Curriculum · Course · Week 1 · فارسی
Static charge is an imbalance of positive and negative charge. In copper, mobile electrons can move through the material. In plastic, charge is much less mobile. A circuit uses a source and a closed path to sustain current. Voltage measures energy transferred per unit charge, and current measures charge passing per unit time. Resistance describes opposition to current.
A rod gains electrons when rubbed. Its net charge becomes negative. A copper wire gives excess electrons a path to move, while an insulating handle helps keep charge from flowing through it.
Consider a battery, switch and lamp in one loop. With the switch open, the gap stops steady current even though the battery still has a voltage. Closing the switch completes the loop. In the external metal wire, electron motion is opposite conventional current.
Battery and lamp with an open switch
If 6 coulombs pass through a wire in 3 seconds at a constant rate, the current is 2 amperes. If the switch opens, steady current drops to zero. Explain why this does not require the battery voltage to become zero.
| Quantity | Question it answers | Unit |
|---|---|---|
| Charge | How much electric charge? | Coulomb (C) |
| Current | How much charge passes a point each second? | Ampere (A), or C/s |
| Voltage | How much energy is transferred per coulomb between two points? | Volt (V), or J/C |
| Resistance | How strongly does a component oppose current? | Ohm (Ω) |
A battery creates a potential difference between its terminals. In an open circuit, there can be a voltage across the switch even when the steady current is zero. Conventional current is drawn from the positive terminal through the external circuit to the negative terminal. Electrons in a metal wire drift in the opposite direction.
A source transfers 12 J to 4 C of charge. Divide energy by charge to get 3 V. If 4 C passes a point in 2 s, divide charge by time to get 2 A. The two divisions describe different properties. Increasing the time to 4 s while keeping the charge fixed halves the average current without changing the given energy per coulomb.
Answers. 1. Positive. 2. 3 V. 3. Only the closed loop has steady current, from battery positive to battery negative through the external circuit.
The line rises by 6 coulombs in 3 seconds, so its slope is 2 amperes. The flat segment shows an interval with zero current. Charge that has already passed the measurement point still counts in the total. A steeper line would mean more charge passing each second.
Read the graph. The graph tracks accumulated charge, so its height and slope answer different questions. At 1.5 s the height is 3 C, and the current is still 2 A. After 3 s the height remains 6 C while the current is zero. If 6 C had passed in 2 s instead, the average current would have been 3 A.
Your goal is to explain what moves, what transfers energy, and what an ideal circuit drawing assumes. A wire does not become empty of electrons when a switch opens. Metals already contain mobile charge carriers. Closing a circuit establishes an electric field that drives a slow net drift superimposed on their microscopic motion. Signal propagation and electron drift are different processes.
Charge is conserved. Rubbing materials transfers electrons rather than creating net charge from nothing. If one object gains a negative charge, another part of the system has lost that charge. In metals, the mobile carriers are electrons. Conventional current is nevertheless defined using the direction of positive charge flow, so its arrow opposes electron drift in the external metal circuit.
Average current over an interval is Iavg=ΔQ/Δt. Here ΔQ is signed charge crossing a chosen section, not the total charge stored in the wire. One ampere means one coulomb per second. Voltage between nodes A and B is VAB=VA−VB. Reversing the meter leads reverses the sign. Energy transferred per charge gives V=ΔE/ΔQ when that energy-per-charge relation is constant over the transfer.
A node is a set of points connected by ideal wire. The lumped DC model assigns one potential to that node and neglects wire resistance. Ground is a chosen zero reference, not necessarily the earth. Adding the same constant to all node potentials changes no voltage difference. An ideal source maintains its stated terminal voltage; a real battery has limitations that this introductory model leaves out.
An object gains 2.0 × 10¹² electrons. Each electron carries −1.602176634 × 10⁻¹⁹ C. Multiplying gives Q≈−3.20 × 10⁻⁷ C=−0.320 μC. The negative sign records excess electrons. Losing the same number would give an equal positive charge. A coulomb is a very large number of elementary charges, so tiny-looking charge values can represent many electrons.
Across a chosen section, 6 C passes in the first 3 s and another 2 C passes in the next 2 s. The interval averages are 6/3=2 A and 2/2=1 A. Over the complete 5 s, Iavg=8/5=1.6 A. Averaging 2 A and 1 A to get 1.5 A would be wrong because the intervals have different durations. On a cumulative-charge graph, these currents are the slopes, while the height at 5 s is 8 C.
Let node A be 9 V and node B be 2 V relative to the same reference. Then VAB=7 V and VBA=−7 V. If 0.40 C of positive charge moves from A to B through a load, the decrease in electric potential energy is 0.40 × 7=2.8 J. Choosing B as the new zero gives A=7 V without changing that energy calculation.
Problems 1–6 are required; 7–8 are optional explanation challenges. Use circuit models or simulation only.
Common mistakes: treating voltage as a quantity at only one point without a reference, confusing accumulated charge with current, and drawing electron drift in the conventional-current direction. Ideal wires and ideal sources are modeling assumptions, not claims about all real components.