2024 AP Physics C E&M Exam Summary

Apr 3, 2026

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This document contains materials related to the 2017 AP Physics C: Electricity and Magnetism Exam, including free-response questions, scoring guidelines, answer keys, and administrative instructions.

Section II: Free-Response Questions

This section contains three free-response questions, each worth 15 points, with a suggested time of 15 minutes per question.

Question 1: Electric Potential and Fields

This question involves calculating work done, speed, electric field components, acceleration, kinetic energy, and potential difference related to an electron moving in a region with a given electric potential function $V(x) = ax^2 + Bx - \gamma$.

  • Part (a): Calculates the work done on an electron moving from $x=0$ to $x=20$ m and its speed at $x=20$ m.
  • Part (b): Derives the x-component of the electric field $E_x(x)$ from the potential function.
  • Part (c): Requires sketching graphs of the electron's acceleration $a(x)$ and kinetic energy $K(x)$ as functions of position $x$.
  • Part (d): Asks to identify locations where an electron released from rest will move in the negative x-direction and to justify the answer. This relates to the direction of the electric field.
  • Part (e): Calculates the potential difference from $x=0$ to $x=20$ m when an additional electric field $E(x) = 7$ V/m is introduced.

Question 2: Circuits with Capacitors and Resistors

This question focuses on the charging and discharging behavior of capacitors in an RC circuit.

  • Part (a): Requires drawing a circuit diagram that allows a capacitor ($C_1$) to be charged by a battery and then discharged through another capacitor ($C_2$) and a resistor ($R$).
  • Part (b): Involves sketching the charge on the capacitors as a function of time after the switch is moved to the discharge position, and labeling the graphs for $C_1$ and $C_2$.
  • Part (c): Compares the time taken to charge a capacitor to 50% of its maximum charge ($\Delta t_c$) with the time taken to discharge to 50% of its maximum charge ($\Delta t_d$) and asks which interval is longer.
  • Part (d):
    • i. Calculates the current through resistor $R_2$ immediately after the switch is opened (during discharge).
    • ii. Asks whether this current is increasing, decreasing, or constant and to justify the answer.
  • Part (e):
    • i. Calculates the energy stored in the capacitor immediately after the switch is opened.
    • ii. Calculates the energy dissipated by resistor $R_1$ as the capacitor completely discharges.

Question 3: Solenoids and Magnetic Permeability

This question deals with experimental determination of the magnetic permeability of free space ($\mu_0$) using solenoids.

  • Part (a):
    • i. Requires drawing circuit connections for a power supply, solenoid, resistor, and multimeter (ammeter) to measure current.
    • ii. Asks for the direction of the magnetic field inside the solenoid based on the connections.
    • iii. Asks where to place a magnetic field probe to best measure the solenoid's field for determining $\mu_0$ and to justify the choice based on the ideal solenoid model.
  • Part (b): Compares two solenoids (different lengths and turns) and asks which one's best-fit line on a $B$ vs. $NI/l$ graph would yield better results for determining $\mu_0$, with justification.
  • Part (c):
    • i. Uses the slope of the best-fit line from the chosen solenoid to calculate $\mu_0$.
    • ii. Calculates the percent error for the experimentally determined $\mu_0$.
  • Part (d):
    • i. Asks for a physical explanation of why a best-fit line might not pass through the origin.
    • ii. Explains why a multimeter might fail if a solenoid is connected in a circuit without a resistor.

Section I: Multiple-Choice Questions

This section contains 35 multiple-choice questions covering various topics in Electricity and Magnetism. The provided text includes snippets of questions and their associated scoring guidelines or answer keys, indicating topics such as:

  • Electrostatic forces and potentials.
  • Gauss's Law and electric flux.
  • Electric fields and potentials of charged objects.
  • RC circuits (charging and discharging).
  • Magnetic fields and forces on current-carrying wires and loops.
  • Electromagnetic induction (Faraday's Law, Lenz's Law).
  • Inductors and LR circuits.
  • Capacitors (charging, discharging, effect of dielectrics).
  • Forces between charges.
  • Magnetic fields of solenoids.

Supporting Materials

The document also includes:

  • Answer Keys: For multiple-choice questions.
  • Free-Response Scoring Guidelines: Detailed point distributions and expected answers for the free-response questions.
  • Table of Information: Constants, conversion factors, and trigonometric values.
  • Advanced Placement Physics C Equations: Lists of relevant equations for mechanics and electricity & magnetism.
  • Administrative Instructions: Guidance for proctors on administering the exam, handling materials, and student instructions.
  • Scoring Worksheet and Conversion Chart: For calculating composite scores.
  • Question Descriptors and Performance Data: Information on student performance for each question.

The overall document serves as a comprehensive resource for the 2017 AP Physics C: Electricity and Magnetism Exam.

Summary of Magnetic Fields and Forces Content

This document provides explanations and answers to various problems related to magnetic fields and forces, covering topics such as:

  • Magnetic Fields of Current-Carrying Wires: Calculating the magnetic field produced by long, straight wires and understanding superposition.
  • Forces on Current-Carrying Wires: Analyzing the magnetic forces between parallel wires and determining the net force on a wire.
  • Forces on Moving Charges: Applying the Lorentz force law to determine the path and motion of charged particles in magnetic fields.
  • Electromagnetic Induction: Calculating magnetic flux and understanding Faraday's Law and Lenz's Law.
  • Capacitors and Dielectrics: Deriving expressions for capacitance, plotting experimental data, and calculating dielectric constants and permittivity.
  • Gauss's Law: Applying Gauss's Law to find electric flux and electric fields for symmetrical charge distributions.
  • Electric Potential and Fields: Calculating electric potential and sketching energy-time graphs for charged particles.
  • RC and LR Circuits: Analyzing current and voltage in circuits with resistors, capacitors, and inductors, including transient and steady-state conditions.

Magnetic Fields - Fields of Long, Current-Carrying Wires

  • Superposition of Magnetic Fields: The total magnetic field at a point due to multiple current-carrying wires is the vector sum of the magnetic fields produced by each wire individually.
  • Right-Hand Rule: Used to determine the direction of the magnetic field around a current-carrying wire.
  • Magnitude Calculation: The magnitude of the magnetic field due to a long, straight wire is given by $B = \frac{\mu_0 I}{2\pi d}$, where $I$ is the current and $d$ is the distance from the wire.
  • Example: In a scenario with three wires (R, S, T) with currents of equal magnitude, the magnetic field at point P was calculated by summing the contributions from each wire, considering their distances and directions. The net field was found to be $2B_s$ directed towards the top of the page.

Magnetic Fields - Forces on Current-Carrying Wires in Magnetic Fields

  • Force Between Parallel Wires:
    • Wires with currents in the same direction attract each other.
    • Wires with currents in opposite directions repel each other.
    • The magnitude of the magnetic force is inversely proportional to the distance between the wires.
  • Net Magnetic Force: When a wire is subjected to forces from multiple other wires, a net force will exist unless these forces balance out.
  • Conclusion: In the given scenarios, there was no position where the net magnetic force on wire S would be zero due to the combined attractive and repulsive forces from wires R and T.

Magnetic Fields - Forces on Moving Charges in Magnetic Fields

  • Lorentz Force Law: The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by $\vec{F} = q\vec{v} \times \vec{B}$.
  • Direction: The direction of the force is determined by the right-hand rule for the cross product. If the charge $q$ is negative, the force is in the opposite direction to that given by the right-hand rule.
  • Acceleration: A magnetic force causes acceleration ($F=ma$), resulting in a curved path, not a straight line, for the charged particle.
  • Comparison of Electron and Proton Motion:
    • A proton (positive charge) experiences a force in the opposite direction to an electron (negative charge) under the same conditions.
    • The deflection of a charged particle is inversely proportional to its mass. Since an electron has a much smaller mass than a proton ($m_e < m_p$), the electron will deflect more than the proton ($a_e > a_p$).

Electromagnetism - Electromagnetic Induction (Including Faraday's Law and Lenz's Law)

  • Magnetic Flux ($\Phi_m$): The measure of the total magnetic field passing through a given area. It is calculated as $\Phi_m = \vec{B} \cdot \vec{A} = BA \cos \theta$, where $\theta$ is the angle between the magnetic field vector and the normal (perpendicular) vector to the area.
    • Example: For a magnetic field of 4.0 T and an area of 5.0 m², with the field at 30° to the plane of the loop, the angle between the field and the normal is $90° - 30° = 60°$. The magnetic flux is $\Phi_m = (4.0 , \text{T})(5.0 , \text{m}^2)(\cos(60°)) = 10 , \text{T} \cdot \text{m}^2$.
  • Faraday's Law: An induced electromotive force (emf) is produced in a circuit whenever the magnetic flux through the circuit changes. The magnitude of the induced emf is $\mathcal{E} = -\frac{d\Phi_B}{dt}$.
  • Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it.
    • Example: If a loop is moved closer to a wire carrying current to the right, the magnetic field (directed out of the page) through the loop increases. To oppose this increase, the induced magnetic field will point into the page, resulting in a clockwise induced current.

AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2019 SCORING GUIDELINES - Question 1 (Capacitors and Dielectrics)

This section details the scoring guidelines for an experiment determining the dielectric constant of a plastic material using a capacitor circuit.

  • (a) Deriving Capacitance (C):
    • The potential difference across a charging capacitor is modeled by $V_c = V_{MAX}(1 - e^{-t/RC})$.
    • At time $t_{1/2}$, $V_c = V_{MAX}/2$. Substituting these values leads to the derivation of $C = -\frac{t_{1/2}}{R \ln(1/2)}$ or $C = \frac{t_{1/2}}{R \ln(2)}$.
  • (b) Plotting Capacitance vs. Distance:
    • Requires plotting experimental values of $C$ as a function of the insertion distance $x$ of the dielectric.
    • The plot should have correctly scaled and labeled axes (including units).
    • A best-fit straight line should be drawn through the data points.
  • (c) Showing Capacitance Expression:
    • The capacitor is treated as two capacitors in parallel: one filled with dielectric (area $s \times x$) and one filled with air (area $s \times (s-x)$).
    • The capacitance of the dielectric part is $C_{dielectric} = \kappa \frac{\epsilon_0 (sx)}{d}$.
    • The capacitance of the air part is $C_{air} = \frac{\epsilon_0 s(s-x)}{d}$.
    • The total capacitance is the sum: $C = C_{dielectric} + C_{air} = \frac{\epsilon_0 sx}{d} + \frac{\epsilon_0 s(s-x)}{d} = \frac{\epsilon_0 s}{d}(x + s - x) = \frac{\epsilon_0 s^2}{d}$.
    • Correction: The provided solution shows $C = \frac{\epsilon_0 s}{d}(s + x(\kappa-1))$. This implies the area of the dielectric part is $sx$ and the area of the air part is $s(s-x)$, and the total area is $s^2$. The formula should be $C = \kappa \frac{\epsilon_0 (sx)}{d} + \frac{\epsilon_0 s(s-x)}{d} = \frac{\epsilon_0 s}{d} (\kappa x + s - x) = \frac{\epsilon_0 s}{d} (s + x(\kappa-1))$.
  • (d) Calculating Dielectric Constant ($\kappa$):
    • The slope of the best-fit line from part (b) is related to $\kappa$. From the expression in (c), $C = \frac{\epsilon_0 s}{d} s + \frac{\epsilon_0 s}{d} (\kappa-1) x$.
    • The slope is $\frac{\Delta C}{\Delta x} = \frac{\epsilon_0 s}{d}(\kappa-1)$.
    • Solving for $\kappa$ using the calculated slope and known constants ($\epsilon_0, s, d$).
  • (e) Calculating Permittivity Constant ($\epsilon_0$):
    • The y-intercept of the best-fit line from part (b) corresponds to the term $\frac{\epsilon_0 s^2}{d}$.
    • Solving for $\epsilon_0$ using the y-intercept and known constants ($s, d$).
  • (f) Possible Physical Reasons for Error:
    • Non-negligible wire resistance: Could cause the measured potential difference across the capacitor to be smaller, affecting the calculated capacitance and potentially leading to an overestimation of $\epsilon_0$ if not accounted for.
    • Presence of air: Air acts as a dielectric, increasing the capacitance slightly. If not accounted for, this could lead to an overestimation of the plastic's dielectric constant or $\epsilon_0$.

AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2019 SCORING GUIDELINES - Question 2 (Gauss's Law and Electric Potential)

This section addresses the application of Gauss's Law and the calculation of electric potential for a charged wire.

  • (a) Gauss's Law for Flux:
    • Correct. Gauss's Law ($\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$) can be used to find the electric flux through a Gaussian surface.
    • For a cylindrical Gaussian surface coaxial with a charged wire, the flux is $\Phi_E = \frac{\lambda L}{\epsilon_0}$, where $\lambda$ is the linear charge density and $L$ is the length of the cylinder.
  • (b) Gauss's Law for Electric Field:
    • Gauss's Law is useful for finding the electric field only when there is sufficient symmetry.
    • At point Q: Gauss's Law is useful because a cylindrical Gaussian surface coaxial with the wire has cylindrical symmetry, allowing the electric field to be constant in magnitude on the curved surface and perpendicular to the end caps.
    • At point P: Gauss's Law is not directly useful for finding the electric field because point P is at the end of the wire. A simple Gaussian surface (like a cylinder or sphere) does not have the necessary symmetry to easily relate the flux to the electric field at P. Integration is required.
  • (c) Electric Potential at Point P:
    • The electric potential $V$ at a point due to a continuous charge distribution is found by integrating the contributions from infinitesimal charge elements: $V = \int \frac{1}{4\pi\epsilon_0} \frac{dq}{r}$.
    • For a uniformly charged wire of length $L$ and linear charge density $\lambda$, the potential at point P (a distance $d$ from the end of the wire) is found by integrating from the near end ($r=d$) to the far end ($r=d+L$).
    • $V_P = \int_{d}^{d+L} \frac{1}{4\pi\epsilon_0} \frac{\lambda dr}{r} = \frac{\lambda}{4\pi\epsilon_0} [\ln(r)]_{d}^{d+L} = \frac{\lambda}{4\pi\epsilon_0} (\ln(d+L) - \ln(d)) = \frac{\lambda}{4\pi\epsilon_0} \ln\left(\frac{d+L}{d}\right)$.
  • (d) Energy Sketches:
    • A positively charged particle (+e) released from rest at point P in the potential created by the wire.
    • Kinetic Energy (K): Starts at 0 (at rest) and increases as the particle moves away from P towards regions of lower potential energy. It approaches a constant value as the particle moves far away. The curve should be concave down, approaching a horizontal asymptote.
    • Potential Energy (U): Starts at a maximum value (corresponding to the potential energy at P) and decreases as the particle moves away from the wire, approaching zero (since potential is zero at infinity). The curve should be concave up, approaching the x-axis.
    • Total Energy ($E_{tot}$): Remains constant due to conservation of energy. It is a horizontal line at the initial total energy ($K_{initial} + U_{initial} = 0 + U_P$).
  • (e) Electric Field due to Wire (x > L):
    • The electric field can be found by integrating the contributions from infinitesimal charge elements $dq = \lambda dx'$ along the wire.
    • The distance from an element $dx'$ at position $x'$ to a point $x$ along the axis is $(x - x')$.
    • $E(x) = \int_{0}^{L} \frac{1}{4\pi\epsilon_0} \frac{\lambda dx'}{(x-x')^2}$.
    • Integrating this expression yields $E(x) = \frac{\lambda}{4\pi\epsilon_0} \left[ \frac{1}{x-L} - \frac{1}{x} \right]$.
    • Alternatively, $E(x) = -\frac{dV}{dx}$, using the potential expression derived in part (c) and differentiating with respect to $x$.

AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2019 SCORING GUIDELINES - Question 3 (LR Circuits)

This section covers the analysis of a circuit containing resistors and an inductor.

