Exercises — Define current (flow of charge) and the ampere

Current is the rate of flow of electric charge past a point in a conductor: $I=\frac{\Delta Q}{\Delta t}.$ The unit is the ampere, and in base terms $1\text{A}=1\frac{\text{C}}{\text{s}}$ (one coulomb per second). Why: current answers "how much charge passed each second," so it is charge ÷ time.

$0.5\text{A}=0.5\text{C/s}$ means half a coulomb of charge passes any point each second. Over $2\text{s}$, exactly $1\text{C}$ has gone by.

False. The ampere (current) is the SI base unit; charge is derived: $1\text{C}=1\text{A}\cdot \text{s}$. This flips the intuition many students carry — see the mistake below.

Steady current → use $Q=It$: $Q=3\text{A}\times 8\text{s}=24\text{C}.$ Why $Q=It$ and not $dQ/dt$? Because $I$ does not change, the "area under the current graph" is just a rectangle of height $3$ and width $8$.

Total charge ÷ charge-per-electron: $N=\frac{Q}{e}=\frac{24}{1.6\times 10^{-19}}=1.5\times 10^{20}\text{electrons}.$

First convert time to seconds (the ampere is per second, not per minute): $t=30\times 60=1800\text{s}.$ $I=\frac{Q}{t}=\frac{4500}{1800}=2.5\text{A}.$ Why convert? $1\text{A}=1\text{C/s}$; leaving minutes would give an answer $60\times$ too small.

Current is the slope of $Q$ vs $t$, so differentiate: $I=\frac{dQ}{dt}=4t+5.$ At $t=3$: $I=4(3)+5=17\text{A}$. Why not $Q=It$? $Q=It$ needs a constant $I$. Here $I=4t+5$ grows with time — the charge curve bends, so the flow rate is different at every instant. Only the derivative captures the slope at that instant.

Charge is the area under the current–time graph ($Q=\int Idt$). Split into two shapes (look at the figure):

• Rectangle (0–3 s): height $2$, width $3$ → area $=2\times 3=6\text{C}$.
• Trapezoid (3–5 s): parallel sides $2$ and $6$, width $2$ → area $=\frac{1}{2}(2+6)\times 2=8\text{C}$. $Q_{\text{total}}=6+8=14\text{C}.$ Why area? $I=dQ/dt$ means $dQ=Idt$; summing all the tiny $Idt$ strips is the area.

Solve for the speed: $v_d=\frac{I}{nqA}$. Here $I$, $n$, $q$ are fixed and $A$ doubles, so $v_d$ halves. Physical picture: a fatter pipe lets the same charge-per-second through while each carrier moves more slowly — there is simply more room, more carriers crossing per slice.

Rearrange the microscopic law: $v_d=\frac{I}{nqA}=\frac{3.4}{(8.5\times 10^{28})(1.6\times 10^{-19})(2\times 10^{-6})}.$ Denominator $=8.5\times 1.6\times 2\times 10^{28-19-6}=27.2\times 10^3=2.72\times 10^4$. $v_d=\frac{3.4}{2.72\times 10^4}=1.25\times 10^{-4}\text{m/s}.$ Time to cross $1\text{m}$: $t=\frac{d}{v_d}=\frac{1}{1.25\times 10^{-4}}=8000\text{s}\approx 2.2\text{hours}$. Lesson: electrons crawl — yet the device responds instantly because the field nudges the whole electron column at once (near light speed).

Why the exponential and why the derivative? Charge here changes smoothly with time, so the instantaneous current is the slope, $I=dQ/dt$. The exponential $e^{-t/\tau }$ is the function whose derivative is a copy of itself scaled by $-1/\tau$ — that is exactly why decaying-charge problems use it. (a) $I(t)=\frac{dQ}{dt}=Q_0\cdot \left(-\frac{1}{\tau }\right)e^{-t/\tau }=-\frac{Q_0}{\tau }{e}^{-t/\tau }.$ (b) At $t=0$, $e^0=1$: $|I(0)|=\frac{Q_0}{\tau }=\frac{0.02}{0.5}=0.04\text{A}.$ (c) The sign is negative: charge is leaving the capacitor, so $Q$ is decreasing, so its slope (the current) points in the discharge direction. The minus sign is bookkeeping for "outflow," not an error.

(a) $I(t)=\frac{10}{4}t=2.5t$ (amperes, with $t$ in seconds). (b) Geometry (look at the figure): the current graph is a triangle with base $T=4$ and height $I_{\text{max}}=10$: $Q=\frac{1}{2}\text{base}\times \text{height}=\frac{1}{2}(4)(10)=20\text{C}.$ Calculus: $Q={\int }_0^{4}2.5tdt=2.5\cdot \frac{t^2}{2}{|}_0^4=1.25\times 16=20\text{C}.$ Both give $20\text{C}$ ✓ — the integral is the area. (c) Average current = total charge ÷ total time: $I_{\text{avg}}=\frac{Q}{T}=\frac{20}{4}=5\text{A}=\frac{I_{\text{max}}}{2}.$ This matches the rule "average of a straight ramp from $0$ = half the peak."

Start from $I=nqA{v}_d$ and solve for area: $A=\frac{I}{nq{v}_d}.$ To keep $v_d$ at most $10^{-4}$, we need $A$ at least this value (smaller $A$ forces faster drift). Plug in: $A=\frac{5}{(8.5\times 10^{28})(1.6\times 10^{-19})(1\times 10^{-4})}.$ Denominator $=8.5\times 1.6\times 10^{28-19-4}=13.6\times 10^5=1.36\times 10^6$. $A=\frac{5}{1.36\times 10^6}\approx 3.68\times 10^{-6}{\text{m}}^2.$ So the wire needs at least about $3.7{\text{mm}}^2$ of cross-section. Why "at least"? $v_d=I/(nqA)$: shrinking $A$ raises $v_d$, so any smaller wire violates the speed cap.