Define electric charge, electron, proton, and the coulomb: Worked examples

Before anything else, let us re-anchor the two symbols we lean on the whole page, so nothing is used unexplained:

Definition The symbols we count with

$Q$ = the total electric charge of a pile of particles, measured in coulombs (C). Think of it as "how much electrical stuff, and which sign."
$N$ = the number of elementary charges — literally how many electrons or protons you are holding. A plain counting number, no units.
$N_{+}$ = the count of positive charges (protons) in a mixed pile.
$N_{-}$ = the count of negative charges (electrons) in a mixed pile.
$N_{net} = N_{+} - N_{-}$ = the leftover count after cancellation — how many unmatched charges remain, and of which sign (positive if protons win, negative if electrons win).
$e = 1.602 \times 1 0^{- 19} C$ = the elementary charge, the charge of exactly one proton ( $+ e$ ) or one electron ( $- e$ ). This is the smallest brick; every real charge is a whole stack of these bricks.

The link between them (from the parent): $Q = N e ⟺ N = \frac{Q}{e}$

The scenario matrix

Every problem this topic can throw at you is one (or a blend) of the cells below. The middle column names the cell; the right column is the example that nails it.

#	Cell (the scenario class)	Covered by
1	Positive pile — count → charge, sign $+$	Example 1
2	Negative pile — count → charge, sign $-$ (electrons)	Example 2
3	Invert — charge → count ( $N = Q / e$ )	Example 3
4	Mixed pile — both signs, find the net charge	Example 4
5	Degenerate / zero — perfectly balanced, $Q = 0$	Example 4 (part b)
6	Ionisation — remove/add electrons, sign flips	Example 5
7	Limit: tiny — a single electron, smallest possible non-zero charge	Example 6
8	Limit: huge — a whole coulomb, how many electrons	Example 7
9	Real-world word problem — charge hidden inside a story	Example 8
10	Exam twist — quantisation trap (is a value even possible?)	Example 9

Figure — Define electric charge, electron, proton, and the coulomb

Reading the figure (text description): it is a horizontal number line for net charge, with $Q = 0$ marked by a single dot in the plum colour at the exact centre — this is the neutral/degenerate case (Example 4b). To the right of zero sit the positive cases in the burnt-orange colour, at three labelled rungs: $+ e$ (one proton), $+ 2 e$ (a cation, as in Example 5a), and a far-right marker for a large positive pile of protons (Example 1). To the left of zero sit the negative cases in the deep-teal colour, again at three labelled rungs: $- e$ (one electron), $- 3 e$ (an anion, Example 5b), and a far-left marker for a large negative pile of electrons (Example 2). An arrow points rightward labelled "more +" and one points leftward labelled "more −", so the further from centre a case sits, the larger its charge magnitude. The single takeaway: every example on this page is just a labelled point on this one line — its side tells you the sign, its distance from centre tells you the size.

Worked examples

Cell 1 — Positive pile (count → charge)

Worked example Example 1 — Charge of a heap of protons

Q: You somehow gather $3 \times 1 0^{15}$ protons. What is their total charge?

Forecast: Protons are positive, so guess a positive, very small number of coulombs. Bigger or smaller than 1 C?

Step 1 — Pick the direction. We know the count $N$ , we want the charge $Q$ , so multiply: $Q = N e$ . Why this step? Given count → charge means "stack $N$ bricks of size $e$ ," which is multiplication.

Step 2 — Plug in. $Q = (3 \times 1 0^{15}) (1.602 \times 1 0^{- 19})$ . Why this step? Each proton contributes exactly one $+ e$ ; total = count × per-particle charge.

Step 3 — Arithmetic. $3 \times 1.602 = 4.806$ , and $1 0^{15} \times 1 0^{- 19} = 1 0^{- 4}$ , so $Q = 4.806 \times 1 0^{- 4} C$ . Why this step? Handle the mantissas and the powers of ten separately — that's how you multiply scientific notation safely.

Step 4 — Attach the sign. Protons are positive, so $Q = + 4.806 \times 1 0^{- 4} C$ . Why this step? $e$ is just the magnitude; the particle's identity decides the sign.