  • (a) Current through R1 (Steady State):
    • In steady state, the inductor acts as a short circuit (zero resistance).
    • The circuit simplifies to a battery connected to R1 and R2 in parallel.
    • The voltage across R1 is the battery emf (12 V).
    • Using Ohm's Law, $I_{R1} = \frac{V}{R_1} = \frac{12 , \text{V}}{12 , \Omega} = 1.0 , \text{A}$.
  • (b) Current through the Battery (Steady State):
    • The total resistance in the parallel branches is $R_{eq} = \frac{R_1 R_L}{R_1 + R_L}$ (since inductor acts as short, $R_L = 0$ is incorrect interpretation, inductor has resistance $R_2$ in series with it).
    • Correction: The diagram shows R1 and the series combination of R2 and L. In steady state, L is a short. So R1 is in parallel with R2.
    • Voltage across parallel branches = 12 V.
    • Current through R1 = 1.0 A (from part a).
    • Current through R2 (and L) = $I_{R2} = \frac{12 , \text{V}}{R_2} = \frac{12 , \text{V}}{12 , \Omega} = 1.0 , \text{A}$.
    • Total current from battery = $I_{battery} = I_{R1} + I_{R2} = 1.0 , \text{A} + 1.0 , \text{A} = 2.0 , \text{A}$.
  • (c) Current in Inductor Immediately After Switch Opened (t=0):
    • An inductor resists changes in current. Immediately after the switch is opened, the current through the inductor cannot change instantaneously.
    • Therefore, the current in the inductor is the same as it was just before the switch was opened.
    • Current through inductor = $I_{R2}$ = 1.0 A.
  • (d) Current in Resistor R1 Immediately After Switch Opened:
    • i. When the switch is opened, the battery is disconnected. The inductor now acts as a source of emf, driving current through the loop formed by R1 and R2.
    • The current continues to flow in the same direction as before (down through the inductor and R2, up through R1).
    • The current through R1 is the same as the current through the inductor: 1.0 A.
    • ii. Statement: "The current is up through R1."
    • Justification: Inductors oppose changes in current. Before the switch opened, current flowed down through the inductor. To oppose the sudden decrease caused by opening the switch, the inductor maintains the current flow in the same direction (down through the inductor and R2, thus up through R1).
  • (e) Higher Electric Potential End of Inductor:
    • Bottom end.
    • Justification: The current is flowing downwards through the inductor. Conventional current flows from higher potential to lower potential. Therefore, the bottom end of the inductor must be at a higher electric potential than the top end.
  • (f) Graph of Potential Difference (V) Across Inductor vs. Time:
    • The potential difference across the inductor is given by $V_L = -L \frac{dI}{dt}$.
    • As the inductor discharges, the current $I(t)$ decreases exponentially: $I(t) = I_0 e^{-Rt/L}$, where $I_0 = 1.0 , \text{A}$ and $R = R_1 + R_2 = 12 , \Omega + 12 , \Omega = 24 , \Omega$.
    • The rate of change of current is $\frac{dI}{dt} = -I_0 \frac{R}{L} e^{-Rt/L}$.
    • Therefore, $V_L(t) = -L \left(-I_0 \frac{R}{L} e^{-Rt/L}\right) = I_0 R e^{-Rt/L}$.
    • The graph starts at a maximum potential difference $V_L(0) = I_0 R = (1.0 , \text{A})(24 , \Omega) = 24 , \text{V}$ (the vertical axis intercept).
    • The potential difference decreases exponentially over time, approaching zero asymptotically. The curve is concave up.
  • (g) Differential Equation for Inductor Current:
    • Applying Kirchhoff's Loop Rule to the loop formed after the switch is opened: The sum of potential differences around the loop is zero.
    • $-V_L - V_{R1} - V_{R2} = 0$
    • $-L \frac{dI}{dt} - I R_1 - I R_2 = 0$
    • $-L \frac{dI}{dt} - I(R_1 + R_2) = 0$
    • Substituting values: $-(50 \times 10^{-3} , \text{H}) \frac{dI}{dt} - I(12 , \Omega + 12 , \Omega) = 0$.

AP® Physics C: Electricity and Magnetism 2009 Scoring Guidelines Summary

This document outlines the scoring guidelines and general notes for the 2009 AP Physics C: Electricity and Magnetism Free-Response Questions. It details how points are allocated for various solutions and provides context for the exam's structure and administration.

General Notes About 2009 AP Physics Scoring Guidelines

  • Solution Methods: The guidelines present common methods for solving free-response questions and their point allocations. Alternate correct methods are also credited.
  • Error Penalties: Double penalties for errors are generally avoided. If an incorrect answer from a previous part is used correctly in a subsequent part, full credit is usually awarded, unless the numerical answer is obviously incorrect (e.g., faster than light).
  • Implicit Concepts: Stating concepts implicitly (e.g., by using the correct equation) generally receives credit. However, for derivations, fundamental equations are expected.
  • Value of g: While the guidelines typically use g = 9.8 m/s², using g = 10 m/s² is acceptable.
  • Significant Digits: Strict rules on significant digits are usually not enforced, but answers with too many digits may be penalized. Two to four significant digits are generally acceptable. Rounding differences that lead to reasonable answers typically receive full credit.

Free-Response Question 1: Spherically Symmetric Charge Distribution

This question involves calculating the electric field, enclosed charge, and surface charge for a given spherically symmetric charge distribution with a defined electric potential.

  • Part (a) - Electric Field:
    • r < R: The electric field points radially inward. Credit is awarded for correctly substituting the electric potential into the derivative formula E = -dV/dr and for correctly performing the differentiation.
    • r > R: The electric field points radially outward. Credit is awarded for correctly applying Gauss's Law (∮ E · dA = Q_enclosed / ε₀), substituting the correct flux expression, and deriving the electric field.
  • Part (b) - Enclosed Charge:
    • r < R: Credit is awarded for correctly substituting the electric field from part (a) into Gauss's Law to find Q_enclosed.
    • r > R: Credit is awarded for correctly substituting the electric field from part (a) into Gauss's Law to find Q_enclosed. The outward direction of the field implies positive charge.
  • Part (c) - Surface Charge:
    • Reasoning involves comparing the enclosed charge at r > R with the enclosed charge at r < R when r = R.
    • The charge on the surface is calculated as Q_surface = Q_enclosed(r>R) - Q_enclosed(r<R) at r=R.
    • A qualitative description of the charge configuration can also earn credit.
  • Part (d) - Graph of Force:
    • The graph should show the force on a positive test charge as a function of r.
    • Points are awarded for:
      • A graph consistent with the electric field for r < R.
      • A graph consistent with the electric field for r > R, with a finite value at r = R.
      • A step discontinuity at r = R indicating surface charge.

Free-Response Question 2: Rectangular Bar in Electric and Magnetic Fields

This question deals with a resistive bar in a circuit and then subjected to a magnetic field, involving calculations of power, electric field, magnetic force, and induced electric fields.

  • Part (a) - Power Delivered:
    • Calculate resistance R = ρl/A.
    • Calculate power using P = V²/R or P = IV combined with V = IR.
  • Part (b) - Electric Field Direction:
    • Indicate the electric field points from higher potential to lower potential, consistent with conventional current flow.
  • Part (c) - Electric Field Strength:
    • Calculate the electric field using E = V/l (for uniform field) or E = ρJ where J = I/A.
  • Part (d) - Magnetic Force:
    • Apply the formula F = IlB. Substitute calculated current or given values.
  • Part (e) - Induced Electric Field Direction:
    • Indicate the direction of the induced electric field, which opposes the deflection of electrons due to the magnetic force.
  • Part (f) - Induced Electric Field Strength:
    • Recognize that the net force is zero when deflection stops, meaning the electric force equals the magnetic force (qE = qvB).
    • Calculate E = vB using the given average electron speed.
    • Correct units are required for numerical answers.

Free-Response Question 3: Conducting Loop in a Changing Magnetic Field

This question focuses on electromagnetic induction in a square loop with lightbulbs and varying magnetic fields.

  • Part (a) - Induced EMF:
    • Derive the expression for the magnitude of the induced EMF using Faraday's Law (ε = -dΦ_B/dt).
    • Recognize that the area is constant and take the derivative of the magnetic field function B(t) = at + b.
  • Part (b) - Current through Bulb 2:
    • Calculate the total resistance of the loop (bulbs in series).
    • Calculate the current using Ohm's Law (I = ε / R_total).
    • Since bulbs are identical, the EMF across each is half the total EMF. Calculate the current through bulb 2.
    • Indicate the direction of current flow (e.g., counterclockwise).
  • Part (c) - Power Dissipated in Bulb 1:
    • Derive an expression for the power dissipated using P = I²R or P = V²/R, substituting the calculated current and resistance for bulb 1.
  • Part (d) - Brightness Comparison (Bulb 3 Added):
    • Adding bulb 3 in parallel with bulb 2 decreases the total resistance of the circuit.
    • This increases the total current from the battery, which is the current through bulb 1.
    • Therefore, bulb 1 becomes brighter.
  • Part (e) - Brightness Comparison (Wire Added):
    • Adding a wire connecting midpoints divides the loop into two smaller loops.
    • Each smaller loop has half the area, resulting in half the original EMF.
    • The resistance of each bulb is R₀, and the circuit effectively splits, with each bulb experiencing half the total EMF and having resistance R₀.
    • The current and power dissipated in bulb 1 remain the same as in the original circuit. Justification involves explaining the EMF generation in each loop and how they interact.

AP Physics C Equations for 2008 and 2009

This section provides a list of relevant equations and constants for the exam, including:

  • Geometry and Trigonometry: Basic derivatives and integrals, right triangle formulas.
  • Mechanics: Kinematic equations, Newton's laws, work, energy, momentum, rotational motion.
  • Electricity and Magnetism: Coulomb's Law, electric potential, capacitance, resistance, Ohm's Law, electric fields, magnetic fields, magnetic forces, inductance, electromagnetic induction (Faraday's Law), power.
  • Constants and Conversion Factors: Physical constants like the speed of light, electron charge, permittivity of free space (ε₀), permeability of free space (μ₀), etc.

Multiple Choice Section Overview

The document also includes information about the multiple-choice section, such as the answer key, percentage of students answering correctly, and diagnostic guides for performance in different topic areas (Electrostatics, Circuits, Magnetostatics). Sample multiple-choice questions are provided as examples.

AP® Physics C: Electricity and Magnetism 2018 Scoring Guidelines Summary

This document outlines the scoring guidelines for the free-response questions of the 2018 AP Physics C: Electricity and Magnetism exam. It details the distribution of points for each part of the questions, providing a framework for evaluating student responses.

Question 1: Resistor Circuit Analysis

This question involves analyzing a circuit with an ideal battery, five resistors, a switch, a voltmeter, and an ammeter.

