Electric charge — properties, quantization, conservation: Worked examples

The scenario matrix

Every problem about charge is really one of these cells. The columns are the three properties; the rows are the "flavour" of input the problem hands you.

Cell	Flavour of input	Which property it tests	Example #
A	Single sign, count the bricks	Quantization ($n=	q
B	A quoted charge that might be illegal	Quantization (integer test)	2
C	Two like signs added	Additivity (same sign)	3
D	A positive and a negative together (cancellation)	Additivity (signed sum)	4
E	Zero / degenerate input (a neutral object, $q = 0$ )	All three at the boundary	5
F	Charge shared by touching identical bodies	Conservation + symmetry	6
G	Charge shared by unequal conductors + repeated contacts	Conservation, no symmetry	7
H	Limiting / macroscopic feel (why $e$ is invisible)	Quantization limit	8
I	Real-world word problem (current = charge per second)	Additivity in time	9
J	Exam twist (pair production bookkeeping)	Conservation with creation	10

We now hit every cell. Watch the sign travel through each one — that is the thread the whole topic hangs on.

Example 1 — Cell A: count the bricks

Worked example How many electrons are in a charge of

$- 3.2 μ C$ ? Forecast: guess the order of magnitude first. A coulomb is $\sim 6 \times 1 0^{18}$ electrons, and we have a millionth of that scale. So expect around $1 0^{13}$ . Hold that number.

Step 1 — Write the magnitude. $∣ q ∣ = 3.2 \times 1 0^{- 6} C$ . Why this step? Counting bricks cares only about how much charge, not its sign. The sign only tells us electrons were added (it's negative), not how many.

Step 2 — Divide total charge by one brick. With $e = 1.6 \times 1 0^{- 19} C$ , $n = \frac{∣ q ∣}{e} = \frac{3.2 \times 1 0 ^{- 6}}{1.6 \times 1 0 ^{- 19}} = 2 \times 1 0^{13}$ Why this step? This is the quantization relation $q = n e$ rearranged. We are literally asking "how many size- $e$ bricks fit into $∣ q ∣$ ?"

Verify: $2 \times 1 0^{13}$ is a whole number ✓ (quantization respected). It sits in our forecast band of $\sim 1 0^{13}$ ✓. Multiply back: $2 \times 1 0^{13} \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 6}$ C ✓.

Example 2 — Cell B: is this charge even legal?

Example 3 — Cell C: adding two like signs

Example 4 — Cell D: positive meets negative (cancellation)

Figure — Electric charge — properties, quantization, conservation

Figure s01 — how signed charges cancel (Cell D). The top row (cyan squares marked $+$ ) is the seven positive bricks $+ 7 e$ ; the bottom row (marked $-$ ) is the ten negative bricks $- 10 e$ . The $x$ -axis is just brick position (left to right); there is no $y$ -axis quantity — the two rows are only "positive shelf" and "negative shelf." Thin white lines join the seven pairs that annihilate. The three amber bricks on the bottom right are the unpaired negatives that survive, so the net is $- 3 e$ . Take-away: signed addition = pair off opposite bricks, and whatever colour is left over sets the final sign.

Example 5 — Cell E: the degenerate case ( $q = 0$ )

Worked example A copper ball is

neutral. Does it contain no charge? What is $n$ ? What force does it feel far from other charges? Forecast: "neutral" feels like "empty," but a metal is packed with charges. Guess: lots of charge, net zero.

Step 1 — Net charge. Neutral means $Q_{total} = 0$ . Why this step? Neutral is a statement about the sum, not the pieces. Protons and electrons are present in equal numbers (see Atomic Structure).

Step 2 — Brick count of the net. $n = \frac{∣ Q _{total} ∣}{e} = \frac{0}{e} = 0$ Why this step? $n = 0$ is a perfectly valid integer — quantization allows the empty rung. Zero is the additive identity: adding it changes nothing.

