-q)V(\vec r_-) \;=\; q[V(\vec r_+) - V(\vec r_-)]. \tag{5}

Why: each charge of the dipole contributes its own qV to the total energy (equation 4), and the two contributions have opposite signs of q but evaluated at different points.

Since \vec r_+ - \vec r_- = \vec d is small, we can Taylor-expand V about the centre of the dipole. For a uniform field \vec E, V(\vec r) = V_0 - \vec E\cdot \vec r (up to a constant V_0 that does not affect differences), so

V(\vec r_+) - V(\vec r_-) \;=\; -\vec E\cdot (\vec r_+ - \vec r_-) \;=\; -\vec E\cdot \vec d.

Substituting into (5):

U \;=\; q(-\vec E\cdot \vec d) \;=\; -\vec E\cdot (q\vec d) \;=\; -\vec p\cdot\vec E. \tag{6}

Writing this in component form where \theta is the angle between \vec p and \vec E:

\boxed{\;U_\text{dipole} \;=\; -\vec p\cdot\vec E \;=\; -pE\cos\theta\;} \tag{7}

Why: the dot product \vec p\cdot\vec E = pE\cos\theta. The minus sign in equation (6) is important — it says the energy is lowest (most negative) when \theta = 0, that is, when \vec p aligns parallel to \vec E. The dipole "wants" to line up with the field.

Three special configurations to internalise:

Dipole energy as a function of orientation angle Plot of U = -pE cos(theta) versus theta from 0 to 2pi. The curve is a cosine with minimum at theta=0 (aligned, U=-pE), zeros at theta=pi/2 and 3pi/2, maximum at theta=pi (anti-aligned, U=+pE). angle θ between p and E U(θ) 0 +pE −pE 0 π/2 π 3π/2 2π stable unstable
The dipole energy is a cosine of the orientation angle. It is minimised when the dipole aligns with the field and maximised when the dipole points opposite to it. Between the two extrema, the torque drives the dipole toward alignment.

The torque that rotates a dipole in a field is the derivative of the energy: \tau = -dU/d\theta = -pE\sin\theta, which is exactly the torque formula \vec\tau = \vec p\times\vec E by a different route. Energy and torque are two faces of the same physics.

Real example: water in a microwave. A water molecule has a permanent dipole moment (the oxygen pulls the electrons toward itself, leaving the two hydrogens slightly positive). In a microwave oven, the 2.45-GHz electric field flips direction 5 billion times a second; each water molecule's dipole tries to align, but before it reaches the aligned state the field has already flipped. The molecule never settles, and the energy stored in U = -\vec p\cdot\vec E is continuously transferred into rotational motion, which through collisions becomes thermal motion — which is how a microwave heats food. The formula U = -pE\cos\theta is directly the reason your chai warms.

Worked examples

Example 1: Energy of three charges at the corners of an equilateral triangle

Three charges are placed at the vertices of an equilateral triangle of side 10.0 cm: q_1 = +3.0 nC, q_2 = +3.0 nC, q_3 = -2.0 nC. Find the potential energy of the configuration. Take k = 9.0 \times 10^9 N·m²/C².

Three charges at the vertices of an equilateral triangle of side 10 cm Vertices labelled q1 (+3nC, bottom left), q2 (+3nC, bottom right), q3 (-2nC, top). All sides 10 cm. Each pair has its own r = 10 cm separation. q₃ = −2 nC q₁ = +3 nC q₂ = +3 nC 10 cm 10 cm 10 cm
Three pairs, each with the same separation $r = 10$ cm. The total configuration energy is the sum of the three pair energies.

Step 1. List the pairs and their contributions.

There are three pairs: (1,2), (1,3), (2,3). Each pair's separation is r = 0.10 m. By equation (2):

U \;=\; \frac{kq_1q_2}{r} + \frac{kq_1q_3}{r} + \frac{kq_2q_3}{r}.

Since all three separations are equal, factor k/r:

U \;=\; \frac{k}{r}(q_1q_2 + q_1q_3 + q_2q_3).

Why: this is the pair-sum rule (equation 2), written out for three charges. Because the triangle is equilateral, the three separations coincide and the factoring works out especially cleanly.

