ARCHETYPE: Assembling & Moving Charges

In 1929 a young physicist named Robert Van de Graaff built a machine whose entire job was to pay an energy bill: a rubber belt carried charge up to a metal dome, and every coulomb delivered cost more work than the one before, because the dome's own growing charge pushed back harder each trip. His generator was, quite literally, an exercise in the work needed to assemble charge — and it bought physics its first million-volt accelerators. Examiners have been running smaller versions of that bill ever since. One family of questions asks for the work to assemble a set of point charges — two charges, a triangle, a square with a diagonal twist. Another asks what happens when a charge is released to move — the kinetic energy gained crossing a potential difference, the speed of an electron leaving an accelerating gun, the closest approach of an alpha particle to a nucleus. They look like different chapters. They are one ledger, read in two directions: build the configuration and you deposit energy pair by pair; let it fly apart and the field pays it back as kinetic energy. This archetype module drills that ledger until it is mechanical. The master pattern has three moves — count the pairs, sign each pair, sum $kq_1q_2/r$ — plus one conversion rate for motion, $W = q\Delta V = \Delta KE$ . Learn the two canonical setups well and an entire cross-board question family (CBSE 3-markers, JEE squares, NEET electron guns, A-Level and AP potential-energy items) becomes fill-in-the-blanks.

Key takeaways

Two canonical setups, one ledger: assembly (charges marched in from infinity, work deposited as $U$ ) and motion (a charge crossing a potential difference, $U$ converted to kinetic energy).
Assembly master result: $W_{\text{ext}} = U = k\sum_{\text{pairs}}\dfrac{q_iq_j}{r_{ij}}$ — every pair billed exactly once, sign per pair, first charge free.
Pair count: 2 charges → 1 pair, 3 → 3 pairs, 4 → 6 pairs. Like pairs cost positive work; unlike pairs pay you back (negative).
Motion master result: $q\Delta V = \Delta KE$ , so an accelerated charge leaves with $v = \sqrt{2qV/m}$ — the electron-gun equation.
Same $\Delta V$ gives the same kinetic energy to an electron and a proton, but speeds differ by $\sqrt{m_p/m_e} \approx 43$ .
Key idea: energy questions about charges are bookkeeping — work to assemble = sum over pairs of $kq_1q_2/r$ , and energy released in motion = charge × potential difference crossed.

The core insight

The work to assemble any set of point charges equals the sum of $kq_iq_j/r_{ij}$ over distinct pairs (each pair once, signs included), and once assembled that stored energy converts to kinetic energy at the fixed rate $W = q\Delta V$ — the two master equations behind every assembly-and-motion exam problem.

Where this lives today

Every silicon chip in your devices was partly manufactured by this module's equations. Ion implantation — the step that dopes transistors — accelerates phosphorus or boron ions through potentials of tens to hundreds of kilovolts and fires them into wafers to a depth set by $qV$ ; the machine's beamline is an industrial electron-gun problem, solved millions of times per second in fabs worldwide. Electron microscopes run the same equation at 100–300 kV to make electrons whose wavelengths resolve individual atoms. The assembly half runs at even larger scale, inside computers. Molecular-dynamics simulations of proteins and materials spend most of their arithmetic summing pair energies $kq_iq_j/r_{ij}$ over hundreds of thousands of partial charges, every femtosecond time-step — drug-binding energies and battery-electrolyte designs are, at bottom, configuration-energy differences computed with the exact ledger you drilled here, signs and all.

The setup

Geometry first, because in an archetype the geometry is the physics. The assembly setup: a bare region of space, a stock of point charges parked at infinity, and a porter (you, or "an external agent") carrying them in one at a time to marked positions. The motion setup: two locations at different potentials — parallel plates, two points near a fixed charge, the electrodes of a gun — and a charge free to travel between them.

Figure

Assembly from infinity — the first charge rides free, each newcomer pays every resident

Why from infinity? Because that is where the meter reads zero: charges so far apart that they no longer feel each other. And why does only the final position matter, not the path the porter walked? Because the electric force is conservative — Electric Potential & Potential Difference established that the work between two points is path-independent. Those two conventions make the energy of a configuration a single well-defined number, and they make your answer independent of the order of assembly too: bring in the charges in any sequence, the same set of pairs gets billed the same amounts. The motion setup is the ledger in reverse. A charge $q$ moving from a point at potential $V_A$ to a point at potential $V_B$ has the field do work $q(V_A - V_B)$ on it — deposit becomes withdrawal. If nothing else touches the charge, that work has exactly one place to go: kinetic energy.