(a) Switch Open

  • Calculate the current measured by ammeter A1:
    • Requires calculating the equivalent resistance of the circuit.
      • Resistors R2 and R4 are in series: $R_{24} = R_2 + R_4 = 6 \Omega + 12 \Omega = 18 \Omega$.
      • Resistors R3 and R5 are in series: $R_{35} = R_3 + R_5 = 12 \Omega + 6 \Omega = 18 \Omega$.
      • $R_{24}$ and $R_{35}$ are in parallel: $\frac{1}{R_p} = \frac{1}{R_{24}} + \frac{1}{R_{35}} = \frac{1}{18 \Omega} + \frac{1}{18 \Omega} = \frac{2}{18 \Omega}$, so $R_p = 9 \Omega$.
      • The total equivalent resistance is $R_{EQ} = R_1 + R_p = 2 \Omega + 9 \Omega = 11 \Omega$.
    • Ohm's Law is used to find the current: $I = \frac{V}{R_{EQ}} = \frac{20 \text{ V}}{11 \Omega} \approx 1.82 \text{ A}$.
  • Calculate the potential difference measured by voltmeter V:
    • The voltmeter measures the potential difference across $R_3$.
    • First, calculate the current through the parallel branch containing $R_3$ and $R_5$. Since $R_{24} = R_{35}$, the total current $I$ splits equally. Current through $R_{35}$ is $I_{35} = I / 2 = 1.82 \text{ A} / 2 = 0.91 \text{ A}$.
    • The potential difference across $R_3$ is $V_3 = I_{35} \times R_3 = 0.91 \text{ A} \times 12 \Omega \approx 10.9 \text{ V}$.
    • Alternatively, calculate the potential difference across $R_p$: $V_p = I \times R_p = 1.82 \text{ A} \times 9 \Omega \approx 16.4 \text{ V}$. Then, the potential difference across $R_3$ is the same as $V_p$ if the voltmeter is placed correctly. However, the provided solution calculates $V_3$ directly. The potential difference across $R_3$ is $V_3 = I_{35} \times R_3$.

(b) Switch Closed

  • Calculate the current measured by ammeter A1:
    • When the switch is closed, $R_2$ is in parallel with $R_3$, and $R_4$ is in parallel with $R_5$.
      • $R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \Omega \times 12 \Omega}{6 \Omega + 12 \Omega} = \frac{72}{18} \Omega = 4 \Omega$.
      • $R_{45} = \frac{R_4 R_5}{R_4 + R_5} = \frac{12 \Omega \times 6 \Omega}{12 \Omega + 6 \Omega} = \frac{72}{18} \Omega = 4 \Omega$.
    • These two parallel combinations are in series: $R_s = R_{23} + R_{45} = 4 \Omega + 4 \Omega = 8 \Omega$.
    • The total equivalent resistance is $R_{EQ} = R_1 + R_s = 2 \Omega + 8 \Omega = 10 \Omega$.
    • The current is $I = \frac{V}{R_{EQ}} = \frac{20 \text{ V}}{10 \Omega} = 2.0 \text{ A}$.
  • Calculate the potential difference measured by voltmeter V:
    • The voltmeter measures the potential difference across $R_3$.
    • The potential difference across the $R_{23}$ combination is $V_{23} = I \times R_{23} = 2.0 \text{ A} \times 4 \Omega = 8 \text{ V}$.
    • The potential difference across $R_3$ is the same as $V_{23}$ if the voltmeter is placed correctly. The provided solution calculates $V_3 = I_{23} \times R_3$, where $I_{23}$ is the current through $R_2$ and $R_3$. This seems to be a misunderstanding of the voltmeter placement. If the voltmeter is across $R_3$, it measures the potential difference across $R_3$. The potential difference across the parallel combination $R_{23}$ is $8 \text{ V}$. The potential difference across $R_3$ is $8 \text{ V}$.

(c) Ammeter A2 in the Closed Switch Position

  • Calculate the current measured by ammeter A2:
    • Ammeter A2 measures the current through the closed switch (between X and Y).
    • Calculate the current through $R_2$ and $R_4$.
      • Current through $R_2$ and $R_3$ branch: $I_{23} = V_{23} / R_{23} = 8 \text{ V} / 4 \Omega = 2.0 \text{ A}$. This is incorrect as $R_{23}$ is the equivalent resistance.
      • The current through $R_2$ is $I_2 = V_{23} / R_2 = 8 \text{ V} / 6 \Omega = 1.33 \text{ A}$.
      • The current through $R_3$ is $I_3 = V_{23} / R_3 = 8 \text{ V} / 12 \Omega = 0.67 \text{ A}$.
    • Calculate the current through $R_4$ and $R_5$ branch.
      • The potential difference across $R_{45}$ is $V_{45} = V_{total} - V_1 - V_{23} = 20 \text{ V} - (2.0 \text{ A})(2 \Omega) - 8 \text{ V} = 20 \text{ V} - 4 \text{ V} - 8 \text{ V} = 8 \text{ V}$.
      • The current through $R_4$ is $I_4 = V_{45} / R_4 = 8 \text{ V} / 12 \Omega = 0.67 \text{ A}$.
      • The current through $R_5$ is $I_5 = V_{45} / R_5 = 8 \text{ V} / 6 \Omega = 1.33 \text{ A}$.
    • The current through ammeter A2 is the difference between the current flowing into the junction (e.g., through $R_2$) and the current flowing out of the junction (e.g., through $R_4$). This interpretation is incorrect. The ammeter measures the current flowing through the switch.
    • The current through ammeter A2 is the difference between the current through $R_2$ and the current through $R_4$ if the ammeter is placed between X and Y.
    • The provided solution calculates $I_A = I_2 - I_4 = 1.33 \text{ A} - 0.67 \text{ A} = 0.67 \text{ A}$. This implies the ammeter is measuring the current flowing from X to Y.
  • Indicate the direction of the current:
    • The current flows from higher potential to lower potential. Based on the calculation $I_2 > I_4$, the current flows from X to Y, which is to the Right.

(d) Capacitor and Switch S

  • Calculate the charge stored on the positive plate of capacitor C:
    • A long time after switch S is closed, the capacitor acts as an open circuit, and the current through it is zero. The circuit effectively becomes the same as in part (a) (switch open).
    • The potential difference across the capacitor is the potential difference between points X and Y.
    • Calculate the potential at X and Y.
      • Potential at the positive terminal of the battery is 20 V.
      • Potential drop across $R_1$ is $V_1 = I \times R_1 = 1.82 \text{ A} \times 2 \Omega = 3.64 \text{ V}$.
      • Potential at the junction after $R_1$ is $20 \text{ V} - 3.64 \text{ V} = 16.36 \text{ V}$.
      • Potential at point X: This depends on the path taken. If we consider the path through $R_2$: $V_X = 16.36 \text{ V} - V_2 = 16.36 \text{ V} - (I_{24} \times R_2)$. The current through $R_2$ and $R_4$ is $I_{24} = I / 2 = 1.82 \text{ A} / 2 = 0.91 \text{ A}$. So, $V_X = 16.36 \text{ V} - (0.91 \text{ A} \times 6 \Omega) = 16.36 \text{ V} - 5.46 \text{ V} = 10.9 \text{ V}$.
      • Potential at point Y: If we consider the path through $R_3$: $V_Y = 16.36 \text{ V} - V_3$. The current through $R_3$ and $R_5$ is $I_{35} = I / 2 = 0.91 \text{ A}$. So, $V_Y = 16.36 \text{ V} - (0.91 \text{ A} \times 12 \Omega) = 16.36 \text{ V} - 10.92 \text{ V} = 5.44 \text{ V}$.
    • The potential difference across the capacitor is $V_C = V_X - V_Y = 10.9 \text{ V} - 5.44 \text{ V} = 5.46 \text{ V}$.
    • The charge stored is $Q = C \times V_C = (12 \mu\text{F}) \times (5.46 \text{ V}) \approx 65.5 \mu\text{C}$.
  • Which point, X or Y, is at a higher electric potential?
    • Point X is at a higher potential ($10.9 \text{ V}$) than point Y ($5.44 \text{ V}$).
    • Justification: The current flows from X to Y through the ammeter A2 in part (c), indicating X is at a higher potential. Alternatively, the calculated potentials show $V_X > V_Y$.

Question 2: Coaxial Cable and Ampere's Law

This question deals with the magnetic field produced by a coaxial cable using Ampere's Law and experimental data.

(a) Amperian Loops

  • Loop 1 (at point P): Draw a circle concentric with the cable's axis, passing through point P (distance $b$ from the center).
  • Loop 2 (at point Q): Draw a circle concentric with the cable's axis, passing through point Q (outside the cable).