Step 2. Compute the product sum.

Convert charges to coulombs: q_1 = q_2 = 3 \times 10^{-9} C, q_3 = -2 \times 10^{-9} C.

q_1q_2 \;=\; (3)(3) \times 10^{-18} \;=\; 9 \times 10^{-18} \text{ C}^2.
q_1q_3 \;=\; (3)(-2) \times 10^{-18} \;=\; -6 \times 10^{-18} \text{ C}^2.
q_2q_3 \;=\; (3)(-2) \times 10^{-18} \;=\; -6 \times 10^{-18} \text{ C}^2.

Sum: 9 - 6 - 6 = -3 \times 10^{-18} C^2.

Why: the two like-charge pairs contribute positive energy (repulsive pairs, energy put in to assemble); the two unlike-charge pairs contribute negative energy (attractive pairs, energy released during assembly). The net outcome — negative — tells you this configuration is bound relative to infinite separation.

Step 3. Plug in k/r.

\frac{k}{r} \;=\; \frac{9.0 \times 10^9}{0.10} \;=\; 9.0 \times 10^{10} \text{ N·m/C}^2.
U \;=\; (9.0 \times 10^{10})(-3 \times 10^{-18}) \;=\; -2.7 \times 10^{-7} \text{ J} \;=\; -0.27 \text{ μJ}.

Result: U = -2.7 \times 10^{-7} J = -0.27 μJ.

What this shows: The negative sign means the configuration has less energy than the reference state (all three charges infinitely apart). If released, the charges would fly apart, but the configuration does not spontaneously gain that 0.27 μJ of energy out of nothing — the energy is released as kinetic energy. To separate the charges back to infinity, an external agent must pump in +0.27 μJ of work.

Example 2: Dipole in a uniform field — work to flip it

A water-like dipole with dipole moment p = 6.2 \times 10^{-30} C·m sits in a uniform external field of magnitude E = 1.5 \times 10^5 V/m. The dipole is initially aligned parallel to the field. Find the work an external agent must do to rotate the dipole so that it points anti-parallel to the field (a 180° flip).

Dipole rotated from aligned to anti-aligned Left panel: dipole arrow pointing right, aligned with field arrow. Right panel: dipole arrow flipped to point left, anti-aligned with field. An external torque has rotated it 180 degrees. Before: θ = 0, U = −pE E field p (dipole moment) After: θ = π, U = +pE E field p (now flipped) external work W = ?
The dipole starts aligned with the field and ends anti-aligned. The work required is the change in potential energy $U_\text{after} - U_\text{before}$.

Step 1. Write the energies in the two configurations using equation (7).

Before: \theta_i = 0, so U_i = -pE\cos 0 = -pE.

After: \theta_f = \pi, so U_f = -pE\cos\pi = +pE.

Why: at \theta = 0 the dipole is aligned with the field (stable, minimum energy). At \theta = \pi it is anti-aligned (unstable, maximum energy). The energy is a simple cosine of the angle.

Step 2. The external work done equals the change in potential energy (for a quasi-static rotation, where the dipole has no kinetic energy at either end):

W_\text{external} \;=\; U_f - U_i \;=\; (+pE) - (-pE) \;=\; 2pE.

Step 3. Substitute the numbers.

W_\text{external} \;=\; 2(6.2 \times 10^{-30})(1.5 \times 10^5) \;=\; 1.86 \times 10^{-24} \text{ J}.

Convert to electron-volts: divide by 1.602 \times 10^{-19} J/eV.

W_\text{external} \;=\; \frac{1.86 \times 10^{-24}}{1.602 \times 10^{-19}} \;\approx\; 1.16 \times 10^{-5} \text{ eV} \;\approx\; 11.6 \text{ μeV}.

Why: the μeV scale is far below thermal energy at room temperature (k_B T \approx 26 meV). Water dipoles thermally flip constantly by themselves, with no external field needed, because the thermal energy exceeds the required flip work by a factor of about 2000. This is why water in a calm glass does not have all its dipoles neatly aligned — entropy wins over the small pE alignment energy.

Result: The external work to flip the dipole is W = 2pE \approx 1.9 \times 10^{-24} J \approx 12 μeV.