The master solution pattern

Assembly. Three moves, always the same. Count the pairs: $n$ charges make $\tfrac{n(n-1)}{2}$ pairs — list them so none is missed or doubled. Sign each pair: like charges → positive (you pushed uphill), unlike → negative (the field helped). Sum:

W_{\text{ext}} = U = k\sum_{\text{pairs}}\frac{q_i q_j}{r_{ij}} \qquad (1)

where: $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ (board-friendly; $8.99\times10^9$ when precision matters); $r_{ij}$ — the final separation of that pair, in metres; the sum runs over each distinct pair once. The pair energy $kq_1q_2/r$ itself belongs to Electrostatic Potential Energy — this module never re-derives it, it deploys it at scale. Signs ride along with the charges automatically: a configuration with $U < 0$ is bound (you would have to pay to disperse it), one with $U > 0$ is spring-loaded and will fly apart if released.

number of charges	pairs to bill
2	1
3 (triangle)	3
4 (square)	6 — four sides + two diagonals
5	10

Motion. One conversion rate. A charge $q$ crossing a potential drop $\Delta V$ gains

\Delta KE = q\,\Delta V \qquad \Rightarrow \qquad v = \sqrt{\frac{2qV}{m}} \;\; \text{(from rest)} \qquad (2)

where: $\Delta V$ — the potential difference actually crossed (volts); $m$ — the particle's mass. Read it physically: the energy handed over depends only on charge and voltage — which is exactly why the electron-volt from Electric Potential & Potential Difference is such a convenient currency — while the speed also cares about the mass under the square root. Potentials of the fixed charges themselves come from Potential of Point Charges & Charge Systems: scalar sums, no components.

Derivation AWhy the assembly bill is a sum over pairs

We derive the master formula (1) directly from the marching-in procedure, and establish the two facts examiners test hardest: the first charge is free, and each pair is billed exactly once. The strategy is to track what each newcomer pays as it arrives, using only $W = q\Delta V$ and the point-charge potential from Potential of Point Charges & Charge Systems. Watch how the growing bill organises itself into pairs without any extra bookkeeping.

Step 1. Charge $q_1$ arrives first. Space is empty — no field, no potential, nothing to fight. $W_1 = 0$ . The first charge always rides free, however large it is.

Step 2. Charge $q_2$ arrives at distance $r_{12}$ from the resident. The resident maintains potential $V = kq_1/r_{12}$ there, so the porter pays $W_2 = q_2 V = kq_1q_2/r_{12}$ — the pair energy of (1,2), and our first ledger line.

Step 3. Charge $q_3$ arrives. Potentials add as scalars, so the two residents jointly maintain $V = kq_1/r_{13} + kq_2/r_{23}$ at the destination, and the porter pays

W_3 = q_3 V = \frac{kq_1q_3}{r_{13}} + \frac{kq_2q_3}{r_{23}}

— exactly the two new pairs (1,3) and (2,3), billed at their final separations. Why residents pay nothing during an arrival: they are held at fixed positions while the newcomer moves, and a stationary charge has zero displacement — no work flows through it.

Step 4. The pattern is now visible: each newcomer pays one toll per resident, and each toll is one pair term. When all charges are in, every pair $(i,j)$ has been billed exactly once — at the moment its later member arrived. Total:

W_{\text{ext}} = \frac{kq_1q_2}{r_{12}} + \frac{kq_1q_3}{r_{13}} + \frac{kq_2q_3}{r_{23}} + \cdots = k\sum_{\text{pairs}}\frac{q_iq_j}{r_{ij}}

Why the assembly order cannot matter: each term depends only on that pair's final separation — path independence again — so any arrival sequence produces the identical set of pair terms. Useful in reverse too: if a question's given order looks awkward, re-order it in your head. *Verified: three equal charges $q = 2.0$ nC at mutual distance $5.0$ cm — Step 2 and Step 3 give $3 \times k q^2/a = 3 \times 7.2\times10^{-7} = 2.16\ \mu\mathrm{J}$ , matching worked example 1 below.*

Go deeper · Derivation B · 1 minFrom potential difference to speed — the gun equation

Derivation BFrom potential difference to speed — the gun equation

We derive the electron-gun result $v = \sqrt{2qV/m}$ and the electron–proton comparison that decorates half the MCQs on this topic. The strategy: energy conservation between the two electrodes, with the field's work computed as $q\Delta V$ — no forces, no kinematics, no field maps needed, which is the entire selling point of the energy method. Watch where the mass enters (speed) and where it does not (energy).

