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AP Physics 2- 11.1 Electric Current- Study Notes- New Syllabus

AP Physics 2- 11.1 Electric Current – Study Notes

AP Physics 2- 11.1 Electric Current – Study Notes – per latest Syllabus.

Key Concepts:

  • Movement of Electric Charges Through a Medium

AP Physics 2-Concise Summary Notes- All Topics

Electric Current:

Current is the rate at which charge passes through a cross-sectional area of a wire.

\( \text{I} = \dfrac{\Delta \text{Q}}{\Delta t} \)

  • Unit: Ampere (A) = Coulomb/second
  • If the current is zero in a section of wire, the net motion of charge carriers is also zero, although individual charge carriers still move randomly.

Driving Force (Potential Difference / EMF):

Electric charge moves in a circuit in response to an electric potential difference, also referred to as electromotive force (emf).

Direction of Current:

  • Although current is not a vector quantity, it does have a direction.
  • The direction of current is defined as the direction in which positive charges would move (conventional current).
  • In reality, in metallic conductors, current results from the flow of electrons, which move in the opposite direction to conventional current.

Movement in Conductors:

In metals, charges are carried by free electrons. Electrons drift slowly under the electric field while colliding with atoms.

Drift velocity:

\( \text{v}_d = \dfrac{\text{I}}{n q A} \)

  • \( n \) = number density of charge carriers
  • \( q \) = charge of each carrier
  • \( A \) = cross-sectional area of conductor

Movement in Insulators:

Charges are tightly bound and cannot flow freely. Only polarization (shift of bound charges) occurs.

Movement in Semiconductors:

Both electrons and holes contribute to current. Conductivity depends on temperature and doping.

Movement in Electrolytes (Liquids):

Charges move as positive and negative ions. Example: In NaCl solution, \( \text{Na}^+ \) moves to cathode, \( \text{Cl}^- \) to anode.

Movement in Gases/Plasma:

Ionization of atoms creates electrons and ions that accelerate under the field. Plasma behaves as a conducting medium.

Example :

A copper wire of radius \(1.0 \, \text{mm}\) carries a steady current of \(3.0 \, \text{A}\). The number density of free electrons in copper is \(n = 8.5 \times 10^{28} \, \text{m}^{-3}\). Find the drift velocity of the electrons in the wire.

▶️ Answer/Explanation

Step 1: Area of cross section

\( \text{A} = \pi r^2 = \pi (1.0 \times 10^{-3} \, \text{m})^2 = 3.1416 \times 10^{-6} \, \text{m}^2 \).

Step 2: Drift velocity formula

\( \text{v}_d = \dfrac{\text{I}}{n \, q \, \text{A}} \)

  • \( \text{I} = 3.0 \, \text{A} \)
  • \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \)
  • \( q = 1.6 \times 10^{-19} \, \text{C} \) (magnitude of electron charge)
  • \( \text{A} = 3.1416 \times 10^{-6} \, \text{m}^2 \)

Step 3: Substitute and compute

\( \text{v}_d = \dfrac{3.0}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(3.1416 \times 10^{-6})} \)

\( \text{v}_d \approx 7.02 \times 10^{-5} \, \text{m/s} \)

Answer: Drift velocity ≈ \( 7.0 \times 10^{-5} \, \text{m/s} \) (very slow — electrons drift slowly while the electric signal propagates quickly).

Example  :

A \(12 \, \text{V}\) battery is connected across a \(4.0 \, \Omega\) resistor.

(a) Find the steady current in the circuit.

(b) How much charge passes through a cross-section in \(10 \, \text{s}\)?

(c) How many electrons does that correspond to?

(d) How much electrical energy is dissipated in the resistor in that time?

▶️ Answer/Explanation

(a) Current using Ohm’s law

\( \text{I} = \dfrac{\text{V}}{\text{R}} = \dfrac{12}{4.0} = 3.0 \, \text{A} \)

(b) Charge transferred in \(10 \, \text{s}\)

\( \Delta \text{Q} = \text{I} \Delta t = 3.0 \times 10 = 30 \, \text{C} \)

(c) Number of electrons

Number \( = \dfrac{\Delta \text{Q}}{q} = \dfrac{30}{1.6 \times 10^{-19}} \approx 1.88 \times 10^{20} \) electrons.

(d) Energy dissipated in resistor

Power \( \text{P} = \text{I}^2 \text{R} = 3.0^2 \times 4.0 = 36 \, \text{W} \).

Energy in \(10 \, \text{s}\): \( \text{E} = \text{P} \Delta t = 36 \times 10 = 360 \, \text{J} \).

Answer: (a) \( 3.0 \, \text{A} \). (b) \( 30 \, \text{C} \). (c) \( 1.88 \times 10^{20} \) electrons. (d) \( 360 \, \text{J} \) dissipated as heat.

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