What this shows: The energy scale of a single dipole in a laboratory field is tiny — much smaller than room-temperature thermal energy. For macroscopic alignment effects (an entire salt flat's ionic crystals orienting, or a dielectric's molecules polarising coherently), either the field must be huge (as inside a capacitor, \sim 10^7 V/m) or the dipoles must be many and cooperative (as in a ferroelectric). The individual dipole formula still applies; the macroscopic behaviour is its statistical sum.

Binding energies — ionic crystals in the Rann of Kutch

The salt flats of the Rann of Kutch are kilometres of white NaCl, the same sodium chloride that is in every pinch of salt on every kitchen table in India. Each crystal grain is a three-dimensional lattice of alternating Na^+ and Cl^- ions. The energy holding that lattice together — the lattice energy of NaCl — is calculated by summing the pair-wise potential energies of every ion pair in the crystal.

For a single Na^+Cl^- pair in isolation at the crystal's spacing r_0 \approx 2.82 Å (= 2.82 \times 10^{-10} m), the binding energy is

U_\text{pair} \;=\; -\frac{kq_1q_2}{r_0} \;=\; -\frac{(9.0 \times 10^9)(1.602 \times 10^{-19})^2}{2.82 \times 10^{-10}}.
U_\text{pair} \;=\; -8.2 \times 10^{-19} \text{ J} \;\approx\; -5.1 \text{ eV per ion pair}.

The full crystal, accounting for all the further neighbours (next-nearest opposite-sign, next-next-nearest same-sign, etc.), gives a Madelung sum that multiplies this pair energy by the Madelung constant M \approx 1.748 for the NaCl structure:

U_\text{lattice} \;=\; -\frac{M e^2}{4\pi\varepsilon_0 r_0} \;\approx\; -7.9 \text{ eV per ion pair}.

This is the electrostatic binding energy per Na^+Cl^- pair in the crystal. To dissolve one kilogram of table salt in water (about 2 \times 10^{25} pairs) would require, if the dissolution were purely electrostatic and the water dielectric contribution were zero,

E_\text{dissolve} \;\sim\; (2 \times 10^{25})(7.9 \text{ eV})(1.602 \times 10^{-19} \text{ J/eV}) \;\approx\; 2.5 \times 10^{7} \text{ J}.

About 25 megajoules — equivalent to 6 kilowatt-hours of electrical energy per kilogram of salt. In reality the dissolution happens easily in water because the high dielectric constant of water (\varepsilon_r \approx 80) screens the electrostatic attraction and lowers the energy cost by nearly that factor. This is why salt dissolves in water but not in oil (\varepsilon_r \approx 2, no screening), and it is why the physics of biological fluids — sea water, blood plasma, cytoplasm — is dominated by the electrostatic energy of ionic hydration. The formulas of this chapter are the foundation of that entire discipline.

Common confusions

If equations (1) through (7) are clear, and you can compute pair-sum energies and dipole-in-a-field energies, you have the working tools. The rest of this section is for readers who want the self-energy of continuous distributions, the connection between the configuration energy and the energy stored in the field itself, and the higher-multipole expansion that generalises the dipole formula.

Self-energy of a continuous distribution

For a continuous charge distribution (a ball of uniform density, a charged shell, a filament), the pair sum (2) becomes a double integral:

U \;=\; \frac{1}{2}\int\int \frac{1}{4\pi\varepsilon_0}\,\frac{\rho(\vec r)\rho(\vec r')}{|\vec r - \vec r'|}\,d^3r\,d^3r',

where the \tfrac{1}{2} corrects the double-counting (here the integrals are over all points, so each pair of volume elements is counted twice), and the |\vec r - \vec r'| = 0 singularity on the diagonal is a divergence that has to be handled carefully for point particles.

For a uniformly charged sphere of total charge Q and radius R, evaluating this integral gives the famous result

U_\text{ball} \;=\; \frac{3}{5}\,\frac{kQ^2}{R}.