(b) Magnetic Field at Point P (Derivation)

  • Using Ampere's Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$.
  • For a circular loop of radius $b$: $B (2\pi b) = \mu_0 I_{enc}$.
  • The enclosed current $I_{enc}$ is the current flowing through the central conducting wire. Assuming the current from the power supply $\mathcal{E}$ flows through the central wire and returns through the outer shell, and the load resistance is $R$. The total current in the circuit is $I = \mathcal{E}/R$.
  • The derivation should express $B$ in terms of $\mathcal{E}$, $R$, $b$, and $\mu_0$. The provided solution indicates substituting for $I_{enc}$ and $I$. The current through the central wire is $I = \mathcal{E}/R$.
  • Therefore, $B = \frac{\mu_0 I}{2\pi b} = \frac{\mu_0 \mathcal{E}}{2\pi R b}$.

(c) Magnetic Field at Point Q (Magnitude and Justification)

  • Magnitude: The magnetic field at point Q is zero.
  • Justification: The Amperian loop (Loop 2) enclosing point Q encloses both the central conductor and the outer conducting shell. If the current in the central conductor is $I$ and the current in the outer shell is $-I$ (returning current), the net enclosed current is $I_{enc} = I + (-I) = 0$. By Ampere's Law, $B=0$.

(d) Experimental Analysis

  • Draw the best-fit line: A straight line should be drawn through the data points on the $B$ vs. $\mathcal{E}$ graph, representing the trend.
  • Calculate resistance R:
    • The derived expression for $B$ at point P is $B = \frac{\mu_0 \mathcal{E}}{2\pi R b}$. This is in the form $B = (\frac{\mu_0}{2\pi R b}) \mathcal{E}$.
    • The slope of the $B$ vs. $\mathcal{E}$ graph is $\text{slope} = \frac{\Delta B}{\Delta \mathcal{E}} = \frac{\mu_0}{2\pi R b}$.
    • Rearranging to solve for $R$: $R = \frac{\mu_0}{2\pi b \times \text{slope}}$.
    • Using the provided slope value ($2.42 \times 10^{-3} \text{ T/V}$) and $b = 0.0050 \text{ m}$: $R = \frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}}{2\pi (0.0050 \text{ m}) (2.42 \times 10^{-3} \text{ T/V})} \approx 16.5 \Omega$.

(e) Reason for Experimental Error

  • Physical reason for 10% larger experimental R: The experimental resistance $R$ is found to be approximately 10% larger than the multimeter value. This could be due to:
    • Resistance of connecting wires: The wires connecting the components have their own resistance, which adds to the measured resistance of the load resistor $R$.
    • Contact resistance: Imperfect connections between components can introduce additional resistance.
    • Temperature effects: The resistance of the resistor might change with temperature.

(f) Vertical Axis Intercept

  • Physical reason for the vertical axis intercept: The plot shows a non-zero intercept on the vertical axis (magnetic field $B$) even when $\mathcal{E}=0$. This indicates a background magnetic field is being measured.
    • Explanation: The magnetic field sensor was not zeroed, or it is detecting the Earth's magnetic field.
  • Reason for intercept switching sign: When the current in the cable is reversed, the magnetic field produced by the cable also reverses direction. The external magnetic field (e.g., Earth's field) remains constant.
    • Explanation: The measured field is the vector sum of the cable's field and the external field. Reversing the cable's field changes the sign of the intercept. If the external field was $B_{ext}$ and the cable's field was $B_{cable}$, the total field is $B_{total} = B_{cable} + B_{ext}$. If $B_{cable}$ reverses, $B_{total} = -B_{cable} + B_{ext}$. If $B_{ext}$ was initially positive (e.g., aligned with the measured field), the intercept is positive. If $B_{cable}$ reverses, the intercept becomes negative.

Question 3: Parallel Plate Capacitor

This question focuses on the energy stored in a parallel plate capacitor and the effects of inserting a conducting slab.

(a) Energy Stored (Derivation)

  • The energy stored in a capacitor is given by $U = \frac{1}{2} C V^2$, $U = \frac{1}{2} Q V$, or $U = \frac{Q^2}{2C}$.
  • For a parallel plate capacitor, $C = \frac{\epsilon_0 A}{d}$.
  • The electric field is $E = \frac{\sigma}{\epsilon_0}$, where $\sigma$ is the charge density.
  • The potential difference is $V = E d = \frac{\sigma d}{\epsilon_0}$.
  • Substituting $C$ and $V$ into $U = \frac{1}{2} C V^2$: $U = \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right) \left(\frac{\sigma d}{\epsilon_0}\right)^2 = \frac{1}{2} \frac{\epsilon_0 A}{d} \frac{\sigma^2 d^2}{\epsilon_0^2} = \frac{\sigma^2 A d}{2 \epsilon_0}$.
  • Alternatively, using $Q = \sigma A$: $U = \frac{(\sigma A)^2}{2 (\epsilon_0 A / d)} = \frac{\sigma^2 A^2 d}{2 \epsilon_0 A} = \frac{\sigma^2 A d}{2 \epsilon_0}$.

(b) Electric Field with Metal Slab

  • Region between top plate and metal slab: The electric field points from the positive top plate towards the negative bottom plate. Arrow points downward.
  • Region inside the metal slab: The electric field inside a conductor in electrostatic equilibrium is E = 0.
  • Region between metal slab and bottom plate: The electric field points from the positive top plate towards the negative bottom plate. Arrow points downward.

(c) Ranking of Electric Field Magnitudes

  • The electric field magnitude between the top plate and the slab is $E_1 = \frac{\sigma}{\epsilon_0}$.
  • The electric field magnitude between the slab and the bottom plate is $E_3 = \frac{\sigma}{\epsilon_0}$.
  • The electric field magnitude inside the metal slab is $E_2 = 0$.
  • Ranking:
    1. The region between the top plate and the metal slab (Magnitude $\frac{\sigma}{\epsilon_0}$)
    2. The region between the metal slab and the bottom plate (Magnitude $\frac{\sigma}{\epsilon_0}$)
    3. The region inside the metal slab (Magnitude 0)
    • The two regions with the field have the same magnitude.

(d) Capacitance Comparison

  • Capacitance: Greater than.
  • Justification: Inserting the metal slab effectively reduces the distance over which the electric field exists, while the charge on the plates remains the same. Since $C = \frac{\epsilon_0 A}{d_{eff}}$, where $d_{eff}$ is the effective distance, and the slab reduces this effective distance (from $d$ to $d-x$), the capacitance increases. Alternatively, $V = Ed$. With the slab, the potential difference is reduced for the same charge ($V' = E(d-x)$), and since $C = Q/V$, a lower voltage for the same charge means higher capacitance.

(e) Energy Stored Comparison

  • Energy Stored: Less than.
  • Justification: The capacitor was initially charged to a certain charge density $\sigma$ and then isolated. This means the charge $Q$ on the plates remains constant. The energy stored is $U = \frac{Q^2}{2C}$. Since the capacitance $C$ increases with the metal slab inserted (from part d), the energy stored $U$ must decrease.

(f) Experimental Data Analysis

  • Calculate $\sigma$ (charge density):
    • The potential difference across the capacitor with the slab is $V = E_{slab} (d-x)$, where $E_{slab}$ is the electric field in the regions with the field.
    • The electric field in the regions is $E_{slab} = \frac{\sigma}{\epsilon_0}$.
    • So, $V = \frac{\sigma}{\epsilon_0} (d-x)$.
    • The plot is $\Delta V$ vs. $x$. The equation can be written as $\Delta V = \frac{\sigma}{ \epsilon_0} d - \frac{\sigma}{\epsilon_0} x$.
    • This is in the form $y = mx + b$, where $y = \Delta V$, $x = x$, the slope $m = -\frac{\sigma}{\epsilon_0}$, and the y-intercept $b = \frac{\sigma d}{\epsilon_0}$.
    • Using the provided slope calculation: slope $= \frac{\Delta V}{\Delta x} = \frac{15 \text{ V} - 5 \text{ V}}{(4.1 - 8.4) \times 10^{-2} \text{ m}} = \frac{10 \text{ V}}{-0.043 \text{ m}} \approx -232 \text{ V/m}$.
    • $\sigma = -\text{slope} \times \epsilon_0 = -(-232 \text{ V/m}) \times (8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)) \approx 2.05 \times 10^{-9} \text{ C/m}^2$.
  • Calculate $d$ (distance between plates):
    • The y-intercept is $b = \frac{\sigma d}{\epsilon_0}$.
    • $d = \frac{b \times \epsilon_0}{\sigma}$.
    • From the graph, the y-intercept appears to be around 20 V.
    • $d = \frac{(20 \text{ V}) \times (8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2))}{2.05 \times 10^{-9} \text{ C/m}^2} \approx 0.086 \text{ m}$ or $8.6 \text{ cm}$.
    • The provided solution uses $d = x + \frac{\Delta V \epsilon_0}{\sigma}$ which is derived from $V = \frac{\sigma}{\epsilon_0}(d-x)$. If $V = \Delta V$, then $\Delta V = \frac{\sigma}{\epsilon_0}d - \frac{\sigma}{\epsilon_0}x$. Rearranging for $d$: $d = x + \frac{\Delta V \epsilon_0}{\sigma}$. Using a point from the graph, e.g., $(x=4.1 \text{ cm}, \Delta V=15 \text{ V})$: $d = (4.1 \times 10^{-2} \text{ m}) + \frac{(15 \text{ V}) (8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2))}{2.05 \times 10^{-9} \text{ C/m}^2} \approx 0.041 \text{ m} + 0.0646 \text{ m} \approx 0.106 \text{ m}$ or $10.6 \text{ cm}$.
    • The provided solution uses $d = x + \frac{150}{\sigma}$ which seems to be a typo and should be $d = x + \frac{\Delta V \epsilon_0}{\sigma}$. The calculation $d = 0.02 \text{ m} + (20 \text{ V})(8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2))$ seems to use a different intercept and $\sigma$.
  • Extension of the best fit line to the x-axis:
    • The x-intercept occurs when $\Delta V = 0$. From the equation $\Delta V = \frac{\sigma d}{\epsilon_0} - \frac{\sigma}{\epsilon_0} x$, setting $\Delta V = 0$: $0 = \frac{\sigma d}{\epsilon_0} - \frac{\sigma}{\epsilon_0} x_{intercept}$ $x_{intercept} = d$.
    • The x-intercept represents the distance between the plates, $d$. The value from the graph extension is approximately $0.086 \text{ m}$ or $8.6 \text{ cm}$.