Step 3 — Force from a distant single charge. The net (monopole) Coulomb force is proportional to the net charge, so with $Q_{total} = 0$ the leading force is zero (see Coulomb's Law). Why this step? Coulomb's law scales with the net charge; zero net charge kills the dominant term.

Step 4 — The edge case (don't over-claim "no force"). In a nonuniform field, a neutral conductor still gets polarised: its own electrons shift so the near face is oppositely charged to the source, giving a small induced-dipole attraction. This is exactly why a charged rod picks up neutral bits of paper. Why this step? "Zero net charge" only kills the monopole force, not the weaker induced force in a nonuniform field. Covering this closes the scenario honestly.

Verify: $n = 0$ is an integer ✓. Equal protons and electrons: e.g. $1 0^{23}$ of each gives $1 0^{23} (+ e) + 1 0^{23} (- e) = 0$ ✓. Degenerate case handled without breaking any rule ✓.

Example 6 — Cell F: identical spheres touch (symmetry)

Worked example Identical metal spheres carry

$+ 12 e$ and $- 4 e$ . They touch, then separate. Charge on each? Forecast: conservation fixes the total; identical geometry splits it evenly. Guess before computing.

Step 1 — Conservation gives the total. $Q_{total} = (+ 12 e) + (- 4 e) = + 8 e (fixed forever)$ Why this step? Touching lets electrons flow between the spheres, but the isolated pair cannot gain or lose net charge — this is conservation in action.

The phrase "stays constant in time" can be written compactly. Let $Q_{total}$ be the total charge and $t$ be time (in seconds). The symbol $\frac{d Q _{total}}{d t}$ means "the rate at which $Q_{total}$ changes as $t$ ticks forward" — how many coulombs per second the total gains or loses. Conservation says that rate is exactly zero: $\frac{d Q _{total}}{d t} = 0 ⟺ Q_{total} never changes.$

Step 2 — Symmetry splits it. $q_{A} = q_{B} = \frac{+ 8 e}{2} = + 4 e$ Why this step? Identical conductors reach equal potential, and equal geometry at equal potential ⇒ equal charge (why they touch: Methods of Charging).

Verify: $q_{A} + q_{B} = + 4 e + 4 e = + 8 e = Q_{total}$ ✓ (conservation closes). $+ 4$ is an integer ✓.

Example 7 — Cell G: unequal conductors, then a repeated contact

Worked example Sphere X (radius

$2 R$ ) holds $+ 9 e$ ; sphere Y (radius $R$ , initially neutral) touches it once, then is removed. Find each charge. Then a second identical-to-Y sphere Z (also radius $R$ , neutral) touches Y and separates. Find Z. Forecast: unequal spheres do not split evenly — the bigger one keeps the larger share, in proportion to radius. So X should end with more than Y. Guess X gets roughly twice Y.

Step 1 — Conservation for the first contact. Total $= + 9 e + 0 = + 9 e$ , and this is fixed while they touch. Why this step? Same conservation rule: the isolated pair's total cannot change.

Step 2 — Unequal split by radius (the key new idea). For spheres in contact, charge distributes so both reach equal potential; a sphere's charge is proportional to its radius. With radii $2 R$ (X) and $R$ (Y), the shares are in ratio $2 : 1$ : $q_{X} = \frac{2}{3} (9 e) = 6 e, q_{Y} = \frac{1}{3} (9 e) = 3 e$ Why this step? This is what breaks the symmetry of Example 6. Only when radii are equal does the split become $50 : 50$ . Here the $2 : 1$ radius ratio forces a $2 : 1$ charge ratio. (Both shares land on whole electrons here, so quantization is not stressed yet.)

