United StatesPhysics C: Electricity and MagnetismSyllabus dot point

What is electric current, and how is it related to drift velocity and current density?

Topic 11.1 Electric Current: define current as the rate of charge flow and relate it to drift velocity, current density and charge carriers.

A calculus-based answer to AP Physics C E&M Topic 11.1, covering current as dQ/dt, conventional versus electron flow, current density, the microscopic model with drift velocity, and conservation of charge in a circuit.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Electric current is the rate at which charge passes a point: $I = \dfrac{dQ}{dt}$ , in amperes (1 A $=$ 1 C/s). When $Q$ varies, differentiate; the total charge over a time is $Q = \displaystyle\int I\,dt$ . Conventional current is defined as the flow of positive charge, opposite to the actual electron drift in a metal. The current density $J = \dfrac{I}{A}$ describes current per unit area. Microscopically, $I = nqv_d A$ , where $n$ is the carrier number density, $q$ the carrier charge, $v_d$ the slow drift velocity (often well under a millimeter per second), and $A$ the cross-section. The current responds almost instantly despite this slow drift, because the electric field that sets the carriers moving propagates through the wire near the speed of light. Charge is conserved: current is the same everywhere in a single unbranched wire.

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What this topic is asking
Current as a rate of charge flow
Conventional current versus electron flow
Current density and the microscopic model
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What this topic is asking

The College Board (Topic 11.1) wants you to define electric current as the rate of charge flow, $I = \dfrac{dQ}{dt}$ , distinguish conventional current from electron flow, and connect the macroscopic current to the microscopic picture of charge carriers drifting with a drift velocity. This opens the circuits unit.

Current as a rate of charge flow

If the current is constant, $I = \dfrac{Q}{t}$ . If it varies, use the derivative, and recover the total charge transported by integrating: $Q = \displaystyle\int_{t_1}^{t_2} I\,dt$ , the area under an $I$ -versus- $t$ graph. The two operations are inverses: differentiate the accumulated charge to get the instantaneous current, integrate the current to get the charge delivered. This calculus relationship matters in RC and LR circuits, where the current changes continuously, and it is the reason a time-varying current question almost always wants a derivative or an integral rather than a plain ratio.

Conventional current versus electron flow

By historical convention, current points in the direction positive charge would move, which in a metal wire is opposite to the actual flow of electrons. All circuit rules are written for conventional current; the electrons drift the other way, but the physics is identical because moving negative charge one way is equivalent to moving positive charge the other.

Current density and the microscopic model

The current density $J = \dfrac{I}{A}$ (A per m squared) describes how concentrated the current is. Relating it to the carriers:

I = nqv_d A \qquad\Longleftrightarrow\qquad J = nqv_d

where $n$ is the number of charge carriers per unit volume, $q$ the charge on each, and $v_d$ the drift velocity, the slow average velocity the carriers acquire along the wire on top of their fast random thermal motion. The product $nq$ is the free-charge density, and multiplying by the drift velocity gives the current density; multiplying again by the cross-sectional area gives the total current. This microscopic picture connects the macroscopic ammeter reading to the actual motion of electrons, and it explains why a thicker wire (larger $A$ ) or a better conductor (larger $n$ ) carries more current for the same drift speed.

Try this

Q1. A steady current of $2.5$ A flows for $40$ s. Find the charge transported. [2 points]

Cue. $Q = It = (2.5)(40) = 100$ C.

Q2. State the direction of conventional current relative to electron flow in a metal. [1 point]

Cue. Opposite: conventional current is the direction positive charge would move.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). The charge that has passed a point in a wire is

Q(t) = 3t^2 + 2t

(coulombs,

t

in seconds). The current at

t = 2

s is (A)

14

A (B)

16

A (C)

8

A (D)

12

A. Justify your reasoning.

Show worked answer →

A 1-point MCQ on current as the derivative of charge. The answer is (A).

Current is $I = \dfrac{dQ}{dt} = \dfrac{d}{dt}(3t^2 + 2t) = 6t + 2$ . At $t = 2$ , $I = 6(2) + 2 = 14$ A. The trap is plugging $t = 2$ into $Q(t)$ itself (giving $16$ ) instead of differentiating first.

AP 2024 (style)4 marksSection II (FRQ, quantitative). A copper wire of cross-sectional area

2.0\times10^{-6}

m squared carries a current of

5.0

A. The free-electron density is

n = 8.5\times10^{28}

per m cubed. (a) Define current density and calculate it. (b) Calculate the drift speed of the electrons. (c) Explain why the drift speed is so small yet the current responds almost instantly.

Show worked answer →

A 4-point FRQ on the microscopic model of current.

(a) Current density (2 points): $J = \dfrac{I}{A} = \dfrac{5.0}{2.0\times10^{-6}} = 2.5\times10^6$ A per m squared.
(b) Drift speed (1 point): $I = nqv_d A$ , so $v_d = \dfrac{I}{nqA} = \dfrac{5.0}{(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6})} = 1.8\times10^{-4}$ m/s.
(c) Explanation (1 point): the electrons drift slowly, but the field that sets them all moving propagates through the wire at near light speed, so the current starts everywhere almost at once.

Markers reward $J = I/A$ , the drift-speed relation, and the field-versus-drift distinction.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)