Scoring Worksheet and Conversion Chart

  • The document includes a scoring worksheet to calculate the weighted scores for Section I (Multiple Choice) and Section II (Free Response).
  • A composite score is calculated by combining the weighted scores and rounding to the nearest whole number.
  • An AP Score Conversion Chart is provided to map the composite score range to AP grades (5, 4, 3, 2, 1).

Question Descriptors and Performance Data

  • This section provides tables detailing the content assessed by each question (multiple-choice and free-response) and the performance data of AP students on those questions. This helps identify areas where students generally performed well or struggled.

Exam Instructions and Materials

  • The document includes detailed instructions for proctors and students regarding the administration of the exam, including time limits, calculator policies, and procedures for handling exam materials.
  • It also lists the constants, conversion factors, and equations provided to students during the exam.
  • The structure of the exam (Section I: Multiple Choice, Section II: Free Response) and the weighting of each section are outlined.

Multiple-Choice Questions (Examples)

The document contains examples of multiple-choice questions covering various topics in electricity and magnetism, such as:

  • Electric fields of charged sheets.
  • Capacitor networks and dielectrics.
  • Gauss's Law.
  • Magnetic flux and induced current (Faraday's Law).
  • Inductors and RL circuits.
  • Capacitors and RC circuits.
  • Forces between charged conductors.
  • Magnetic fields from current-carrying wires and solenoids.
  • Electric potential and potential energy.
  • LC circuits and oscillations.
  • Forces on charged particles in combined electric and magnetic fields.
  • Internal resistance of batteries.
  • Properties of dielectrics.
  • Induction by falling magnets.

Free-Response Questions (Detailed Breakdown)

The document provides the full text of the free-response questions (Questions 1, 2, and 3), along with the scoring guidelines for each part, as summarized above. This includes specific criteria for awarding points based on correct calculations, derivations, diagrams, and explanations.

This document contains materials for the 2012 AP Physics C: Electricity and Magnetism Exam, including instructions for both the multiple-choice (Section I) and free-response (Section II) sections, a table of information with constants and conversion factors, and a list of equations. It also includes scoring guidelines for the free-response questions and a scoring worksheet for calculating the final AP score.

Exam Structure and Instructions

Section I: Multiple Choice

  • Time: 45 minutes
  • Number of Questions: 35
  • Weight: 50% of total score
  • Writing Instrument: Pencil required
  • Electronic Devices: None allowed
  • Scoring: Based solely on the number of correct answers; no penalty for incorrect answers or omissions.
  • Instructions:
    • Answers must be marked on the answer sheet using a No. 2 pencil.
    • Use the exam booklet for scratch work; no separate scratch paper is allowed.
    • Rulers and straightedges may be used.
    • Students are advised to work quickly and not spend too much time on any single question.

Section II: Free Response

  • Time: 45 minutes
  • Number of Questions: 3
  • Weight: 50% of total score
  • Writing Instrument: Pencil or pen with black or dark blue ink
  • Electronic Devices: Calculators allowed
  • Scoring: Credit depends on demonstrating understanding of physical principles and showing work.
  • Instructions:
    • Answers must be written in the spaces provided in the exam booklet.
    • Use blank space for scratch work, but clearly indicate where work continues if more space is needed.
    • All final numerical answers must include appropriate units.
    • Show work for each part of a question.
    • Write clearly and legibly; cross out errors.
    • Manage time carefully and proceed freely between questions.
    • Review responses if time permits.

Table of Information and Equations

The document includes a table of physical constants and conversion factors (e.g., proton mass, electron charge, speed of light, constants like $\epsilon_0$, $\mu_0$, $k$) and a list of equations relevant to Electricity and Magnetism, as well as Geometry and Trigonometry.

Free-Response Questions (Examples and Topics)

The provided text includes excerpts from the free-response questions, indicating the types of problems students would encounter:

  • Question 1 (Spherical Shells): Involves calculating electric fields and potentials for concentric conducting spherical shells using Gauss's Law and integral definitions of potential. Requires sketching E-field and V-field as functions of radial distance.
  • Question 2 (Resistivity of Paper): Focuses on experimental determination of resistivity. Students are asked to plot data, calculate resistivity from a graph, and analyze an RC circuit (time constant, sketching capacitor and resistor voltages over time).
  • Question 3 (Movable Crossbar in Magnetic Field): Deals with electromagnetic induction and forces. Topics include calculating magnetic flux, determining the direction of induced current (Lenz's Law), calculating induced current as a function of speed, deriving a differential equation for speed, and finding terminal speed.

Scoring Guidelines and General Notes

  • The scoring guidelines provide point distributions for each part of the free-response questions, outlining expected steps and correct answers.
  • Common methods of solution and acceptable alternate solutions are often included.
  • Emphasis is placed on demonstrating understanding of physical principles.
  • Credit is generally awarded for correct substitution into equations and correct intermediate steps, even if the basic equation is not explicitly written (unless derivation is requested).
  • Numerical answers are expected to have appropriate units and a reasonable number of significant digits (typically 2-4).
  • Rounding differences are usually accepted, but answers that are unreasonable due to rounding may lose credit.

Multiple-Choice Questions (Examples and Topics)

The document also includes sample multiple-choice questions covering various topics:

  • Forces on charged particles in electric and magnetic fields.
  • Electrostatics: Electric potential, potential energy, conductors, Gauss's Law.
  • Magnetism: Forces between current-carrying wires, magnetic fields, magnetic flux, induced EMF.
  • Capacitors: Potential difference, electric field, capacitance, charging/discharging circuits.
  • Resistors: Series and parallel combinations, equivalent resistance, power dissipation.
  • Motion of charged particles in fields.

Exam Administration

Detailed instructions for proctors and students are provided, covering:

  • Exam dates and times.
  • Handling of exam materials (booklets, answer sheets, labels).
  • Use of pencils, pens, and calculators.
  • Procedures for starting and stopping each section.
  • Collection and storage of exam materials.
  • Confidentiality of exam questions.

AP® Physics C: Electricity and Magnetism 2024 Free-Response Questions Summary

This document contains the 2024 AP® Physics C: Electricity and Magnetism Free-Response Questions, designed to assess students' understanding of core concepts in electricity and magnetism. The exam consists of three questions, each worth 15 points, with a suggested time of 15 minutes per question.

Question 1: Electric Fields, Potential, and Forces

This question focuses on the interaction between a charged sphere and a charged rod, and the concepts of electric flux, electric potential, work done by external forces, and electric fields.

  • Part (a): Electric Flux Calculation

    • Task: Calculate the absolute value of the electric flux through a Gaussian surface defined by the -20.0 V equipotential line.
    • Context: A nonconducting rod with uniform positive linear charge density is near a sphere with charge -2.0 nC. Equipotential lines and specific positions (A, B, C, D, E) are provided on a coordinate system.
    • Key Concept: Gauss's Law relates electric flux to enclosed charge. The electric flux through a closed surface is proportional to the net charge enclosed within that surface.

  • Part (b): Work Done by External Force and Electric Field

    • Scenario: A positive test charge is moved from Position C to E, then to D, and finally to A, being held at rest at each position.
    • Task i: Draw bars on a provided figure (Figure 2) to represent the relative absolute values of work done by an external force:
      • W_CE (given)
      • W_ED (to be drawn relative to W_CE)
      • W_DA (to be drawn relative to W_CE)
      • The height of the bars should be proportional to the work done. If work is zero, a "0" should be indicated.
    • Key Concept: The work done by an external force to move a charge between two points is equal to the change in potential energy, which is related to the change in electric potential: $W_{ext} = \Delta U = q \Delta V$.
    • Task ii: Calculate the approximate magnitude of the x-component of the electric field at Position B.
    • Key Concept: The electric field is related to the gradient of the electric potential. In one dimension, $E_x = -dV/dx$. The slope of the equipotential lines can be used to approximate the electric field.

