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

Answer: (b).

• (a) has the wrong sign on the exponent — that would be a huge charge, not a tiny one.
• (c) is the number of electrons in one coulomb ($1/e$), a pure count, not a charge in C.
• (d) is roughly the electron's mass in kg — a different quantity entirely. One electron carries $-e$; the magnitude of that is $e=1.602\times 10^{-19}\text{C}$.

Carriers are electrons; they flow from the negative terminal to the positive terminal (opposite to "conventional current" direction). The protons stay locked in the nuclei and do not move. See Conductors and Insulators for why electrons are free in metals.

Step 1 — pick the tool. We are counting elementary charges, so use $Q=Ne$. Why this and not $N=Q/e$? We were given the count $N$ and want the charge $Q$, so multiply. Step 2 — substitute. $Q=(3.0\times 10^{13})(1.602\times 10^{-19}\text{C})$ Step 3 — evaluate. $Q=4.806\times 10^{-6}\text{C}=4.81\mu \text{C}$. Step 4 — sign. They are electrons, each $-e$, so the net charge is negative. Answer: $Q\approx -4.81\mu \text{C}$.

Step 1 — pick the tool. We know $Q$, we want the count $N$, so invert: $N=Q/e$. Step 2 — substitute & evaluate. $N=\frac{0.50}{1.602\times 10^{-19}}\approx 3.12\times 10^{18}$ Answer: about $3.12\times 10^{18}$ electrons — roughly half the number in one full coulomb ($6.242\times 10^{18}$), which makes sense since $0.5\text{C}$ is half a coulomb.

Step 1 — convert units. $-8.0\text{nC}=-8.0\times 10^{-9}\text{C}$. Why? $e$ is in coulombs, so $Q$ must be too. Step 2 — count with $N=|Q|/e$. We use the magnitude because $N$ is a count (always positive). $N=\frac{8.0\times 10^{-9}}{1.602\times 10^{-19}}\approx 4.99\times 10^{10}$ Answer: about $5.0\times 10^{10}$ extra electrons. A tiny-sounding "nanocoulomb" is still ~50 billion electrons.

Step 1 — start neutral. $13$ protons give $+13e$; $13$ electrons give $-13e$; they cancel to $0$. Why? Equal magnitude, opposite sign — that is the definition of neutral. See Atomic Structure. Step 2 — remove 3 electrons. Protons unchanged at $+13e$; electrons now $-10e$. Step 3 — net. $+13e-10e=+3e$. Step 4 — in coulombs. $+3e=3(1.602\times 10^{-19})=+4.806\times 10^{-19}\text{C}$. Answer: $+3e\approx +4.81\times 10^{-19}\text{C}$ — an ${\text{Al}}^{3+}$ ion. Losing electrons makes an atom positive, because the (unchanged) positive protons now win.

Step 1 — total charge (conserved). $Q_{\text{total}}=(+2.0)+(-3.0)=-1.0\mu \text{C}$. Why add signed? Charge is additive with sign — a positive and a negative partly cancel. Step 2 — share equally. Each object gets half: $\frac{-1.0}{2}=-0.50\mu \text{C}$. Answer: A ends at $-0.50\mu \text{C}$, B ends at $-0.50\mu \text{C}$. Sanity check: total after $=-0.50+(-0.50)=-1.0\mu \text{C}$ ✓ — matches before, so charge was conserved.

Step 1 — connect current to charge. Current means "charge per second," $I=Q/t$, so $Q=It$. Why this tool? We're given a rate (A) and a time (s); their product is a total (C). This is the bridge to Electric Current and the Ampere. Step 2 — (a) total charge. $Q=(2.0\text{A})(30\text{s})=60\text{C}$ Step 3 — (b) count electrons with $N=Q/e$. $N=\frac{60}{1.602\times 10^{-19}}\approx 3.75\times 10^{20}$ Answer: (a) $60\text{C}$; (b) about $3.75\times 10^{20}$ electrons pass in half a minute.

