Exponential decay is a process where a quantity decreases at a rate proportional to its current value, following Q(t) = Q₀e^(-t/τ); in AP Physics C: E&M, it describes how charge, voltage, and current fall off when a capacitor discharges through a resistor in an RC circuit (Topic 11.8).
Exponential decay happens whenever something shrinks at a rate proportional to how much of it is left. In math terms, dQ/dt = -Q/τ, which solves to Q(t) = Q₀e^(-t/τ). The bigger the quantity, the faster it drops. As it gets smaller, the decay slows down, so the quantity approaches zero but never quite gets there.
In AP Physics C: E&M, this shows up in RC circuits (Topic 11.8). When a charged capacitor discharges through a resistor, Kirchhoff's loop rule gives you exactly that differential equation, with time constant τ = RC. Charge, the voltage across the capacitor, and the current in the circuit all decay with the same form: q(t) = Q₀e^(-t/τ). The time constant τ is the natural clock of the circuit. After one τ, the charge has fallen to about 37% (1/e) of its starting value. After about 5τ, the discharge is essentially done.
Exponential decay is the heart of Topic 11.8 (Resistor-Capacitor Circuits) in Unit 11. This is one of the few places on the E&M exam where you're expected to actually solve a differential equation, not just quote a result. The exam loves this because it ties together circuit analysis (Kirchhoff's rules), calculus (separable differential equations), and energy reasoning (where does the capacitor's stored energy go? It dissipates as heat in the resistor). If you can set up dq/dt = -q/RC from a loop equation and solve it, you've demonstrated exactly the kind of derivation skill Physics C FRQs reward. The same e^(-t/τ) structure also describes current decay in RL circuits, so learning it once pays off twice.
Keep studying AP® Physics C: E&M Unit 11
Transient response (Unit 11)
Exponential decay IS the transient response of a discharging RC circuit. 'Transient' just means the temporary behavior between flipping the switch and settling down, and for RC circuits that temporary behavior is always an exponential with time constant τ = RC.
Steady state (Unit 11)
Steady state is where exponential decay ends up. Once roughly 5τ has passed, the current is effectively zero and nothing changes anymore. Exam questions often ask you to analyze a circuit at t = 0, during the decay, and at steady state as three separate snapshots.
Conservation of electric charge (Unit 11)
The charge leaving the capacitor doesn't vanish. It flows through the circuit as current, I = -dq/dt, which is why current decays with the exact same e^(-t/τ) shape as the charge does.
Equivalent capacitance (Unit 11)
When a circuit has multiple capacitors or resistors, you reduce them to a single equivalent C and R first. The time constant is then τ = R_eq·C_eq, which is a classic way exams add a layer of difficulty to a decay problem.
On multiple choice, expect stems like 'a capacitor charged to Q₀ begins discharging at t = 0; what is q(t)?' where you pick Q₀e^(-t/RC) from a lineup of look-alike expressions (watch for sign errors and 1 - e^(-t/τ) imposters). Other common asks include what fraction of charge remains after one time constant (about 37%), whether the voltage across the capacitor decays the same way as the charge (it does, since V = q/C), and what happens to the stored energy during discharge (it's dissipated as heat in the resistor, and since U = q²/2C, energy decays as e^(-2t/τ), twice as fast). On FRQs, the high-value skill is the derivation. Write the loop equation, get dq/dt = -q/RC, separate variables, integrate, and apply the initial condition. You should also be able to sketch the decay curve and identify τ graphically as the time to reach 1/e of the initial value.
Discharging follows q(t) = Q₀e^(-t/τ), starting at Q₀ and falling toward zero. Charging follows q(t) = Q_max(1 - e^(-t/τ)), starting at zero and rising toward Q_max. Both use the same time constant τ = RC, but they're mirror images. A quick sanity check on any answer choice is to plug in t = 0 and t → ∞ and see if the limits match the physical situation.
Exponential decay means the rate of decrease is proportional to the current value, which gives Q(t) = Q₀e^(-t/τ) for a discharging capacitor.
The time constant of an RC circuit is τ = RC, and after one time constant the charge has dropped to about 37% (1/e) of its initial value.
Charge, capacitor voltage, and current all decay with the same e^(-t/τ) form during discharge, because V = q/C and I = -dq/dt.
The energy stored in the capacitor is dissipated as heat in the resistor, and since U = q²/2C, the energy decays as e^(-2t/τ), faster than the charge itself.
On FRQs, be ready to derive the decay equation from Kirchhoff's loop rule by separating variables and integrating, not just memorize the result.
Charging is the mirror image of decay: q(t) = Q_max(1 - e^(-t/τ)) rises toward a maximum instead of falling toward zero.
It's a process where a quantity decreases at a rate proportional to its current value, following Q(t) = Q₀e^(-t/τ). On the E&M exam it describes how charge, voltage, and current fall off when a capacitor discharges through a resistor (Topic 11.8).
Mathematically, no. The exponential e^(-t/τ) approaches zero but never reaches it. Practically, after about 5 time constants the charge is under 1% of its starting value, so we treat the circuit as fully discharged.
About 37% of the initial charge, since e^(-1) ≈ 0.368. This is a favorite multiple-choice fact, and it also gives you a way to read τ off a graph: find the time where the curve hits 1/e of its starting value.
Discharging is pure exponential decay, q(t) = Q₀e^(-t/τ), falling from Q₀ to zero. Charging is q(t) = Q_max(1 - e^(-t/τ)), rising from zero to Q_max. Same time constant τ = RC, opposite direction. Always check the t = 0 and t → ∞ limits to tell them apart.
No, it decays twice as fast. Since U = q²/2C and the charge goes as e^(-t/τ), the energy goes as e^(-2t/τ). That energy doesn't disappear; it's dissipated as heat in the resistor.
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