Answer: $Q \approx + 4.81 \times 1 0^{- 4} C = + 0.481 mC$ .

Verify: Units — (pure count)(C) = C. ✓ Magnitude — much less than 1 C, as forecast, because $3 \times 1 0^{15}$ is far below the $6.24 \times 1 0^{18}$ needed for a whole coulomb. ✓

Cell 2 — Negative pile (count → charge, sign $-$ )

Worked example Example 2 — Charge of a heap of electrons

Q: What is the total charge of $5 \times 1 0^{12}$ electrons?

Forecast: Electrons are negative → answer is negative and tiny (fewer particles than Example 1).

Step 1 — Direction. Count → charge, so $Q = N e$ . Why this step? Same tool as before; only the sign at the end differs.

Step 2 — Plug in. $Q = (5 \times 1 0^{12}) (1.602 \times 1 0^{- 19})$ . Why this step? Every electron carries one brick of charge magnitude $e$ .

Step 3 — Arithmetic. $5 \times 1.602 = 8.01$ , $1 0^{12} \times 1 0^{- 19} = 1 0^{- 7}$ , so magnitude $= 8.01 \times 1 0^{- 7} C$ . Why this step? Mantissa × mantissa, then add the exponents.

Step 4 — Sign. Electrons → negative: $Q = - 8.01 \times 1 0^{- 7} C$ .

Answer: $Q \approx - 0.801 μ C$ (that's $- 8.01 \times 1 0^{- 7} C$ ).

Verify: $1 0^{- 7} C = 1 0^{- 1} μ C$ , so $8.01 \times 1 0^{- 7} C = 0.801 μ C$ . ✓ Sign negative as forecast. ✓

Cell 3 — Invert (charge → count)

Worked example Example 3 — How many electrons make 2 C?

Q: A wire delivers a total charge of $2 C$ (carried by electrons). How many electrons passed?

Forecast: A whole coulomb is huge, and this is 2 of them → expect a number bigger than $1 0^{19}$ .

Step 1 — Direction. We know $Q$ , want $N$ , so divide: $N = \frac{Q}{e}$ . Why this step? Going charge → count means "how many bricks of size $e$ fit inside $Q$ ?" — that's division.

Step 2 — Use magnitude only for the count. $N = \frac{2}{1.602 \times 1 0 ^{- 19}}$ . Why this step? $N$ is a plain count (can't be negative); we handle the electron's sign separately in words.

Step 3 — Arithmetic. $\frac{2}{1.602} = 1.248$ , and dividing by $1 0^{- 19}$ means $\times 1 0^{+ 19}$ , so $N = 1.248 \times 1 0^{19}$ . Why this step? Dividing by a small power of ten flips its sign to make the answer large — the physical reason a coulomb is so many particles.

Answer: About $1.25 \times 1 0^{19}$ electrons.

Verify: Multiply back: $(1.248 \times 1 0^{19}) (1.602 \times 1 0^{- 19}) = 1.248 \times 1.602 = 2.00 C$ . ✓ Bigger than $1 0^{19}$ as forecast. ✓

Cells 4 & 5 — Mixed pile & the degenerate zero

Worked example Example 4 — Net charge of a mixed pile (and when it vanishes)

Q(a): A region holds $8 \times 1 0^{9}$ protons and $6 \times 1 0^{9}$ electrons. What is the net charge? Q(b): What if instead it held $6 \times 1 0^{9}$ of each?

Forecast: In (a) protons win (more of them) → net positive. In (b) they perfectly cancel → net zero (the degenerate case).

Step 1 — Count the imbalance, not the total. Here $N_{+} = 8 \times 1 0^{9}$ (protons) and $N_{-} = 6 \times 1 0^{9}$ (electrons), so the net count is $N_{net} = N_{+} - N_{-} = 8 \times 1 0^{9} - 6 \times 1 0^{9} = 2 \times 1 0^{9}$ excess protons. Why this step? Equal-and-opposite bricks annihilate in the sum: every electron cancels one proton. Only the leftover particles — $N_{net}$ — carry net charge.