Step 1. A particle of charge $q$ starts at rest at the low-speed electrode and crosses an accelerating potential difference $V$ . Work done by the field: $W = qV$ . This is the definition of potential difference from Electric Potential & Potential Difference cashing itself in.

Step 2. Energy conservation with no other forces acting:

qV = \tfrac12 mv^2

Why no other terms appear: gravity is laughably weak here (for an electron crossing centimetres, smaller than the electric term by more than ten orders of magnitude), and the gun operates in vacuum — no friction, no collisions.

Step 3. Solve for the exit speed:

v = \sqrt{\frac{2qV}{m}}

Step 4. Compare species at the same $V$ . Kinetic energies are identical — $qV$ each, mass-blind. Speeds are not:

\frac{v_e}{v_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1836} \approx 43

Why examiners love this: "same energy, different speed" catches everyone who equates energy with speed. The electron leaves 43 times faster, carrying the same joules. *Verified: electron through $V = 500$ V: $KE = 8.0\times10^{-17}$ J $= 500$ eV, $v = \sqrt{2\times1.6\times10^{-19}\times500/9.1\times10^{-31}} = 1.33\times10^{7}$ m/s; proton: $3.1\times10^{5}$ m/s; ratio $42.8 \approx \sqrt{1836}$ ✓.*

Figure

The electron gun — a potential difference converted to kinetic energy

Limiting cases

case	result	meaning
single charge brought in	$W = 0$	the first charge is free — no field to fight
unlike pair	$U = -kq_1q_2/r < 0$	bound: energy must be supplied to separate it
all like charges	$U > 0$	spring-loaded: released, it converts all of $U$ to KE
every charge doubled	$U \to 4U$	pair bills scale as charge × charge
every distance doubled	$U \to U/2$	all pair terms carry $1/r$
electron vs proton, same $V$	same $KE$ , speeds differ by $\sqrt{1836} \approx 43$	energy is mass-blind; speed is not
head-on approach to a fixed charge	stops where $\tfrac12mv_0^2 = kq_1q_2/d$	closest approach: all KE converted back to pair energy

Common trap“Each of the n charges costs qV of work, so multiply by n”

This double-counts every pair. If you go around a finished configuration and add up $q_i \times V(\text{at } i)$ for all charges, each pair gets reported twice — once from each end — so that census gives exactly double the assembly work. The honest versions: either bill pairs once (equation 1, the safe route), or take half the census, $U = \tfrac12\sum q_iV_i$ . On a three-charge triangle the census gives $3 \times qV_{\text{corner}} = 4.32\ \mu\mathrm{J}$ while the true energy is $2.16\ \mu\mathrm{J}$ — options love to offer both. When in doubt, retreat to listing pairs; the pair list cannot express the mistake.

Worked examples

Board level — assembling a triangle (CBSE/ISC pattern, 3 marks)

Three point charges of $+2.0\ \mathrm{nC}$ each are placed at the corners of an equilateral triangle of side $5.0\ \mathrm{cm}$ . (a) How much work is needed to assemble this arrangement from infinity? (b) How much of that work is spent bringing in the third charge alone? Take $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ .

Step 1. Count and sign the pairs: 3 charges → 3 pairs, all like → all positive.

Step 2. One pair's bill: $\dfrac{kq^2}{a} = \dfrac{(9.0\times10^9)(2.0\times10^{-9})^2}{0.050} = 7.2\times10^{-7}\ \mathrm{J}$ .

Step 3. Total: $U = 3 \times 7.2\times10^{-7} = \boxed{2.16\ \mu\mathrm{J}}$ .