Step 3 — Second contact (Y with identical neutral Z). Now Y (radius $R$ ) and Z (radius $R$ ) are equal, so the earlier symmetry rule applies. Total $= 3 e + 0 = 3 e$ . The ideal continuous-fluid answer would give $1.5 e$ each — but that is not a legal charge ( $1.5$ is not an integer). Reality resolves this by handing out whole electrons: one sphere ends with $2 e$ , the other with $1 e$ . Which sphere gets the extra electron is not fixed by symmetry — it is decided by the random microscopic details of the moment of contact (thermal jiggling, tiny asymmetries), so over many trials it is a coin-flip. We simply label one outcome: say $q_{Y} = 2 e, q_{Z} = 1 e$ ; the mirror outcome $q_{Y} = 1 e, q_{Z} = 2 e$ is equally valid. Why this step? Two lessons collide here: equal radii want an equal share, but an odd electron count cannot be halved. Quantization overrides the smooth split, and the leftover single electron goes to one side at random — there is no physical rule that prefers Y over Z.

Verify: first split conserves: $6 e + 3 e = 9 e$ ✓, and the ratio $6 : 3 = 2 : 1$ matches the radius ratio ✓. Second split conserves for either labelling: $2 e + 1 e = 3 e$ ✓. Every charge ( $6 e, 3 e, 2 e, 1 e$ ) is an integer multiple of $e$ ✓.

Example 8 — Cell H: why quantization hides (the limit)

Worked example Your phone battery pushes about

$Q = 1 C$ through the circuit. By what fraction does one electron change this total? Why does charge feel "smooth"? Forecast: one electron out of a coulomb — expect an absurdly tiny fraction.

Step 1 — Fractional step size. $\frac{e}{Q} = \frac{1.6 \times 1 0 ^{- 19}}{1} = 1.6 \times 1 0^{- 19}$ Why this step? Quantization is only visible when one brick is a big slice of the whole. Here one brick is $\sim 1 0^{- 19}$ of the total — utterly negligible.

Step 2 — Interpret the limit. As the number of electrons grows huge, the relative gap between rung $n$ and $n + 1$ is $1/ n \to 0$ . The staircase looks like a ramp. Why this step? This is exactly why the parent note says $e$ is "invisible in daily life" — same reason sand looks like a smooth fluid from far off.

Verify: the number of electrons in one coulomb (call it $N$ , the brick-count of a whole coulomb) is $N = Q / e = 1/ e \approx 6.24 \times 1 0^{18}$ ✓ (same $q = n e$ relation as Example 1, just with $q = 1$ C). Adding one brick gives a fractional change of $1/ N \approx 1.6 \times 1 0^{- 19}$ ✓ — undetectable.

Figure s02 — quantization hides at large $n$ (Cell H). Both panels plot the same thing: the horizontal axis is $n$ , the number of bricks (an integer count); the vertical axis is the charge in units of $e$ . The amber line is the true staircase $q = n e$ (charge jumps by one whole brick at each step); the cyan dots (left) and cyan dashed line (right) mark the ideal $q = n$ trend. Left panel, small $n$ : the amber steps are tall relative to the values, so quantization is obvious. Right panel, large $n$ : the same amber staircase, zoomed out to $n$ up to $200$ , flattens into a near-smooth ramp because each step's relative size $1/ n$ shrinks toward zero. The physics is identical in both panels — only the scale changed.

Example 9 — Cell I: real-world word problem (charge in time)

Worked example A wire carries a steady current of

$2 A$ (amperes) for $5 seconds$ . How much charge flows, and how many electrons? Forecast: current is "charge per second," so total charge should be current $\times$ time $= 10$ C.

Step 1 — Name the current symbol. Let $I$ stand for the current: the amount of charge passing a point of the wire each second, measured in amperes (A). By definition $1 A = 1 C per second$ . So here $I = 2 A$ means two coulombs cross every second. Why this step? The letter $I$ is the standard symbol for current; we must attach it to its meaning before using it in a formula.

Step 2 — Charge from current. Since each second delivers $I$ coulombs and $t$ seconds pass, $Q = I t = 2 A \times 5 s = 10 C$ Why this step? Additivity in time: each second delivers $2$ C, and $5$ separate seconds simply add: $2 + 2 + 2 + 2 + 2 = 10$ . Same signed-sum rule, spread over time.

