The questions are on the gauss's law test. Take it under real conditions before reading the mark scheme.
Quick answer: Electric flux is Φ = ∮E · dA, the field passing through a surface. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics C test on gauss's law, from flux through a tilted area to limits of Gauss’s law, with the equivalent answers to accept.
The model answers show where each mark is earned. Derivations are marked on the symmetry argument, the flux expression and the enclosed charge, so all three must appear. Numerical answers need a unit and, where relevant, a direction.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 8
Question 1 Flux through a tilted area (2 marks)
Φ = E A cosθ = 800 × 0.30 × cos 30° = 2.1 × 10² N m²/C.Marking: formula with cosθ (1); answer with unit (1).
Question 2 Net flux (3 marks)
a. Q_enc = +3.0 nC, so Φ = 3.0 × 10⁻⁹ / 8.85 × 10⁻¹² = 3.4 × 10² N m²/C.b. No change, because outside charge adds zero net flux.
Marking: (a) Q_enc (1), flux (1); (b) 1 mark.
Question 3 Inside a conductor (2 marks)
If E were not zero, free charges would experience a force and move, so equilibrium requires E = 0. A Gaussian surface just inside the metal then has zero flux, so it encloses no net charge. All net charge is on the outer surface.Marking: charges move until E = 0 (1); Gauss’s law gives charge on surface (1).
Question 4 Line of charge (3 marks)
By symmetry E is radial and constant in size on the curved side, and the end faces carry no flux. Φ = E (2πrL). The enclosed charge is λL, so E (2πrL) = λL/ε₀. Hence E = λ / (2πε₀r).Marking: symmetry argument (1); flux and Q_enc (1); result (1).
Question 5 Insulating sphere (3 marks)
Inside: E = kQr/R³ = 8.99 × 10⁹ × 9.0 × 10⁻⁹ × 0.030 / (0.060)³ = 1.1 × 10⁴ N/C. Outside: E = kQ/r² = 8.99 × 10⁹ × 9.0 × 10⁻⁹ / 0.0144 = 5.6 × 10³ N/C.Marking: Q_enc or formula inside (1); inside value (1); outside value (1).
Question 6 Charged ring (3 marks)
E = kQx/(x² + R²)^(3/2) = 8.99 × 10⁹ × 6.0 × 10⁻⁹ × 0.30 / (0.25)^(3/2) = 16.18 / 0.125 = 1.3 × 10² N/C along the axis. Opposite elements of the ring give perpendicular components that are equal and opposite, so they cancel.Marking: formula (1); value (1); symmetry explanation (1).
Question 7 Large sheet (2 marks)
E = σ/(2ε₀) = 6.0 × 10⁻⁹ / (2 × 8.85 × 10⁻¹²) = 3.4 × 10² N/C, directed away from the sheet on both sides.Marking: formula (1); value and direction (1).
Question 8 Limits of Gauss’s law (2 marks)
A short rod has no spherical, cylindrical or planar symmetry, so there is no closed surface on which E is constant and perpendicular. The law is still true, but E must be found by integrating dE = k dq/r².Marking: lack of symmetry (1); integration needed (1).