Step 4. The third charge paid one toll per resident — two tolls: $W_3 = 2 \times 7.2\times10^{-7} = \boxed{1.44\ \mu\mathrm{J}}$ . (Equivalently: the two residents maintain $V = 2kq/a = 720$ V at the empty corner, and $W_3 = qV = 2.0\times10^{-9} \times 720 = 1.44\ \mu\mathrm{J}$ ✓.) The first two charges account for the remaining $0.72\ \mu\mathrm{J}$ — and the first alone cost nothing.

Board level — the electron gun (NEET/ISC pattern)

An electron, initially at rest, is accelerated through a potential difference of $500\ \mathrm{V}$ . Find its final kinetic energy in joules and electron-volts, and its final speed. ( $e = 1.6\times10^{-19}\ \mathrm{C}$ , $m_e = 9.1\times10^{-31}\ \mathrm{kg}$ .)

Step 1. Energy from the conversion rate: $KE = eV = (1.6\times10^{-19})(500) = \boxed{8.0\times10^{-17}\ \mathrm{J}}$ .

Step 2. In the natural currency: an electron crossing 500 volts gains exactly $\boxed{500\ \mathrm{eV}}$ — the electron-volt is defined to make this step mental arithmetic.

Step 3. Speed: $v = \sqrt{\dfrac{2KE}{m_e}} = \sqrt{\dfrac{2(8.0\times10^{-17})}{9.1\times10^{-31}}} = \boxed{1.33\times10^{7}\ \mathrm{m/s}}$ — about 4% of light speed, from a battery-sized voltage. (Fast enough that beyond a few kV, examiners quietly switch to relativity-safe wording.)

Competitive level — the square with a twist (JEE-Main pattern, ~3 minutes)

Four charges, each $+1.0\ \mu\mathrm{C}$ , sit at the corners of a square of side $10\ \mathrm{cm}$ . (a) Find the energy of the configuration. (b) One of the four charges is replaced by $-1.0\ \mu\mathrm{C}$ . Find the new configuration energy.

Step 1. Pair inventory for a square: 6 pairs — 4 sides at separation $a$ , 2 diagonals at $a\sqrt2$ . Base bill: $\dfrac{kq^2}{a} = \dfrac{(9.0\times10^9)(10^{-6})^2}{0.10} = 0.090\ \mathrm{J}$ .

Step 2. All-positive square: $U = \dfrac{kq^2}{a}\left(4 + \dfrac{2}{\sqrt2}\right) = 0.090 \times (4 + 1.414) = \boxed{0.49\ \mathrm{J}}$ .

Step 3. Now flip one corner to $-q$ and re-sign the pairs, don't recompute them. The negative charge has two side-neighbours and one diagonal partner: sides become $+,+,-,-$ (sum zero) and diagonals become $+\tfrac{1}{\sqrt2}, -\tfrac{1}{\sqrt2}$ (sum zero).

Step 4. $U = \boxed{0}$ — exactly. The four attractions and four repulsions... six pair terms cancel to the last joule. A configuration with zero net assembly cost: it took no net work to build, though individual charges were pushed and pulled plenty along the way. Timing note: part (b) untimed is a 6-term recomputation; with the re-sign trick it is 30 seconds. JEE plants "0" among plausible non-zero options precisely to reward the pair-ledger view. Budget 3 minutes total.

Competitive level — closest approach (NEET/JEE stretch, ~3 minutes)

An alpha particle (charge $+2e$ ) with kinetic energy $5.0\ \mathrm{MeV}$ is fired head-on at a gold nucleus (charge $+79e$ ), which stays fixed. Find the distance of closest approach. ( $e = 1.6\times10^{-19}\ \mathrm{C}$ , $k = 9.0\times10^9\ \mathrm{N\,m^2/C^2}$ .)

Step 1. Recognise the ledger in reverse: the alpha spends kinetic energy buying pair energy with the nucleus. At closest approach (head-on), it is momentarily at rest — all KE converted.

Step 2. Energy balance: $KE = \dfrac{k(2e)(79e)}{d}$ , so $d = \dfrac{158\,ke^2}{KE}$ .

Step 3. Convert the energy: $KE = 5.0\times10^6 \times 1.6\times10^{-19} = 8.0\times10^{-13}\ \mathrm{J}$ .