Step 3 — Convert to electrons. $n = \frac{Q}{e} = \frac{10}{1.6 \times 1 0 ^{- 19}} = 6.25 \times 1 0^{19}$ Why this step? Count the bricks that passed, using quantization again.

Verify: units of $I t$ are $A \cdot s = C$ ✓. The written form $6.25 \times 1 0^{19}$ is a whole number of electrons: $6.25 \times 1 0^{19} = 625 \times 1 0^{17} = 62, 500, 000, 000, 000, 000, 000$ (a 20-digit integer; the decimal $6.25$ is only the $\times 1 0^{19}$ shorthand, not a fractional electron) ✓. It is also $10 \times$ the " $6.25 \times 1 0^{18}$ per coulomb" fact from the parent note ✓.

Example 10 — Cell J: the exam twist (pair production)

Worked example A gamma-ray photon (charge

$0$ ) near a nucleus converts into an electron and a positron. What is the total charge after, and why can't this happen in empty vacuum? Forecast: the photon has zero charge; whatever appears must also sum to zero. So expect $- e$ and $+ e$ together.

Step 1 — Charge bookkeeping. Before: photon, $q = 0$ . After: electron $(- e)$ + positron $(+ e)$ : $Q_{after} = (- e) + (+ e) = 0 = Q_{before}$ Why this step? Conservation says the net charge cannot change; charge may only be created in equal-and-opposite pairs. Zero in, zero out.

Step 2 — The twist: why a nucleus is needed. Charge balances anywhere, but in free space a single photon cannot conserve energy and momentum simultaneously when making two massive particles. A nearby nucleus absorbs the recoil momentum. Why this step? This is the classic trap: students think charge conservation alone permits it. It's necessary but not sufficient — momentum conservation is the extra gatekeeper.

Verify: $(- e) + (+ e) = 0$ ✓ matches the photon's $0$ ✓. Net electron-type count changed by $0$ (one electron, one anti-electron) ✓ — conservation intact.

Recall Scenario checklist (test yourself)

Given a single charge, how do you count electrons? ::: $n = ∣ q ∣/ e$ ; the sign only says added/removed. Charge $2.5 e$ appears in an answer — what do you conclude? ::: It's illegal for a free particle; $n$ must be an integer. Two identical spheres $+ 12 e$ and $- 4 e$ touch — each ends with? ::: $+ 4 e$ (total $+ 8 e$ shared equally by symmetry). Sphere X ( $2 R$ , $+ 9 e$ ) touched once by neutral Y ( $R$ ) — how do they split? ::: By radius ratio $2 : 1$ , giving $6 e$ and $3 e$ . After Y ( $3 e$ ) touches identical neutral Z, why isn't it $1.5 e$ each? ::: $1.5$ isn't a legal charge; whole electrons force $2 e$ and $1 e$ , the extra one landing on a random side. $2 A$ for $5 s$ delivers how much charge? ::: $Q = I t = 10 C$ . Why can't a lone photon pair-produce in vacuum? ::: Charge balances but energy AND momentum can't both be conserved without a recoil body. Does a neutral conductor feel any force near a charged rod? ::: Yes — a small induced-dipole attraction in the nonuniform field, even though net charge is zero.

Active-recall flashcards

How many electrons make

- 3.2 μ

2 \times 1 0^{13}

q = 4.0 \times 1 0^{- 19}

C allowed?

No;

n = 2.5

is not an integer.

Net of

+ 7 e

and

- 10 e

- 3 e

+ 12 e

and

- 4 e

identical spheres touch — each?

+ 4 e

Unequal spheres

2 R

(

+ 9 e

) and

R

(neutral) touch — split?

6 e

and

3 e

(by radius ratio

2 : 1

Y with

3 e

touches identical neutral Z — result?

2 e

and

1 e

(not

1.5 e

each; the odd electron goes to a random side).

Charge from

2

A over

5

10

C (

6.25 \times 1 0^{19}

electrons).

Total charge after pair production from a photon?

0

(an

- e

and a

+ e