Electrostatic Potential Energy

Chapter one of electrostatics was a chapter about pushes: Coulomb's law, field vectors, force diagrams with arrows to resolve. This chapter opens the second set of books — the energy accounts. Squeeze two positive charges together and you have stored something real, exactly as real as the energy in a compressed spring: let go, and the charges fly apart, cashing the stored energy as motion. Pull a positive charge away from a negative one and you are lifting a ball out of a well — the system holds your investment and will pay it back the moment you release the charge. That stored quantity is electrostatic potential energy, and it is the single most bankable idea in this chapter. Boards love it because the formula is one line; JEE and NEET love it because energy bookkeeping solves problems that force diagrams make miserable — no components, no angles, just signed numbers that add. The energy view is also where electrostatics finally connects to the mechanics you already own: the same "work done against a conservative force gets banked" logic that ran Potential Energy: Gravitational & Elastic now runs on charges, with Coulomb's law supplying the force. One warning before the ledger opens: signs are not decoration here. A positive potential energy and a negative one describe two physically different worlds — one spring-loaded, one bound — and almost every mark lost on this topic is a sign lost, not a formula forgotten. This article builds the sign logic into the pictures from the start, so the formulas never stand alone.

Key takeaways

Electrostatic potential energy $U$ is the work an external agent does to assemble charges quasi-statically from far apart — banked in the system, repaid as kinetic energy on release.
For a pair: $U = \dfrac{kq_1q_2}{r}$ with $U = 0$ at infinite separation. Like charges: $U > 0$ (spring-loaded). Unlike charges: $U < 0$ (bound).
Work bookkeeping: $\Delta U = -W_{\text{field}} = W_{\text{ext}}$ (when the charge starts and ends at rest). The field and the external agent are always on opposite sides of the ledger.
For several charges, add the energy of every pair, each counted once: three charges make three pairs, four make six.
Energy is a scalar — signs matter, directions do not. This is why the energy route beats the force route in multi-charge problems.
Key idea: potential energy belongs to the configuration, not to any single charge — it is the assembly bill, sign included, and release converts it to kinetic energy.

The core insight

The electrostatic potential energy of a charge arrangement is the total work needed to assemble it from infinite separation — $U = kq_1q_2/r$ per pair, positive for like charges, negative for unlike — and because the Coulomb force is conservative, that number depends only on the final arrangement, never on the route taken.

Where this lives today

The pair-sum is not a school formula that retires after the exam — it is the workhorse of computational chemistry and drug discovery. Simulating a protein in water means booking the Coulomb pair energy of hundreds of thousands of partial charges, millions of time steps in a row: the binding of a drug candidate to its target is, to leading order, a change in exactly the quantity this article defined. Closer to your bench: every ionic solid you handle is a negative- $U$ configuration. Salt grains do not spontaneously puff into ions because each Na⁺Cl⁻ pair sits about $8$ electron volts below dispersal — a number you can estimate with nothing but $kq_1q_2/r$ and a table of ionic spacings. And when the next module hands you the per-charge version of this idea (potential), the pair here becomes the battery there: same ledger, new column.

The physical picture

Start with empty space and bring in one charge. Nothing resists you — there is no field to fight yet — so the first charge rides in free. Now bring in a second. If it has the same sign as the first, every centimetre of approach is uphill work: the Coulomb repulsion pushes back harder and harder as the gap closes, and every joule you spend against it is banked in the pair. Stop at separation $r$ , pin the charges, and the account reads $U = kq_1q_2/r$ — positive, because you paid in. Cut the pins and the field refunds everything: the charges accelerate apart, and by the time they are far away the entire deposit has become kinetic energy. Now rerun the movie with opposite signs. The field pulls the newcomer in — you do not spend energy, you harvest it, holding the charge back so it arrives gently. The account goes negative: $U = kq_1q_2/r < 0$ because $q_1q_2 < 0$ . A negative balance means the pair is bound — to separate the charges again, someone must pay the debt. That is not an accounting curiosity; it is why sodium chloride is a solid and why electrons stay attached to atoms.