Step 4. $d = \dfrac{(9.0\times10^9)(158)(1.6\times10^{-19})^2}{8.0\times10^{-13}} = \boxed{4.6\times10^{-14}\ \mathrm{m}}$ — about 46 femtometres. This is Rutherford's own 1911 arithmetic: the alpha never touched the nucleus, and the fact that scattering stayed "Coulomb-clean" down to tens of femtometres set the first upper bound on nuclear size. Timing note: the only decision is spotting "closest approach = all KE becomes pair energy". Convert MeV to joules first — mixing eV and joules mid-formula is where this question kills. Budget 3 minutes.

Exam traps

Double counting — the ½ ambush. Summing $qV$ over all charges counts every pair twice; the census route needs the factor $\tfrac12$ . The pair list (equation 1) is immune. If two candidate answers differ by a factor of 2, this trap is on the paper.

A missed diagonal. Squares have six pairs, not four. Rectangles, rhombuses and "three charges on a line" all hide a long pair. Write the count $\tfrac{n(n-1)}{2}$ first and tick pairs off against it.

Sign amnesia. Every unlike pair enters with a minus sign, and the final $U$ can legitimately be negative (bound) or zero (example 3). Options with only positive values are betting you'll drop signs.

Work by whom? $W_{\text{ext}} = +\Delta U$ when the porter moves charges quasi-statically; the field's work is $-\Delta U$ . "Work done in assembling" means the external agent's. A sign flip here inverts the whole answer — the sign discipline drill is its own technique module coming in the E2 gym.

Centimetres in $r$ . Pair separations go into equation (1) in metres. With $a = 5.0$ cm, forgetting the conversion inflates every pair by 100 — answers in the "hundreds of microjoules" for nano-coulomb problems are the symptom.

eV ↔ joule slips. $qV$ with $q$ in coulombs gives joules; "an electron through 500 V" gives 500 eV by definition. Convert once, at a marked line, and never mix currencies inside one formula — MeV questions (example 4) are graded on exactly this.

Forgetting $\sqrt{\phantom{x}}$ or the 2 in the gun equation. $v = \sqrt{2qV/m}$ : dropping the 2 costs $\sqrt2$ , forgetting the square root gives absurd speeds. Sanity check: 500 V must land an electron near $10^7$ m/s, never near $10^{14}$ .

Constants per board. CBSE/ISC/NEET: $k = 9.0\times10^9$ , $e = 1.6\times10^{-19}$ C as given here. JEE papers state their constants — use exactly what is printed. A-Level (9702) data sheets give $\varepsilon_0$ ; remember $k = 1/4\pi\varepsilon_0$ . AP supplies the constant sheet — no memorisation credit, no excuse for a wrong $k$ .

Origins · 1 minRobert J. Van de Graaff and how this was discovered

Robert J. Van de Graaff

1901–1967 · Princeton & MIT

Van de Graaff was an Alabama-born engineer who turned this module's ledger into a building-sized instrument. His 1929 generator used a silk (later rubber) belt to ferry charge continuously onto a hollow metal dome — each increment of charge carried in against the potential of all the charge already delivered, the assembly bill paid mechanically, coulomb by coulomb. His 1931 machine reached 1.5 million volts; the giant twin-dome version built for MIT in 1933 could fling sparks metres long and became the front end of early nuclear physics, accelerating protons and ions through precisely the $qV \to \tfrac12mv^2$ conversion of this module's second derivation. School versions still make hair stand on end in every physics lab — and working tandem Van de Graaff accelerators still run today in labs and hospitals, quietly converting assembled charge into fast particles.

History

1785

— Coulomb measures the force between charges: the integrand of every assembly bill to come.

1911

— Rutherford's closest-approach arithmetic on alpha particles bounds the size of the nucleus (worked example 4's original run).

1929

— Van de Graaff demonstrates his first belt generator: assembly work industrialised.

1933

— The MIT twin-dome Van de Graaff reaches millions of volts and opens accelerator physics.

1950

s — Tandem Van de Graaff accelerators double the payoff by making ions cross the same potential twice.

Going further

Texts. NCERT Physics Class 12, Ch. 2 (§2.8, Potential Energy of a System of Charges) · HC Verma, Concepts of Physics Vol. 2, Ch. 29 · SL Arora, Ch. 2 solved problems (the square and triangle families).