Step 2 — Turn the leftover into charge. $Q = N_{net} e = (2 \times 1 0^{9}) (1.602 \times 1 0^{- 19})$ . Why this step? After cancellation you're back to a single-sign pile — Example 1's situation.

Step 3 — Arithmetic. $2 \times 1.602 = 3.204$ , $1 0^{9} \times 1 0^{- 19} = 1 0^{- 10}$ , so $Q = + 3.204 \times 1 0^{- 10} C$ . Why this step? Excess is protons, so the sign is $+$ .

Answer (a): $Q \approx + 3.20 \times 1 0^{- 10} C$ .

Step 4 — Part (b), the degenerate case. Now $N_{+} = N_{-} = 6 \times 1 0^{9}$ , so $N_{net} = 6 \times 1 0^{9} - 6 \times 1 0^{9} = 0$ , giving $Q = 0 \times e = 0 C$ . Why this step? This is the zero/degenerate input cell — the balanced, neutral state. Not a special new rule, just $N_{net} = 0$ dropped into the same formula.

Answer (b): $Q = 0 C$ — the object is electrically neutral (the centre of the number line in the figure).

Verify: (a) sign positive because protons outnumber electrons ✓; magnitude between Example 2 and 1's scale, consistent with $2 \times 1 0^{9}$ particles ✓. (b) equal counts must cancel — matches the neutral-atom rule from the parent. ✓

Cell 6 — Ionisation (sign flip)

Worked example Example 5 — Ionising an atom in two directions

Q: A neutral sodium atom has 11 protons and 11 electrons. (a) It loses 2 electrons. (b) A different neutral atom gains 3 electrons. Net charge in each?

Forecast: Lose electrons → left with excess protons → positive ion. Gain electrons → excess electrons → negative ion.

Step 1 — Start from neutrality. Neutral means $+ 11 e$ and $- 11 e$ cancel to $0$ . Why this step? We only ever track the change from neutral, so we need the neutral baseline first.

Step 2(a) — Remove 2 electrons. Now $N_{+} = 11$ , $N_{-} = 9$ , so $N_{net} = 11 - 9 = + 2$ . Why this step? Removing negative bricks leaves the positive bricks uncancelled.

Step 3(a) — Charge. $Q = N_{net} e = + 2 e = 2 (1.602 \times 1 0^{- 19}) = + 3.204 \times 1 0^{- 19} C$ . Why this step? Excess $= 2$ protons' worth → multiply by $e$ .

Answer (a): $Q = + 2 e = + 3.204 \times 1 0^{- 19} C$ (a $+ 2$ cation).

Step 2(b) — Add 3 electrons. Now $N_{+} = 11$ , $N_{-} = 14$ , so $N_{net} = 11 - 14 = - 3$ . Step 3(b) — Charge. $Q = N_{net} e = - 3 e = - 3 (1.602 \times 1 0^{- 19}) = - 4.806 \times 1 0^{- 19} C$ .

Answer (b): $Q = - 3 e = - 4.806 \times 1 0^{- 19} C$ (a $- 3$ anion).

Verify: Signs match the forecast (lose → +, gain → −). ✓ Magnitudes are exact small multiples of $e$ : $2 e$ and $3 e$ , as ionisation must be (you can only move whole electrons). ✓

Cell 7 — Limiting case: the smallest possible non-zero charge

Worked example Example 6 — Can charge get smaller than one electron?

Q: What is the smallest amount of free charge (other than zero) that can ever exist, and what is its value?

Forecast: Guess it's the single elementary charge $e$ — you can't split an electron.

Step 1 — Recall quantisation. Free charge is always $Q = N e$ with $N$ a whole number. Why this step? This limiting case is really a question about the smallest allowed $N$ .

Step 2 — Find the smallest non-zero $N$ . Whole numbers non-zero start at $N = 1$ . Why this step? $N = 0$ gives $Q = 0$ (Example 4b); the next step down the ladder is one brick.