Figure

Two assembly jobs — like charges bank your work, unlike charges go into debt

The gravity analogy you already trust makes this quantitative. Lifting a mass $m$ by height $h$ banks $mgh$ ; the field pays it back when the mass falls. The upgrade electrostatics brings: gravity only attracts, but charges come in two signs, so electrical accounts run both positive and negative. Keep the slogan from Conservative & Non-Conservative Forces in your pocket: work done against a conservative force is never lost, only stored. Coulomb's force is conservative — the stored amount depends only on where the charges end up, never on the path or the order of assembly — and that single fact is what makes a clean function $U(r)$ possible at all. A graph makes the two worlds unmistakable. Plot $U$ against separation $r$ for each case: the like-pair curve is a $1/r$ wall that towers as $r \to 0$ and dies to zero at large $r$ ; the unlike-pair curve is its mirror image, a well that deepens without limit as the charges approach.

Figure

The pair energy curve — a wall for like charges, a well for unlike

Read the far end of the graph carefully, because it is a choice: we declare $U = 0$ when the charges are infinitely far apart, where they do not interact at all. Every value of $U$ is measured against that agreed zero. It is the same move as choosing the ground floor as $h = 0$ in mechanics — free, conventional, and absolutely load-bearing for signs.

Work: the field and the agent keep opposite books

Before the formula, nail down who pays whom — this is where the marks actually live. When a charge moves, two different parties can do work on it: the electric field (via the Coulomb force) and an external agent (your hand, quasi-statically, so the charge never picks up stray kinetic energy). Suppose the field does work $W_{\text{field}}$ during a move. Energy conservation then fixes the change in the stored account:

\Delta U = U_{\text{final}} - U_{\text{initial}} = -\,W_{\text{field}} \qquad (1)

where: $\Delta U$ — the change in the system's potential energy (joules); $W_{\text{field}}$ — the work done by the electrostatic force during the move. The minus sign is the entire content: when the field works for you (unlike charges pulled together, $W_{\text{field}} > 0$ ), the account drains ( $\Delta U < 0$ ); when you fight the field, the account fills. For a charge carried gently between rest and rest, the external agent's work is exactly the deposit: $W_{\text{ext}} = \Delta U$ . Field and agent always sit on opposite sides of the ledger, and exam questions switch which one they ask about mid-problem precisely to catch students who memorised one equation without the sign logic.

The pair formula

Now price a specific job: carry charge $q_2$ in from very far away and park it a distance $r$ from a pinned charge $q_1$ . The force you fight is Coulomb's, $F = kq_1q_2/r'^2$ at running separation $r'$ — a force that changes along the trip, so this is a variable-force work problem, the kind Work by a Variable Force & F–x Graphs taught you to read as an area under the F–r curve. Adding up those area strips all the way in from infinity gives a beautifully compact total:

U = \frac{kq_1q_2}{r} \qquad (2)

where: $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ (board-friendly; use $8.99\times10^9$ when the options are tight); $q_1, q_2$ — the charges with their signs; $r$ — their separation. Read the structure: the product $q_1q_2$ carries the sign of the whole account, and the single power of $r$ — not the $r^2$ of the force — is the fingerprint of energy, because work accumulates force over distance and that accumulation trades one power of $r$ . At $2.0\times10^{-10}\ \mathrm{m}$ — an atomic-scale gap — two protons hold $U = 1.15\times10^{-18}$ J. Tiny in joules, but convert to the natural currency of atoms (about $7.2$ electron volts — a unit the next module makes official) and it is exactly the energy scale of chemical bonds. The formula knows chemistry.

Derivation APricing the approach of two charges

We want equation (2): the work needed to bring $q_2$ from infinity to a distance $r$ from a pinned $q_1$ , done quasi-statically. The honest tool is the area under the force–distance graph, because Coulomb's force strengthens as the gap closes — no single value of force represents the whole trip. We will read the area conceptually, check the answer against every limit we know, and let the calculus-free logic carry the result. Watch the sign logic ride along automatically in the product $q_1q_2$ .

Step 1. Set the zero. At infinite separation the charges do not interact, so we declare $U = 0$ there. Whatever work the trip costs is the potential energy at the destination.

Step 2. Identify the force to fight: $F = kq_1q_2/r'^2$ at separation $r'$ , along the line joining the charges. For like charges this opposes the approach the whole way; for unlike charges it assists.