Step 1 — charge from count. $Q=Ne=(4.5\times 10^{13})(1.602\times 10^{-19})=7.209\times 10^{-6}\text{C}\approx 7.21\mu \text{C}$. Step 2 — convert time. $0.20\text{ms}=0.20\times 10^{-3}\text{s}=2.0\times 10^{-4}\text{s}$. Step 3 — current from $I=Q/t$. $I=\frac{7.209\times 10^{-6}}{2.0\times 10^{-4}}\approx 3.60\times 10^{-2}\text{A}=36\text{mA}$ Answer: about $36\text{mA}$. We used both $Q=Ne$ (count → charge) and $I=Q/t$ (charge → current) — that's the synthesis.

Step 1 — (a) electrons per second. Charge per second is $I=0.75\text{C/s}$. Convert to a count using $N=Q/e$ with $Q=0.75\text{C}$ (the charge in one second): $N=\frac{0.75}{1.602\times 10^{-19}}\approx 4.68\times 10^{18}\text{electrons per second.}$ Step 2 — (b) set up drift speed. We're given $I=nAve$, so solve for $v$: $v=\frac{I}{nAe}$. Why this tool? It links the bulk current to the microscopic motion of the sea of electrons. Convert area: $A=1.0{\text{mm}}^2=1.0\times 10^{-6}{\text{m}}^2$. $v=\frac{0.75}{(8.5\times 10^{28})(1.0\times 10^{-6})(1.602\times 10^{-19})}$ Denominator $=8.5\times 10^{28}\times 1.0\times 10^{-6}\times 1.602\times 10^{-19}=1.362\times 10^4$. $v=\frac{0.75}{1.362\times 10^4}\approx 5.51\times 10^{-5}\text{m/s}\approx 0.055\text{mm/s}.$ Step 3 — (c) interpret. The electrons crawl at ~0.05 mm/s (slower than a snail), yet a whole ampere flows. Why? Because the wire is packed with charges everywhere at once ($8.5\times 10^{28}$ per m³). Nudge the whole sea slightly and an enormous number still passes per second. Current = (how many) × (how fast); the huge count compensates for the crawl. This is why a light turns on instantly even though individual electrons barely move. Answers: (a) $\approx 4.68\times 10^{18}$ e⁻/s; (b) $v\approx 5.5\times 10^{-5}\text{m/s}$; (c) huge carrier density compensates for tiny speed.

Step 1 — (a) current. $t=0.30\text{ms}=3.0\times 10^{-4}\text{s}$. $I=\frac{Q}{t}=\frac{15}{3.0\times 10^{-4}}=5.0\times 10^4\text{A}=50,000\text{A}.$ Step 2 — (b) electrons. $N=Q/e=\frac{15}{1.602\times 10^{-19}}\approx 9.36\times 10^{19}$. Step 3 — (c) slow leak. $t=1.0\text{hour}=3600\text{s}$. $I=\frac{15}{3600}\approx 4.17\times 10^{-3}\text{A}=4.17\text{mA}.$ Interpretation: Same amount of charge (15 C, same $9.36\times 10^{19}$ electrons) but delivered $10^7$ times faster gives a current $10^7$ times bigger. "How much charge" and "how fast it flows" are independent — a coulomb is huge, but whether it's dangerous depends on the time it takes to move. Answers: (a) $5.0\times 10^4\text{A}$; (b) $\approx 9.36\times 10^{19}$ electrons; (c) $\approx 4.17\text{mA}$.

• Tool to go from a count $N$ to charge $Q$? → $Q=Ne$
• Tool to go from charge $Q$ and time $t$ to current? → $I=Q/t$
• Does losing electrons make an atom positive or negative? → positive
• Do $+2\mu \text{C}$ and $-3\mu \text{C}$ cancel to zero when combined? → no, they sum to $-1\mu \text{C}$
• Why is drift speed tiny yet current large? → gigantic number of carriers per m³