Step 3 — Evaluate. $Q = 1 \times e = 1.602 \times 1 0^{- 19} C$ . Why this step? One electron or one proton — the indivisible unit.

Answer: The smallest non-zero free charge is $\pm e = \pm 1.602 \times 1 0^{- 19} C$ .

Verify: Any smaller would need $N$ between 0 and 1 (a fraction of an electron), which quantisation forbids. ✓ This is the floor of the number line's spacing in the figure.

Cell 8 — Limiting case: a whole coulomb (the huge end)

Worked example Example 7 — Feeling how big 1 coulomb really is

Q: Exactly how many electrons carry a charge of $1 C$ ? Then: if a device leaked $1$ electron per second, how long to lose $1 C$ ?

Forecast: From the parent, expect $\approx 6.24 \times 1 0^{18}$ electrons — and a leak time so long it's basically forever.

Step 1 — Direction. Charge → count, so $N = \frac{Q}{e} = \frac{1}{1.602 \times 1 0 ^{- 19}}$ . Why this step? We want how many bricks fill one coulomb — division.

Step 2 — Arithmetic. $\frac{1}{1.602} = 0.6242$ , times $1 0^{+ 19}$ gives $N = 6.242 \times 1 0^{18}$ . Why this step? This is the definition of the coulomb read backwards.

Step 3 — Time at 1 electron/second. $t = N \times 1 s = 6.242 \times 1 0^{18} s$ . Why this step? One electron per second means the count equals the number of seconds.

Step 4 — Put that in human terms. $6.242 \times 1 0^{18} s \div 3.156 \times 1 0^{7} s/year \approx 1.98 \times 1 0^{11}$ years — about 200 billion years. Why this step? Dividing by seconds-per-year converts the count into a graspable timescale.

Answer: $6.24 \times 1 0^{18}$ electrons; leaking one per second it would take $\sim 2 \times 1 0^{11}$ years — far longer than the age of the universe.

Verify: $(6.242 \times 1 0^{18}) (1.602 \times 1 0^{- 19}) = 6.242 \times 1.602 \times 1 0^{- 1} = 1.000 C$ . ✓ Confirms the "1 coulomb is enormous" mistake-fix from the parent. ✓

Cell 9 — Real-world word problem

Worked example Example 8 — Charge hidden inside a story

Q: A camera flash capacitor releases its charge, and during the flash $4.0 \times 1 0^{16}$ electrons move through the bulb. How much charge flowed? Bonus: how does this compare to the $2 C$ of Example 3?

Forecast: $4 \times 1 0^{16}$ is well under $6 \times 1 0^{18}$ , so expect a fraction of a coulomb — milli-coulomb scale — much less than 2 C.

Step 1 — Extract the numbers from the words. "electrons move" → $N = 4.0 \times 1 0^{16}$ , and we want $Q$ . Why this step? Word problems bury $N$ or $Q$ in a sentence; first job is to spot which one you're given.

Step 2 — Direction. Count → charge, so $Q = N e$ . Why this step? We have the count of electrons, we want the charge.

Step 3 — Arithmetic. $Q = (4.0 \times 1 0^{16}) (1.602 \times 1 0^{- 19}) = 4.0 \times 1.602 \times 1 0^{- 3} = 6.408 \times 1 0^{- 3} C$ . Why this step? Multiply the mantissas ( $4.0 \times 1.602 = 6.408$ ) and add the exponents ( $1 0^{16} \times 1 0^{- 19} = 1 0^{- 3}$ ) separately — the safe way to multiply scientific notation.

Step 4 — Convert to a human-friendly unit. $6.408 \times 1 0^{- 3} C = 6.41 mC$ , since $1 mC = 1 0^{- 3} C$ . Sign is negative (electrons carry it). Why this step? Milli-coulombs (mC) are the natural everyday unit for capacitor discharges; a bare $\times 1 0^{- 3}$ hides how ordinary this number is.

Step 5 — Do the bonus comparison to 2 C. $\frac{2 C}{6.408 \times 1 0 ^{- 3} C} = \frac{2}{6.408 \times 1 0 ^{- 3}} \approx 312$ . Why this step? The bonus asks "how much less?" — a ratio answers that directly, no units needed since both are charges.

