The International System of Units — Question 9

Question 9

What does a graph slope actually carry?

A device reports accumulated electric charge $Q$ as a function of time $t$. The coulomb satisfies $1\ \mathrm{C}=1\ \mathrm{A\,s}$. For constant current, the model is $Q=Q_0+It$, where $Q_0$ is initial accumulated charge and $I$ is current. Treat the plotted values as exact within this exercise.

The axis label $Q/\mathrm{C}$ means the numerical value of $Q$ when expressed in coulombs; similarly $t/\mathrm{s}$ is a numerical value. A student writes “$Q=1+2t$, so $I=2$ with no unit.”

1. Write a correct equation between numerical values and a correct equation between quantities. Identify $Q_0$ and $I$ with units.

2. At $t=3\ \mathrm{s}$, is $Q/t$ the current? Explain using the graph and the definition of the model.

3. Distinguish the slope of the numerical-value plot from the physical quantity $\Delta Q/\Delta t$. Why does a dimensionless plotted slope not make electric current dimensionless?

Question 9 - Solution

What does a graph slope actually carry?

(a) Restore the units carried by the axes.

A correct numerical-value equation is $$\frac{Q}{\mathrm{C}}=1+2\frac{t}{\mathrm{s}}.$$ Multiplying through by coulomb gives the quantity equation $$\boxed{Q=(1\ \mathrm{C})+(2\ \mathrm{C\,s^{-1}})t}.$$ Thus $Q_0=1\ \mathrm{C}$ and $\boxed{I=2\ \mathrm{A}}$. The student’s bare expression mixes a quantity on the left with unspecified numerical terms on the right.

(b) Use a change, not an absolute reading.

$$\frac{Q(t)}{t}=I+\frac{Q_0}{t},\qquad\frac{Q(3\ \mathrm s)}{3\ \mathrm s}=2\ \mathrm A+\frac13\ \mathrm A.$$

At $3\ \mathrm{s}$, $Q/t=7\ \mathrm{C}/(3\ \mathrm{s})=(7/3)\ \mathrm{A}$, not $2\ \mathrm{A}$. It includes the initial charge. In contrast, $$\frac{Q(3\ \mathrm{s})-Q(0)}{3\ \mathrm{s}-0} =\frac{(7-1)\ \mathrm{C}}{3\ \mathrm{s}}=2\ \mathrm{A}.$$

(c) Numerical coordinates are ratios to units.

The plotted slope is $\Delta(Q/\mathrm{C})/\Delta(t/\mathrm{s})=2$, a pure number. Restoring the axis units gives $\Delta Q/\Delta t=2\ \mathrm{C/s}=2\ \mathrm{A}$. Dividing each plotted quantity by a unit makes the coordinates numbers; it does not remove dimensions from the original physical quantities.