Provenance. Archetype module built on the pair-sum machinery of the UG module "Work and Energy of Charge Configurations" (triangle example downshifted from there; all numbers recomputed). A verifier should recompute: pair bill $kq^2/a = 7.2\times10^{-7}$ J and triangle total $2.16\ \mu\mathrm{J}$ , third-charge share $1.44\ \mu\mathrm{J}$ (also as $q \times 720$ V); electron at 500 V → $8.0\times10^{-17}$ J, $1.33\times10^7$ m/s, proton ratio $\sqrt{1836} = 42.8$ ; square $0.090\times5.414 = 0.49$ J and the flipped-corner cancellation to 0; alpha-on-gold $d = 158ke^2/(8.0\times10^{-13}) = 4.6\times10^{-14}$ m.

Prerequisites. Electrostatic Potential Energy · Potential of Point Charges & Charge Systems · Kinetic Energy & the Work–Energy Theorem. Leads to. Sign Discipline in V, U and W (coming in the E2 gym) · Work and Energy of Charge Configurations (UG deep dive).

Why is the first charge free? It still had to be carried all that way.

Work is force times displacement, and for the first charge there is no electric force anywhere along the trip — space is empty of fields until it arrives. Carrying it "all that way" costs nothing electrically (we always ignore gravity and friction in these idealised problems). The bill starts with the second charge, which must be pushed through the first one's field. This is also why the answer scales as pairs: work only ever arises from interactions, and interactions need two participants.

Does the order in which I assemble the charges change the total work?

No — and this is worth trusting deeply, because it lets you re-order any awkward question. Each pair's term depends only on the two charges and their final separation, not on who arrived first: path independence (the conservative-field property from Electric Potential & Potential Difference) guarantees the toll for a pair is the same whether it was paid in one trip or split across two. Assemble a square corner-by-corner in any of the 24 possible orders; all give the same six pair terms.

If the assembled energy of the flipped square is zero, does that mean the charges don't interact?

They interact furiously — every pair still attracts or repels with full Coulomb force. Zero total energy only means the positive pair terms and negative pair terms happen to cancel in the sum: assembling it took exactly as much pushing as pulling. The configuration is not in equilibrium, and released, the charges would immediately accelerate. Energy is a single number describing the whole arrangement; forces are vectors acting on each charge — zero total energy implies nothing about individual forces.

When one of three assembled charges is released, how much energy does it take away?

Exactly the pair terms it participates in. In the triangle of worked example 1, the escaper owns two of the three pairs, so it departs with $2 \times 0.72 = 1.44\ \mu\mathrm{J}$ of kinetic energy at infinity, leaving the pinned pair holding its single term, $0.72\ \mu\mathrm{J}$ . The audit always balances: each pair's energy leaves with whichever member departs (or stays banked if both stay). What you must not do is divide the total $U$ equally among the charges — pairs, not charges, own the energy.

Is qΔV = ½mv² still fine for the megavolt machines Van de Graaff built?

The energy side, $\Delta KE = q\Delta V$ , is always exact — it is bookkeeping. What fails at high voltage is converting that energy to speed with $\tfrac12mv^2$ : an electron gains its rest-mass-equivalent energy (511 keV) after only half a million volts, and relativistic kinetic energy must take over; the classical formula would soon predict speeds beyond light. Protons, 1836 times heavier, stay classical up to hundreds of megavolts. For board and JEE problems at kilovolt scales, the classical form is safe — and examiners choose their voltages to keep it so.

For closest approach, why can I ignore the force and just use energy?

Because the question asks only where the particle stops, not when or with what deceleration profile. Energy conservation connects the start (all kinetic) to the turning point (all pair energy) without touching anything in between — the force varies wildly along the path, and the energy method never needs to know. If a question asks for the force at closest approach, compute $d$ by energy first, then apply Coulomb's Law once at that distance. Two tools, used in the right order.

Do I include the pairs among the fixed charges when a question asks for the work to bring in one more?

No. "Work to bring in charge X" bills only the new tolls — X against each resident. The residents' mutual pairs were paid earlier and don't change while X arrives (they sit at fixed separations). That is exactly worked example 1(b): the third charge pays two tolls, $1.44\ \mu\mathrm{J}$ , not the full $2.16\ \mu\mathrm{J}$ . Read assembly questions with the ledger in mind: which pairs are newly created by this move? Those, and only those, get billed.