Answer: $Q \approx - 6.41 mC$ . This is about $\frac{1}{312}$ of the $2 C$ in Example 3 — the flash moves a tiny sliver of a coulomb. On the number-line figure it would be a teal point close to (but not at) the centre — negative, and far nearer zero than the large-pile marker.

Verify: $6.41 mC$ is milli-coulomb scale, far less than 2 C, exactly as forecast. ✓ Ratio check: $312 \times 6.408 \times 1 0^{- 3} = 2.0 C$ . ✓ Milli-coulomb is a realistic flash-capacitor charge — the physics is sane. ✓

Cell 10 — Exam twist: the quantisation trap

Worked example Example 9 — Is this charge even possible?

Q: An exam claims an isolated object carries a charge of $Q = 2.40 \times 1 0^{- 19} C$ . Is this physically possible? If not, what is the nearest allowed value?

Forecast: Smells like a trap — this looks like it wants a fractional number of electrons. Guess: not allowed.

Step 1 — Test with the count. Compute $N = \frac{Q}{e} = \frac{2.40 \times 1 0 ^{- 19}}{1.602 \times 1 0 ^{- 19}}$ . Why this step? Free charge is legal only if $N$ comes out a whole number. So the test is the division.

Step 2 — Arithmetic. The $1 0^{- 19}$ cancels top and bottom, leaving $\frac{2.40}{1.602} = 1.498 \dots \approx 1.50$ . Why this step? When the powers of ten match, dividing scientific notation reduces to dividing the mantissas.

Step 3 — Judge. $N = 1.50$ is not a whole number → you'd need one-and-a-half electrons. Impossible. Why this step? Quantisation forbids fractional free charge (this is the exam's whole point).

Step 4 — Nearest allowed values. Round $N$ to the nearest integers: $N = 1 \Rightarrow Q = e = 1.602 \times 1 0^{- 19} C$ , or $N = 2 \Rightarrow Q = 2 e = 3.204 \times 1 0^{- 19} C$ . Why this step? Legal charges sit only on the integer rungs; the claimed value falls between rungs 1 and 2.

Answer: Not possible — it implies $1.50$ electrons. The nearest allowed charges are $+ e$ ( $1.602 \times 1 0^{- 19} C$ ) and $+ 2 e$ ( $3.204 \times 1 0^{- 19} C$ ). On the number-line figure the claimed value would fall in the gap between the $+ e$ and $+ 2 e$ rungs — a spot no real object can occupy.

Verify: $1.498$ rounds to neither 1 nor 2 exactly, confirming it's strictly between rungs. ✓ Both quoted allowed values are exact integer multiples of $e$ . ✓

Active recall

Recall Which direction of the formula for each cell?

Count given, want charge? ::: Multiply — $Q = N e$ . Charge given, want count? ::: Divide — $N = Q / e$ . Mixed pile of + and −? ::: Cancel first — use net count $N_{net} = N_{+} - N_{-}$ , then $\times e$ . How to test if a charge value is physically allowed? ::: Compute $N = Q / e$ ; it must be a whole number.

Recall Fast numeric recall

Charge of $5 \times 1 0^{12}$ electrons? ::: $- 0.801 μ C$ . Electrons in $2 C$ ? ::: $\approx 1.25 \times 1 0^{19}$ . Net charge after losing 2 electrons? ::: $+ 2 e = + 3.204 \times 1 0^{- 19} C$ . Smallest non-zero free charge? ::: $\pm e = \pm 1.602 \times 1 0^{- 19} C$ .

Connections

Parent: Charge, electron, proton, coulomb — the master formula $Q = N e$ these examples exercise.
Electric Current and the Ampere — Example 7's "electrons per second" is exactly current, $I = Q / t$ .
Voltage and Potential Difference — what pushes the piles of charge to move.
Conductors and Insulators — whether the counted electrons are free to flow.
Coulomb's Law — the force between the positive and negative piles.
Atomic Structure — the source of the protons and electrons we count in ionisation.