This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to find total flux through closed surfaces. Gauss's Law states that the total electric flux through any closed surface equals Q_enclosed/ε₀, regardless of surface shape or symmetry. In this scenario, a point charge Q = +4.0 μC is at the center of a cube, so all charge is enclosed and Φ_E = Q/ε₀ = (4.0×10⁻⁶)/(8.85×10⁻¹²) = 4.52×10⁵ N·m²/C. Choice A is correct because it gives the right magnitude with proper units - the cube's size and shape don't affect the total flux, only the enclosed charge matters. Choice B is incorrect because it wrongly claims zero flux due to lack of symmetry - Gauss's Law works for any closed surface, symmetric or not. To help students: Emphasize that while symmetry helps in calculating fields, total flux depends only on enclosed charge for any closed surface. Watch for: confusion between needing symmetry to find field versus finding total flux.

This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric fields due to infinite charged planes using Gauss's Law. For an infinite plane with uniform surface charge density σ, the electric field has constant magnitude E = |σ|/(2ε₀) at any distance from the plane, with direction determined by the sign of σ. In this scenario, we have σ = -4.0×10⁻⁶ C/m², so the field magnitude is E = (4.0×10⁻⁶)/(2×8.85×10⁻¹²) = 2.26×10⁵ ≈ 2.3×10⁵ N/C. Choice A is correct because it has the right magnitude and, since σ is negative, the field points toward the plane (negative charges attract positive test charges). Choice C is incorrect because it has the wrong direction - negative surface charge creates fields pointing toward the plane, not away. To help students: Emphasize that for infinite planes, field magnitude is independent of distance and depends only on σ. Watch for: sign errors in determining field direction and forgetting the factor of 2 in the denominator.

This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding Gauss's Law and calculating electric flux through closed surfaces. Gauss's Law states that the total electric flux through any closed surface equals Q_enclosed/ε₀, regardless of the surface shape or charge location within it. In this scenario, a spherical surface encloses Q = -2.0 μC, so Φ_E = Q/ε₀ = (-2.0×10⁻⁶)/(8.85×10⁻¹²) = -2.26×10⁵ N·m²/C. Choice B is correct because it has the right magnitude, correct units (N·m²/C for flux), and negative sign (negative charge produces inward flux, counted as negative). Choice A is incorrect because it has the wrong units - flux is not measured in N/C (that's field strength). To help students: Emphasize that flux depends only on enclosed charge, not on surface size or shape, and reinforce proper units for flux. Watch for: unit confusion between field (N/C) and flux (N·m²/C), and sign errors.

This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating electric fields from point charges using Coulomb's law (which is equivalent to applying Gauss's Law to spherical symmetry). For a point charge Q in vacuum, the electric field at distance r is E = kQ/r², where k = 1/(4πε₀) = 8.99×10⁹ N·m²/C². In this scenario, Q = +3.0 μC and r = 0.40 m, so E = (8.99×10⁹)(3.0×10⁻⁶)/(0.40)² = 1.69×10⁵ ≈ 1.7×10⁵ N/C. Choice A is correct because it correctly calculates the magnitude and, since Q is positive, the field points radially outward from the charge. Choice B is incorrect because it has the wrong direction - positive charges create outward-pointing fields. To help students: Reinforce that field direction depends on charge sign: positive charges create outward fields, negative charges create inward fields. Watch for: calculation errors with scientific notation and confusion about field direction.

This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric fields for spherical shells and applying Gauss's Law. For a thin spherical shell with all charge on its surface at radius R, the electric field is zero everywhere inside (r < R) due to the shell theorem. In this scenario, we have a shell of radius 0.40 m and need the field at r = 0.20 m, which is inside the shell. Choice A is correct because it recognizes that E = 0 for any point inside a uniformly charged spherical shell, regardless of the charge magnitude or the specific location inside. Choice B is incorrect because it calculates the field as if the point were outside the shell or as if all charge were at the center. To help students: Emphasize the shell theorem - inside any spherical shell of charge, the net field from all charge elements cancels to zero. Watch for: students incorrectly applying the point charge formula inside shells or confusing this with solid spheres.