The \tfrac{3}{5} is pure geometry. This formula is why the classical "electron as a uniformly charged ball of radius R" breaks down: to reproduce the electron's rest-mass energy m_ec^2, you would need R \approx 1.4 \times 10^{-15} m, of the order of the classical electron radius — but no direct measurement ever sees any structure down to that scale. Classical electrostatics of point charges contains a built-in, unavoidable divergence in the self-energy, and removing it (renormalisation) is one of the foundational problems of modern quantum field theory.

Energy stored in the field — u = \tfrac{1}{2}\varepsilon_0 E^2

An equivalent way of bookkeeping the same configuration energy: instead of summing pair interactions, integrate an energy density over all space.

U \;=\; \int u\,d^3r, \qquad u(\vec r) \;=\; \tfrac{1}{2}\varepsilon_0 \vec E^{\,2}(\vec r).

The derivation (using Poisson's equation and integration by parts) shows that summing pair energies and integrating the field's energy density give the same U for any finite charge distribution. This changes the philosophical picture. In the pair-sum view, the energy is stored "in" the relative positions of the charges. In the field-density view, the energy is stored "in" the electric field itself, distributed through all space. The two views are mathematically equivalent; physicists tend to prefer the field-density view because it generalises to radiation (where the charges have moved on but the field carries energy, travelling outward as a light wave).

For the parallel-plate capacitor, the field between the plates is uniform, and the total field energy is

U \;=\; \int u\,d^3r \;=\; \tfrac{1}{2}\varepsilon_0 E^2 (\text{volume between plates}) \;=\; \tfrac{1}{2}CV^2,

where C is the capacitance. That is the standard capacitor-energy formula, derived from the field-energy picture.

Beyond the dipole — the multipole expansion

For a localised charge distribution viewed from far away, the potential can be expanded in a series of multipole moments:

V(\vec r) \;=\; \frac{k Q_\text{tot}}{r} + \frac{k\vec p \cdot \hat r}{r^2} + \frac{k}{2r^3}\sum_{ij} Q_{ij}\hat r_i \hat r_j \hat + \ldots

The first term is the monopole, proportional to the total charge Q_\text{tot} and falling as 1/r.

The second is the dipole term, proportional to the dipole moment \vec p and falling as 1/r^2.

The third is the quadrupole term, proportional to the quadrupole tensor Q_{ij} and falling as 1/r^3.

For a neutral system (Q_\text{tot} = 0), the dipole term is the leading contribution. For a system with \vec p = 0 but non-zero quadrupole (two equal dipoles placed back-to-back, or a rod of positive charge between two rods of negative), the quadrupole dominates. Higher-order moments (octupole, hexadecapole) appear when the charge distribution is increasingly complex.

The energy of a general multipole in a general external field generalises equation (7): for a quadrupole in a field with gradient \nabla\vec E, U = -\tfrac{1}{2}\sum_{ij}Q_{ij}(\partial_i E_j), and so on. The whole sequence is the backbone of atomic and molecular electrostatics — and it is what, in its nuclear-physics version, describes the oblate deformation of rotating nuclei studied at the Variable Energy Cyclotron Centre in Kolkata.

Work done by the field vs work done by external agents — the clean accounting

One last clarification often causes confusion, so let us state it one more time.

  • Work done by the electric force on the charge: W^\text{field} = -\Delta U. The field does positive work when the charge moves to lower potential energy (downhill), and negative work when it moves to higher (uphill).
  • Work done by the external agent (you, a battery, a motor) on the charge: W^\text{ext} = +\Delta U (for quasi-static motion with no change in KE). You do positive work when raising the charge uphill in U; negative work when letting it fall downhill.
  • Work-energy theorem: W^\text{net} = W^\text{field} + W^\text{ext} = \Delta\,\text{KE}. For quasi-static processes, \Delta\,\text{KE} = 0, so W^\text{ext} = -W^\text{field} = \Delta U and all external work is converted to potential energy.

If the motion is not quasi-static, some external work goes into kinetic energy. Conservation of total energy says W^\text{ext} = \Delta U + \Delta\,\text{KE}, and everything else follows. The formulas of this chapter are the \Delta U piece of that equation; the kinetic piece is the concern of the next step in the reader's trajectory: circuits, where currents flow and potential energy is continuously converted into other forms.

Where this leads next