In Physics 2 you learned that a capacitor charges through a resistor with a time constant $\tau = RC$. Now we can see where the exponential comes from: the loop rule turns into a differential equation that separates neatly.
💡 Why it works
Close the switch at $t = 0$ with the capacitor empty. Around the loop: $\varepsilon - IR - \dfrac{q}{C} = 0$. The current is the rate the capacitor's charge grows, $I = dq/dt$, so
$$R\frac{dq}{dt} = \varepsilon - \frac{q}{C} \quad\Longrightarrow\quad \frac{dq}{C\varepsilon - q} = \frac{dt}{RC}.$$
Integrate from $q = 0$ at $t = 0$: $-\ln\dfrac{C\varepsilon - q}{C\varepsilon} = \dfrac{t}{RC}$, which gives the solution below. The more charge already on the capacitor, the more it pushes back, so the charging slows down as it goes.
🔑 Key idea
$$q = C\varepsilon\left(1 - e^{-t/RC}\right), \qquad I = \frac{dq}{dt} = \frac{\varepsilon}{R}e^{-t/RC}, \qquad V_C = \varepsilon\left(1 - e^{-t/RC}\right).$$
After one time constant $\tau = RC$ the charge is 63% of its final value; after $5\tau$ it is over 99%.
The time constant sets the pace and nothing else: double $R$ or double $C$ and every stage of the charging takes twice as long, but the curves keep the same shape. Circuits use this to measure out time, from the blink rate of a flashing light to the delay before a camera flash is ready again.
Charging: the capacitor's voltage climbs toward $\varepsilon$ while the current starts at its largest value and dies away. Change $R$ and $C$: only the time scale $RC$ changes.The same equation in scaled form, $dy/dx = 1 - y$, with $y = q/(C\varepsilon)$ and $x = t/RC$. Every solution heads for $y = 1$; drag the starting point to see a partly charged capacitor finish the job.
Worked examples
✏️ Worked example: Charge and current over time
A $500\ \mu\text{F}$ capacitor charges through $2.0\ \text{k}\Omega$ from a $12\ \text{V}$ battery. Find $\tau$, the starting current, and $q$ and $I$ at $t = 2.0\ \text{s}$.
For the same circuit, when does the charge reach 90% of its final value?
$1 - e^{-t/RC} = 0.90$, so $e^{-t/RC} = 0.10$ and $t = RC\ln 10$.
$t = (1.0\ \text{s})(2.303) = 2.30\ \text{s}$.
Answer: $t \approx 2.30\ \text{s}$
✏️ Worked example: Where the energy goes
While the capacitor charges fully, how much energy does the battery supply, and how much ends up stored?
The battery moves charge $Q = C\varepsilon$ up through its emf: work $Q\varepsilon = C\varepsilon^2$.
The capacitor stores $\tfrac{1}{2}C\varepsilon^2$, half of it.
The other half heats the resistor: $\int_0^\infty I^2R\,dt = \dfrac{\varepsilon^2}{R}\cdot\dfrac{RC}{2} = \tfrac{1}{2}C\varepsilon^2$, whatever the value of $R$.
"The current is zero at the start, since the capacitor has no charge to push it." The battery does the pushing. At $t = 0$ the empty capacitor has $V_C = 0$, so it acts like a plain wire, and the whole battery voltage lands on the resistor: $I_0 = \varepsilon/R$, the largest current of the whole process. It is the end of charging, when $V_C$ has risen to $\varepsilon$, that the current is zero. Check with the loop rule at $t = 0$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.