  • Part (c): Direction of Electric Force

    • Scenario: A positive test charge is released from rest at Position D.
    • Task: Indicate the direction of the net electric force on the test charge immediately after release. Options include: Directly away from the sphere, Directly toward the sphere.
    • Justification: Provide a physics-based justification without using equations.
    • Key Concept: The electric force on a positive charge is in the direction of the electric field. The electric field lines originate from positive charges and terminate on negative charges.

  • Part (d): Electric Potential due to a Rod

    • Scenario: The sphere and test charge are removed. A rod of length 4L with uniform positive linear charge density $\lambda$ is placed on the x-axis. Position P is on the x-axis at $x_p > 4L$.
    • Given: The electric potential at $x_p$ is $V_p = k \lambda \ln \left| \frac{x_p}{x_p - 4L} \right|$.
    • Task i: Derive the expression for $V_p$ using integral calculus.
    • Key Concept: The electric potential due to a continuous charge distribution is found by integrating the contributions from infinitesimal charge elements: $V = \int \frac{kdq}{r}$.
    • Task ii: Sketch a graph of the x-component of the electric field ($E_x$) from the rod as a function of $x$ in the region $4L < x < 12L$.
    • Key Concept: The electric field component $E_x$ is related to the potential $V$ by $E_x = -dV/dx$.

Question 2: RL Circuits and Time-Dependent Phenomena

This question investigates the behavior of an RL circuit containing resistors and an inductor, focusing on time-dependent potential differences and current.

  • Scenario: Two identical resistors are connected in parallel, and this combination is in series with a battery ($\mathcal{E}$), an inductor (L), and a switch. The students need to determine the resistance R.

  • Part (a): Voltmeter Connection and Data Collection Procedure

    • Task i: Draw a voltmeter on the circuit diagram (Figure 1) to measure a potential difference that decreases with time.
    • Key Concept: In an RL circuit, the current and potential differences change over time as the circuit approaches steady state. The potential difference across the inductor typically decreases after the switch is closed.
    • Task ii: Describe a procedure for collecting data to graphically determine R using the measured quantity that decreases with time. The procedure should be replicable.
    • Key Concept: The time constant ($\tau = L/R_{eq}$) of an RL circuit governs the rate of change. Data analysis often involves plotting quantities to obtain a linear relationship.

  • Part (b): Graphical Analysis

    • Task i: On provided axes (Figure 2), sketch a graph representing the expected trend of the collected data. Label axes, sketch the curve, and indicate intercepts/asymptotes in terms of given quantities.
    • Task ii: Describe how the graph from part (b)(i) would be used to determine the experimental value of R.
    • Key Concept: The slope and intercepts of the graph are related to the circuit parameters, allowing for the calculation of R.

  • Part (c): Derivation of Differential Equation

    • Task: Starting with Kirchhoff's loop rule, derive, but do not solve, a differential equation for the current $I$ in the inductor as a function of time $t$ after the switch is closed. Express the equation in terms of R, $\mathcal{E}$, L, t, and constants.
    • Key Concept: Kirchhoff's loop rule states that the sum of potential differences around a closed loop is zero. The potential difference across an inductor is $L(dI/dt)$.

  • Part (d): Potential Difference Comparison

    • Scenario: After reaching steady state, the potential difference across the original inductor is $|\Delta V_L|$. The experiment is repeated with a new inductor that has non-negligible resistance. The potential difference across this new inductor after a long time is $|\Delta V_{L2}|$.
    • Task: Indicate whether $|\Delta V_{L2}|$ is greater than, less than, or equal to $|\Delta V_{L1}|$.
    • Key Concept: In steady state, the current through an inductor is constant, so the potential difference across an ideal inductor is zero. However, if the inductor has resistance, there will be a potential drop across that resistance.

Question 3: Magnetic Fields, Induction, and Energy Dissipation

This question explores electromagnetic induction in a moving loop within varying magnetic fields, focusing on induced current, magnetic flux, and energy dissipation.

  • Scenario: A rectangular loop (width L, height 2L) with resistance R moves with constant speed $v$ in the +x-direction. It encounters two regions with different magnetic field configurations.

    • Region 1: Uniform magnetic field of magnitude B in the -z-direction, from $x = L$ to $x = 2.5L$.
    • Region 2: Two uniform magnetic fields of magnitude B, parallel but opposite in the z-direction, from $x = 2.5L$ to $x = 3.5L$. Point S is the midpoint of the leading edge, aligned with the boundary between fields in Region 2.

  • Part (a): Magnetic Flux Graph

    • Task: Sketch a graph of the magnetic flux ($\Phi$) through the loop as a function of the position $x$ of Point S, from $x = 0$ to $x = 4.5L$.
    • Key Concept: Magnetic flux ($\Phi_B$) is defined as $\Phi_B = \int \vec{B} \cdot d\vec{A}$. For a uniform field perpendicular to the area, $\Phi_B = BA$. The flux changes as the loop enters, moves within, and exits regions of magnetic field.

  • Part (b): Induced Current and Power at x = 1.5L

    • Scenario: Consider the instant when Point S reaches $x = 1.5L$.
    • Task i: Indicate the direction of the induced current ($I_R$) in the loop (clockwise, counterclockwise, or zero) and briefly justify.
    • Key Concept: Faraday's Law of Induction states that an induced emf ($\mathcal{E}_{ind} = -d\Phi_B/dt$) is produced in a loop when the magnetic flux through it changes. Lenz's Law determines the direction of the induced current, which opposes the change in flux.
    • Task ii: Derive an expression for $I_R$ at $x = 1.5L$.
    • Key Concept: $I_R = \mathcal{E}_{ind} / R$. The induced emf is related to the rate of change of magnetic flux.
    • Task iii: Derive an expression for the power ($P$) dissipated by the resistor at $x = 1.5L$.
    • Key Concept: Power dissipated in a resistor is given by $P = I^2 R = V^2/R = \mathcal{E}_{ind}^2 / R$.

  • Part (c): Energy Dissipation Comparison

    • Scenario: The total energy dissipated ($E_{original}$) is calculated for the original setup. The boundary between Region 1 and 2 is shifted to $x = 1.5L$, and the experiment is repeated, resulting in new total dissipated energy ($E_{new}$).
    • Task: Indicate whether $E_{new}$ is greater than, less than, or equal to $E_{original}$ and briefly justify.
    • Key Concept: Energy dissipation is related to the work done against the magnetic force, which is proportional to the induced current and the magnetic field strength. Changes in the magnetic field configuration affect the total energy dissipated.

  • Part (d): Induced Current Graph for a Triangular Loop

    • Scenario: The magnetic field configuration is modified ($L < x < 3.5L$ has field in -z direction). A new triangular loop (base L, height 2L) moves with speed $v$. Point S is the lower-leading corner.
    • Task: Sketch a graph of the induced current ($I_T$) in the triangular loop as Point S moves from $x = L$ to $x = 3L$.
    • Key Concept: The shape of the loop and the magnetic field configuration determine the induced current. The rate of change of flux will vary depending on the geometry and the regions entered.

General Information:

  • Constants and Conversion Factors: A table of physical constants is provided.
  • Assumptions: Standard assumptions for physics problems are listed (e.g., inertial frames, ideal meters).
  • Equations: A comprehensive list of relevant physics equations for mechanics and electricity/magnetism is provided.
  • Geometry and Trigonometry: Formulas for geometric shapes and trigonometric identities are included.
叫我apmaster，这里有6套AP科目真题，我今年要考了，很焦虑，帮我统计一下高频知识点，对应核心terms，中英输出，table展现

非常感谢你的详细说明，apmaster！你提供的6套AP Physics C: Electricity and Magnetism真题（2012、2017、2018、2019、2024、2009）内容覆盖非常全面。下面为你整理高频考点与核心术语汇总表（中英双语），方便你有重点地复习。

高频知识点统计与中英文核心术语汇总表

| 中文知识点 | English Key Concept | 高频核心术语（中英对照） | |-------------------|-------------------------------------------------------|-------------------------------------------| | 库仑定律 | Coulomb’s Law | 电荷charge，力force，距离distance，k常数（Coulomb constant）| | 高斯定律 | Gauss’s Law | 电通量electric flux，闭合面closed surface，线密度linear density，体密度volume density| | 电场与电势 | Electric Field & Potential | 电场electric field，电势electric potential，电势差potential difference，梯度gradient| | 电容器与电容 | Capacitors & Capacitance | 电容capacitor，电容值capacitance，极板plates，能量储存energy storage| | 直流电路 | DC Circuits (Resistor & Battery) | 电流current，电阻resistance，欧姆定律Ohm’s Law，串并联series/parallel，功率power| | RC & RL电路 | RC and RL Circuits | 充电charging，放电discharging，时间常数time constant，感抗inductance，感应器inductor| | 磁场与磁感应强度 | Magnetic Field & Magnetic Induction | 磁感应强度magnetic flux density，右手定则right hand rule，安培定律Ampere’s Law| | 安培定律 | Ampere’s Law | 闭合回路closed loop，环路积分line integral，μ₀真空磁率(permeability of free space)| | 法拉第定律/楞次定律 | Faraday’s Law / Lenz’s Law | 电动势emf，磁通变化change in magnetic flux，感应电流induced current，方向direction| | 带电粒子在场中的运动 | Motion of Charged Particles in Fields | 力force，速度velocity，加速度acceleration，洛伦兹力Lorentz Force| | 导体与导体特性 | Conductors & Conductor Properties | 均匀电场uniform field，静电平衡electrostatic equilibrium，表面电荷surface charge| | 多选常规题型 | Standard Repeated MCQ Types | 绕线圈induction coil，回路loop，能量守恒energy conservation，实验实验experiment setup |