Step 3. Read the work as area. On a graph of $F$ against $r'$ , the work done between $r' = \infty$ and $r' = r$ is the area under the curve — the same move as the spring-energy area in mechanics, but with a $1/r'^2$ curve instead of a straight line. Summing the strips of that curve from infinity down to $r$ gives a total that behaves as one power of $r'$ better than the force:

W_{\text{ext}} = \frac{kq_1q_2}{r}

Why one power better: each strip contributes force times a small step of distance; piling up $1/r'^2$ strips produces a running total that shrinks like $1/r'$ — exactly as the area under the spring's straight line $kx$ piles up to the quadratic $\tfrac12kx^2$ . Boards that allow calculus write this as an integral; the result is the same.

Step 4. Bank it: $U(r) = kq_1q_2/r$ , with the sign of $q_1q_2$ deciding whether the account is a credit (like charges — you paid) or a debt (unlike — the field paid you). Why the path never mattered: Coulomb's force is conservative, so any detour's sideways excursions cost and refund equal amounts; only the start and end separations survive. That is what lets $U$ be a function of $r$ alone. *Verified: limits check — $U \to 0$ as $r \to \infty$ ✓ (matches the declared zero); $U$ grows without bound as $r \to 0$ for like charges ✓ (infinitely hard to merge them); differentiating the $1/r$ curve steepness recovers a $1/r^2$ force ✓; for two protons at $2.0\times10^{-10}$ m, $U = (9.0\times10^9)(1.6\times10^{-19})^2/(2.0\times10^{-10}) = 1.15\times10^{-18}$ J, the number used in the text.*

Two consequences are worth owning before any example. First, released charges cash the account: pin two protons at $2.0\times10^{-10}$ m, release both, and momentum conservation splits the refunded $1.15\times10^{-18}$ J equally — each departs at $v = \sqrt{U/m_p} = 2.6\times10^4$ m/s. Second, the account is quadratic in charge: double both charges and $U$ quadruples, because the product $q_1q_2$ picks up both factors. Energy never scales as innocently as the fields did.

Many charges: bill every pair once

Real problems rarely stop at two charges, and here the energy method wins outright. Assemble three charges and count the bills: the first arrives free; the second pays its interaction with the first; the third pays two bills, one against each resident. Total: three bills, one per pair:

U_{\text{total}} = k\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right) \qquad (3)

where: $r_{12}, r_{13}, r_{23}$ — the three pair separations, each pair appearing exactly once, with its own sign. For $n$ charges the count is $n(n-1)/2$ pairs — three charges make 3, four make 6, five make 10. The formula is pure arithmetic: no vectors, no angles, no resolving — which is why a four-charge energy question fits in three marks while the matching force question needs a full page.

Go deeper · Derivation B · 1 minWhy three charges make exactly three bills

Derivation BWhy three charges make exactly three bills

We want the pair-sum rule (3) and, just as importantly, the two ways students break it: double-counting pairs, and inventing bills that do not exist. The strategy is to run the assembly explicitly for three charges and audit the ledger line by line. Watch that the answer cannot depend on the order of arrival — that is the conservative force at work again, and it is also the fastest self-check you own in the exam hall.

Step 1. Bring in $q_1$ : empty space, no field, no work. Ledger: nothing.

Step 2. Bring in $q_2$ : it fights (or rides) only $q_1$ 's field. Ledger gains one line: $kq_1q_2/r_{12}$ .

Step 3. Bring in $q_3$ : it now moves through the combined field of both residents, and because work against a sum of forces splits into a sum of works, it pays two separate bills: $kq_1q_3/r_{13} + kq_2q_3/r_{23}$ . Why the residents pay nothing during this step: they are pinned. Only the mover's displacement generates work. The interaction $q_1$ – $q_2$ was already billed in Step 2 and is not billed again.

Step 4. Close the ledger: three lines, one per unordered pair, giving equation (3). Rerun the assembly in any other order — bring $q_3$ first, or $q_2$ last — and the individual steps change but the final three lines are identical, because each line depends only on the final separation of its pair. Why there is no factor of one-half in (3): the half appears only in the alternative "census" bookkeeping used at university level, where every pair is deliberately counted from both ends and then halved. In the pair-sum, each pair is written once. If you list pairs with $i < j$ , never divide by two. *Verified: for the worked example below — charges $+1.0\ \mu$ C, $+2.0\ \mu$ C, $-3.0\ \mu$ C on an equilateral triangle of side 0.20 m — the ledger reads $k(2 - 3 - 6)\times10^{-12}/0.20 = -0.315$ J by both assembly orders tried; recomputed numerically.*