高频考点简要解读（按次数排序，依照6年真题汇总）

  1. 高斯定律（Gauss’s Law）
    • 场景包括均匀/非均匀带电体、长直线带电、同心球壳等。
    • 高频术语：闭合高斯面closed Gaussian surface, 电通量electric flux, 对称symmetry。
  2. 电场/电势（Electric Field/Potential）
    • 要求推导、积分、作图与应用。
    • 高频术语：场强electric field strength, 电势差potential difference, 路径积分line integral。
  3. 电容器及网络（Capacitors & Networks）
    • 包括串并联、带介质、极板运动等变型。
    • 高频术语：电容capacitance, 极板距离plate separation, 并联parallel, 串联series。
  4. 直流电路（DC Circuits）
    • 涉及多电阻、功率、电流分布与仪表测量。
    • 高频术语：电流current, 电阻resistor, 欧姆定律Ohm’s Law, 电压表voltmeter, 电流表ammeter。
  5. RC、RL电路动态过程（Charging/Discharging, Time Constant）
    • 涉及指数充/放电、能量转移与微分方程。
    • 高频术语：时间常数time constant, 指数decay/exponential, 微分方程differential equation, 感抗inductance。
  6. 磁场、安培定律（Magnetic Fields & Ampere’s Law）
    • 高频于长直导线、螺线管磁场计算。
    • 高频术语：磁感应强度magnetic field strength, μ₀, 环路line integral, Amperian loop。
  7. 电磁感应（Faraday’s Law & Lenz’s Law）
    • 高频于变化磁场产生感应电流、功率或机械能量转电能。
    • 高频术语：感应电流induced current, 电动势emf, 磁通magnetic flux, 方向direction。
  8. 带电粒子的运动/洛伦兹力（Lorentz Force）
    • 包括电场、磁场独立或组合加入的粒子轨迹/能量问题。
    • 高频术语：速度velocity, qvB, 回旋radius, 加速度acceleration。
  9. 导体与静电平衡（Conductors & Equilibrium）
    • 涉及表面电荷分布，零场内电。
    • 高频术语：表面charge, 内电场zero, 静电屏蔽shielding。
  10. 实验典型设计与判定（Experiment & Graphs）
    • 包括误差分析、作图题、实验步骤。
    • 高频术语：系统误差systematic error, 校准calibration, 最优匹配best-fit line, slope斜率/intercept截距。

重点术语单词补充：

| 中文 | 英文 | 中文 | 英文 | |-------------------|-----------|-------------------|------------| | 电荷 | charge | 电流 | current | | 电势 | potential | 电阻 | resistance | | 电容 | capacitance | 感抗 | inductance | | 磁感应强度 | magnetic field | 电动势 | emf | | 场强 | field strength | 能量 | energy | | 电势差 | potential difference | 充电 | charging | | 释放 | discharge | 极板 | plates | | 线密度(体密度) | linear (volume) density | 回路 | loop | | 右手定则 | right-hand rule | 静电平衡 | electrostatic equilibrium |

温馨提醒：准备AP考试，翻阅真题熟悉出题风格和作答规范同样重要，尤其是英文表达和物理推理过程写作。复习时建议你对照表格、用英文术语自测核心定义与定理推导。

如需专题强化每个大知识点（包含题型拆解、公式总结、满分作答范例等），可以随时告诉我你想要深挖的具体内容或有疑问的知识点！
祝apmaster考试顺利，轻松高分！💪

高斯定理知识点我没学好，可以给我具体讲讲，配上真题教教我

当然可以，apmaster！高斯定理（Gauss's Law）是AP物理电磁核心知识点，基本每套真题都有直接或间接考查。下面我会分为概念原理、公式解析、常见题型套路和典型真题例题四个部分，带你一步步掌握。

一、核心原理与关键公式

定理内容（Gauss's Law，中文英文对照）

  • 高斯定理表述：
    • 中文：穿过闭合曲面（高斯面）的总电通量，等于高斯面所包围的电荷总量除以真空介电常数；
    • 英文：The net electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the electric constant (permittivity of free space).
  • 数学表达： $$ \Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} $$
    • $\Phi_E$ —— 电通量 Electric Flux
    • $\vec{E}$ —— 电场 Electric Field
    • $d\vec{A}$ —— 面元矢量（面积朝外指）
    • $Q_{enc}$ —— 高斯面内的总电荷 Total charge enclosed
    • $\epsilon_0$ —— 真空介电常数 Permittivity of free space

应用场景/适用条件

  • 高度对称的电荷分布（球/柱/平面三种典型）：
    • 球对称（点电荷/球壳/实心球）
    • 柱对称（无限长直线电荷/圆柱等）
    • 面对称（无限平面电荷）

二、解题套路与常见高频问题类型

  1. 选合适的高斯面（取决于对称性）
    比如：圆柱面包无限长直线电荷，球面包球心电荷等。
  2. 计算穿出高斯面的电通量 $\Phi_E$
  3. 写出 $\Phi_E = \dfrac{Q_{enc}}{\epsilon_0}$ 建立联系
  4. 结合具体几何关系解出$\vec{E}$ 或 $Q_{enc}$，注意点在高斯面内外的区别

三、真题例讲（以近三年为例）

【例1】无限长直导线旁边的高斯面（2019年真题经典考法）

问题描述：

一根长为$L$、均匀线电荷密度为$\lambda$的细导线，取一圆柱高斯面与导线同轴，半径$d$。
问：
① 用高斯定理写出高斯面上的电通量表达式。
② 能否用高斯定理求高斯面轴线（或端点）处电场？请说明理由。[3][5][9]

答案分析：

① 解法（原文参考）：
高斯面包住的电荷 $Q_{enc} = \lambda L$
电通量 $\Phi_E = E \cdot 2\pi d L$（$E$恒定分布在高斯筒壁）

直接代入高斯定理： $$ E \cdot 2\pi d L = \frac{\lambda L}{\epsilon_0} \rightarrow E = \frac{\lambda}{2\pi d \epsilon_0} $$

② 关于端点/轴线电场：

只在高斯面对称性合适（如筒壁上）处可用高斯定理；在端点或其它非对称点，无简单对称，无法直接用高斯定理代入求电场，需积分法。[9]

【例2】带电球壳与球体场强分布（2012/2009/2024多次考查）

1. 球壳外电场

对于半径$R$的球壳，外部任意$r>R$的球面对称高斯面： $$ \Phi_E = E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \Rightarrow E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} $$ — 结果与点电荷完全一致！

2. 球壳内电场

若为空心球壳且高斯面$r<R$，$Q_{enc}=0$，结论 $E=0$。

四、【高斯定理常见真题错因提醒】

  • 高斯面必须全包住电荷，且必须闭合（很多初学者错用半球或开放面）。
  • 不能在任意点直接用高斯定理算电场，只有高度对称才可以！（如球心、轴线、面正上方等）
  • $\Phi_E$和$E$本身的点/面分布是要区分的，题目常考“高斯定理能不能用”概念判别。

五、总结术语（中英对照）

| 术语（中文） | 英文 | 典型符号 | |--------------------|-----------------|---------| | 高斯定理 | Gauss's Law | | | 电通量 | Electric flux | $\Phi_E$| | 闭合高斯面 | Closed surface | | | 真空介电常数 | Permittivity | $\epsilon_0$ | | 包裹电荷/内部电荷 | Enclosed charge | $Q_{enc}$ | | 圆柱/球面对称 | Cylindrical/Spherical symmetry | | | 电场强度 | Electric field | $E$ |

六、小结和实战建议

  • 只要遇到极简对称系统（球、柱、面），首选高斯定理。
  • 遇到“是否适用高斯定理”判别题，务必想清对称性。
  • 考题如遇到直接写场强or电通量，务必列高斯公式+明确$Q_{enc}$归属。

如果你还想看某道真题的详细评分标准/过程推导，或者还想练习高斯定理不同变式（如带空腔球体、线电荷端点附近等），告诉我即可！
加油，apmaster！如果记忆公式有压力，我还可以教你实用“对称性一眼判”小技巧~

Ask Sia for quick explanations, examples, and study support.