Limiting cases

case	result	meaning
$r \to \infty$	$U \to 0$	the agreed zero: non-interacting charges hold nothing
like pair, $r$ halved	$U$ doubles	the $1/r$ account grows as the gap closes — walls steepen
unlike pair	$U < 0$ always	bound system: separating it costs $\lvert U\rvert$ of external work
both charges doubled	$U \to 4U$	quadratic in charge — energy never superposes like fields do
one charge is zero	its pairs vanish	no interaction, no bill: only genuine pairs are billed
$n$ charges	$n(n-1)/2$ pair terms	three charges → 3 bills, four → 6; count before you compute

Common trapA negative potential energy means the system has negative energy stored — something must be wrong

Nothing is wrong; the account is simply in debt, and the debt is physical. $U$ is measured from an agreed zero (charges dispersed to infinity), so a negative value says the assembled arrangement holds less energy than the dispersed one — the field paid out during assembly, and an external agent must pay $\lvert U \rvert$ to take the system apart. Negative $U$ is the signature of binding: hydrogen's electron–proton pair, salt's Na⁺Cl⁻, every stable structure in chemistry. The related trap is treating $U$ as belonging to one charge ("the energy of $q_3$ "): potential energy belongs to the configuration, and only when one charge moves while the rest stay pinned may you loosely speak of "its" energy, meaning the pairs that involve it.

Worked examples

Board level — the triangle ledger (CBSE/ISC pattern)

Charges $q_1 = +1.0\ \mu\mathrm{C}$ , $q_2 = +2.0\ \mu\mathrm{C}$ and $q_3 = -3.0\ \mu\mathrm{C}$ sit at the corners of an equilateral triangle of side $a = 20\ \mathrm{cm}$ . (a) Find the electrostatic potential energy of the system. (b) How much work must an external agent do to remove $q_3$ to infinity? Take $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ .

Step 1. Count the pairs before touching the calculator: three charges, so exactly three bills, all at the same separation $a = 0.20$ m.

Step 2. Write the ledger with signs riding on the products: $q_1q_2 = +2.0\times10^{-12}\ \mathrm{C^2}$ , $q_1q_3 = -3.0\times10^{-12}\ \mathrm{C^2}$ , $q_2q_3 = -6.0\times10^{-12}\ \mathrm{C^2}$ . Sum: $-7.0\times10^{-12}\ \mathrm{C^2}$ .

Step 3. One multiplication finishes (a): $U = \dfrac{k}{a}\sum q_iq_j = \dfrac{9.0\times10^9}{0.20}\times(-7.0\times10^{-12}) = \boxed{-0.315\ \mathrm{J}}$ . Negative — the two attractions outvote the one repulsion; the trio is bound.

Step 4. For (b), removal is a change of configuration, so compute the final account. With $q_3$ gone, only the $q_1$ – $q_2$ pair remains: $U_{\text{final}} = \dfrac{(9.0\times10^9)(2.0\times10^{-12})}{0.20} = +0.090$ J.

Step 5. The agent pays the difference: $W_{\text{ext}} = U_{\text{final}} - U_{\text{initial}} = 0.090 - (-0.315) = \boxed{+0.405\ \mathrm{J}}$ . Positive, as it must be — you are pulling a negative charge away from the positives that bind it. The marks-scheme sentence: "work by external agent equals the increase in the system's potential energy (start and end at rest)."

Competitive level — release and run (JEE-Main/NEET pattern)

A small ball of mass $10\ \mathrm{g}$ carrying charge $+2.0\ \mu\mathrm{C}$ is held $10\ \mathrm{cm}$ from a fixed charge $+8.0\ \mu\mathrm{C}$ and released from rest. Find its speed when it is $40\ \mathrm{cm}$ from the fixed charge. ( $k = 9.0\times10^9$ in SI units; ignore gravity.)

Step 1. Route choice — this is an energy problem, not a force problem: the force changes along the way, so kinematics with constant acceleration is illegal. Energy conservation: $KE_{\text{gained}} = U_{\text{initial}} - U_{\text{final}}$ .

Step 2. Account at start: $U_i = \dfrac{kq_1q_2}{r_i} = \dfrac{(9.0\times10^9)(2.0\times10^{-6})(8.0\times10^{-6})}{0.10} = 1.44$ J.

Step 3. Account at 40 cm: $U_f = \dfrac{0.144}{0.40} \times 1 = 0.36$ J — quick ratio: four times the distance, one quarter the energy, $1.44/4 = 0.36$ J.

Step 4. Refund becomes motion: $\tfrac12mv^2 = 1.44 - 0.36 = 1.08$ J, so $v = \sqrt{\dfrac{2(1.08)}{0.010}} = \sqrt{216} = \boxed{14.7\ \mathrm{m/s}}$ .

Step 5. Option-check bonus: the speed at infinity would be $v_\infty = \sqrt{2(1.44)/0.010} = 17.0$ m/s — any option above that for any finite distance is impossible physics, and spotting that kills two options in seconds. Timing note: 90 seconds. The only decisions are (i) energy not kinematics, (ii) the $1/r$ ratio shortcut in Step 3. If you catch yourself computing acceleration, restart — that route needs calculus you should not spend.

Exam traps

Work by the field vs work by you. The single biggest mark-loser. $W_{\text{field}} = -\Delta U$ but $W_{\text{ext}} = +\Delta U$ (rest to rest). Read which one the question asks, write the ledger equation first, then compute. If your "work to separate a bound pair" comes out negative, you answered the field's question, not the examiner's.

Dropping a sign inside the pair sum. The products $q_iq_j$ carry signs; the separations $r_{ij}$ never do. Write the sign of each product before multiplying by $k/r$ — sign errors caught at the ledger stage cost seconds, caught after the arithmetic they cost the question.

Double-counting pairs. The pair $q_1$ – $q_2$ and the pair $q_2$ – $q_1$ are the same bill. List pairs systematically ( $12, 13, 23$ ) and stop at $n(n-1)/2$ terms. There is no factor of $\tfrac12$ in the pair-sum — that belongs to a university-level rewrite you will meet later.

Using $r^2$ . Force goes as $1/r^2$ ; energy goes as $1/r$ . Under time pressure the muscle memory of Coulomb's law substitutes the square — the limiting-cases table is your antidote: energy halves when distance doubles.

Treating $U$ as a vector. No components, no angles, ever. If your solution to an energy question contains $\cos\theta$ , you have wandered into a force calculation (the one exception: the geometry may set the separations $r_{ij}$ , as the triangle's side did).

Constants and units. All boards accept $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ here; JEE options are usually built from it. Convert cm to m before anything else — a stray $10^{-2}$ inside a $1/r$ is a factor-of-100 disaster. Energies land in joules; the electron-volt belongs to the next module.

Origins · 1 minGaspard-Gustave de Coriolis and how this was discovered

Gaspard-Gustave de Coriolis

1792–1843 · École Polytechnique, Paris

The word this chapter runs on — work — had to be invented, and Coriolis is the physicist who nailed it down. In his 1829 book on machine mechanics he defined travail as force accumulated through distance, argued it was the honest currency for comparing engines, water wheels and human labour, and fixed the factor of one-half in the "living force" $\tfrac12mv^2$ so that work and kinetic energy would balance exactly. That bookkeeping — pay work in, bank it, cash it out as motion — is precisely the machinery this article runs on charges: Coulomb's force replaces the crane, but the ledger is Coriolis's. He is more famous for the deflection force that bears his name (the one from the rotation chapter), yet his quieter gift is the one every energy equation you will ever write inherits.

History

1785

— Coulomb measures the inverse-square force between charges: the force this article's ledger prices.

1829

— Coriolis publishes the modern definition of work and fixes the half in kinetic energy — the currency of every energy account.

1847

— Helmholtz states the conservation of energy, putting electrical, mechanical and thermal accounts in one ledger.

1853

— William Thomson (Lord Kelvin) computes the energy of a charged system, founding electrostatic energetics.

1932

— Cockcroft and Walton split the lithium nucleus with electrostatically accelerated protons — assembly energy run in reverse, and the first experimental check of E = mc² into the bargain.

Going further

Texts. NCERT Physics Part 1 (Class 12), Ch. 2 §2.7–2.8 — the CBSE/NEET canonical treatment of the pair energy and the system sum · H.C. Verma, Concepts of Physics 2, Ch. 29 · Halliday, Resnick & Walker Ch. 24 for the AP-2/A-Level voice · ISC Nootan Physics 12, Electrostatics unit.

Provenance. All formulas are standard (NCERT Ch. 2, HRW Ch. 24). A verifier should recompute: proton pair $U = (9.0\times10^9)(1.6\times10^{-19})^2/(2.0\times10^{-10}) = 1.15\times10^{-18}$ J and release speed $v = \sqrt{U/m_p} = 2.6\times10^4$ m/s each; triangle ledger $k(2-3-6)\times10^{-12}/0.20 = -0.315$ J and removal work $0.090 + 0.315 = +0.405$ J; release problem $U_i = 1.44$ J, $U_f = 0.36$ J, $v = \sqrt{2\times1.08/0.010} = 14.7$ m/s, $v_\infty = 17.0$ m/s.

Prerequisites. Coulomb's Law · Potential Energy: Gravitational & Elastic · Conservative & Non-Conservative Forces. Leads to. Electric Potential & Potential Difference (the per-charge version of this account) · ARCHETYPE: Assembling & Moving Charges (the gym built on this module) · Work and Energy of Charge Configurations (UG deep dive).

Is potential energy stored in each charge, or where exactly does it live?

In the configuration — the arrangement as a whole. The cleanest evidence: one charge alone has no interaction energy at all; the account opens only when a second charge gives the first something to interact with. In practice, when a problem pins all charges except one, it is harmless shorthand to call the mover's pair terms "its" energy, because those are the only accounts that change. But keep the honest picture for conceptual questions: $U$ belongs to the system, and deeper physics (which you will meet at university) relocates it into the field filling the space between the charges.

Why do we set U = 0 at infinity and not somewhere convenient like r = 1 m?

Any zero is legal — only changes in $U$ are ever measured, exactly as any floor can be $h=0$ in mechanics. Infinity wins because it is the one choice with physical meaning built in: charges infinitely far apart genuinely do not interact, so "zero stored energy" describes a real state of affairs rather than a bookkeeping whim. It also makes the formula clean: $U = kq_1q_2/r$ with no added constant. Watch for the rare problem that redefines the zero (some graphs do); the physics never changes, but every quoted value shifts by the same constant.

If the field does positive work, why does the potential energy go down? I keep flipping this.

Anchor it to gravity, which you never flip: when a ball falls, gravity does positive work and the gravitational account is spent — the stored energy became motion. Same here: positive $W_{\text{field}}$ means the field converted stored energy into kinetic energy, so the store must shrink: $\Delta U = -W_{\text{field}}$ . The confusion usually comes from silently switching to the external agent mid-thought — your work fills the account, the field's work drains it. Before any calculation, write down whose work the question wants, and the sign takes care of itself.

For three charges, why don't I divide the sum by 2? My cousin's university notes have a half in the formula.

Both formulas are right — they count differently. The pair-sum you use lists each pair once ( $12, 13, 23$ ): no half. University notes often use a "census" version that asks every charge for its energy against all the others and then halves the total, because that census counts each pair twice (once from each end). Same number, two bookkeepings. At HS level, stick to the pair-sum and the rule "each pair exactly once"; if you ever see the $\tfrac12$ version, know that its potential term already includes both partners' viewpoints.

The triangle example gave U = −0.315 J. Does the system explode or collapse if I unpin the charges?

Not "explode" — negative total $U$ means bound, so the trio cannot fly apart to infinity on its own (that would need the account to reach zero, and it is 0.315 J short). Released, the charges accelerate — unlike pairs pulling together — trading potential for kinetic energy, and in the idealised point-charge world the attractions deepen without limit as separations shrink; real ions are rescued by quantum effects. The exam-relevant reading of $U < 0$ : someone must pay 0.315 J (plus any kinetic energy present) to disperse this system.

When a released charge speeds up, the fixed charge also feels a force. Doesn't it get energy too?

It feels a force but does not move, and work is force through distance — no displacement, no work, no energy delivered (whatever pins it absorbs the force without yielding). That is why the released ball in the worked example collected the entire 1.08 J refund. Free both charges and the story changes: the refund splits by momentum conservation, the lighter partner taking the larger share — a favourite twist in the archetype module Assembling & Moving Charges, later in this chapter.

How is this different from gravitational potential energy — the formulas look identical?

The structure is identical — both are conservative inverse-square forces, so both give $1/r$ energy accounts, and your intuition transfers wholesale. Two differences matter. Sign: gravity only attracts, so gravitational $U$ is always negative; charges repel or attract, so electrostatic $U$ runs both signs — walls and wells. Strength: for two protons, electric repulsion outguns gravitational attraction by about $10^{36}$ , which is why electrostatics runs chemistry while gravity only runs astronomy. The $mgh$ of mechanics is the same account linearised over small heights.