3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A flat surface is in a region where there is an electric field. Which statement about the electric flux through the surface is correct?
Answer and reasoning
AIt is the amount of electric charge that flows through the surface in each second. A student who takes ‘flux’ to mean a flow of material picks this. Nothing flows through the surface: electric flux describes the field passing through it, and it exists where no charges move at all.
BIt is just the strength of the electric field at the surface, measured in newtons per coulomb. A student who treats flux and field as the same quantity picks this. The field is defined at each point, in N/C; the flux belongs to a whole surface, depends on its area and orientation, and is measured in N·m²/C.
CIt is the field strength times the surface’s area, whatever the surface’s orientation. A student who ignores the surface’s orientation picks this. EA is the flux only when the field is perpendicular to the surface; a tilted surface has less flux, and a surface parallel to the field has none.
DIt is a measure of how much electric field passes through the surface, in N·m²/C.Correct Electric flux describes how much electric field passes through a surface; it is a property of the whole surface, in N·m²/C. For a uniform field it is ΦE = E⃗ · A⃗ = EA cosθ, so it depends on the field, the area and the angle between the field and the area vector.
The figure shows a cross section of a closed box in a uniform electric field. Which statement gives the direction of the area vector of face L and the sign of the electric flux through face L?
Answer and reasoning
AToward +x, along the field, so the flux through face L is positive A student who points the area vector along the field picks this. The area vector is set by the surface, not the field: it is perpendicular to the face and, for a closed surface, outward, which for face L is toward −x.
BToward −x, out of the box, so the flux through face L is negativeCorrect For a closed surface, each face’s area vector is perpendicular to the face and points outward: for face L, toward −x. The field points toward +x, opposite to that area vector, so E⃗ · A⃗ < 0 and the flux through face L is negative, as it is wherever field enters a closed surface.
CToward +y, along the face, so no field passes through face L A student who draws the area vector along the face picks this. The area vector is perpendicular to the surface; the field crosses face L head-on, so this is where the flux per unit area is largest in magnitude, not zero.
DToward −x, out of the box, yet the flux through face L is still positive A student who thinks flux cannot be negative picks this. Flux is the dot product E⃗ · A⃗; here the field and the outward area vector point in opposite directions, so the flux is negative.
An open hemispherical surface of radius R, shaped like a bowl with no flat base, is in a uniform electric field of magnitude E. The field is parallel to the hemisphere’s axis of symmetry. What is the magnitude of the electric flux through the hemispherical surface?
Answer and reasoning
AπR²ECorrect Split the bowl into ring-shaped strips at angle θ from the axis, each of area 2πR² sinθ dθ, whose area vectors make angle θ with the field. ΦE = ∫0π/2 E cosθ (2πR² sinθ) dθ = πR²E. As a check, every field line that crosses the flat circle spanning the rim, of area πR², crosses the bowl once.
B2πR²E A student who multiplies the field by the whole curved area, ignoring the surface’s orientation, picks this. Over most of the bowl the field meets the surface at a slant, and near the rim it runs almost along it, so the flux is less than E times the area.
C0 A student who thinks the flux through any curved surface in a uniform field is zero picks this. Zero net flux holds for a closed surface, which field both enters and leaves; each field line that meets the open bowl crosses it once, all in the same direction.
Dπ²R²E/2 A student who uses the angle between the field and the surface, rather than the area vector, picks this: ∫E sinθ (2πR² sinθ) dθ = π²R²E/2. The dot product E⃗ · dA⃗ uses the angle between the field and the normal to the surface.
Working Measure the polar angle θ from the axis. A ring-shaped strip of the surface has dA = 2πR² sinθ dθ, and the angle between E⃗ and dA⃗ is θ. ΦE = ∫0π/2 E cosθ (2πR² sinθ dθ) = 2πR²E[sin²θ/2]0π/2 = πR²E, the flux through the flat circle spanning the rim. Distractors (sympy-checked): E times the curved area, 2πR²E; flux assumed zero for a curved surface in a uniform field; angle measured from the surface, ∫E sinθ (2πR² sinθ dθ) = π²R²E/2.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.5.A.1 Flux Fix
Flux
A measure of how much of some quantity passes through a given area. For a fluid it is the volume crossing the area per unit time; for a field, it is the amount of field passing through the area.
Electric flux, ΦE
A measure of the electric field passing through a surface. It depends on the field, the area of the surface and the surface’s orientation relative to the field. SI unit: N·m²/C (equivalently V·m).
Students often think Electric flux is the amount of charge (or of some material) flowing through a surface. In fact No. Electric flux measures how much electric field passes through a surface. Nothing moves through the surface; there is flux wherever a field crosses a surface, even where no charges are moving.
Students often think The electric flux through a surface is the strength of the electric field at the surface. In fact No. The field is defined at each point, in N/C. The flux is a property of a whole surface: the field combined with the surface’s area and orientation, in N·m²/C.
8.5.A.2 Area vector, A⃗ Fix
Area vector, A⃗
A vector whose magnitude is the area of a flat surface and whose direction is perpendicular to the surface. For a closed surface it points outward from the enclosed region; for an open surface either perpendicular direction may be chosen, and the choice must be stated.
Flux of a uniform field through a flat surface
ΦE = E⃗ · A⃗ = EA cosθ, where θ is the angle between the field and the area vector (the normal to the surface), not the angle between the field and the surface itself. The flux is largest when the field is perpendicular to the surface and zero when the field is parallel to it.
Sign of electric flux
Given by the dot product E⃗ · A⃗: positive when the field has a component along the area vector, negative when it has a component opposite to it, and zero when the field is parallel to the surface. For a closed surface, flux is positive where field leaves and negative where it enters.
Net flux through a closed surface
The sum of the fluxes through all parts of a closed surface, each found with its outward area vector. It is positive when more field leaves than enters and negative when more enters than leaves.
Students often think The angle in ΦE = EA cosθ is the angle between the field and the surface itself. In fact No. θ is the angle between the field and the area vector, which is perpendicular to the surface. When the field is perpendicular to the surface, θ = 0 and the flux is largest; when the field runs along the surface, θ = 90° and the flux is zero.
Students often think The flux through a surface is the field strength times the surface’s area, whatever the surface’s orientation. In fact Only when the field is perpendicular to the surface. In general only the component of the field along the area vector passes through the surface: ΦE = EA cosθ. A tilted surface has less flux, and a surface parallel to the field has none.
8.5.A.3 Area element, dA⃗ Fix
Area element, dA⃗
A small patch of a surface, with magnitude dA and direction perpendicular to the surface at that point. Useful shapes: a thin ring of a flat disk, dA = 2πr dr; a thin ring-shaped strip of a sphere of radius R, at angle θ from its axis, dA = 2πR² sinθ dθ.
Flux as a surface integral
ΦE = ∫ E⃗ · dA⃗: the sum of E⃗ · dA⃗ over every area element of the surface. It is needed whenever the field changes in magnitude or direction relative to the surface from one part of the surface to another.
Students often think The flux through a surface in a non-uniform field is the field at the center of the surface (or at one convenient point) times the area. In fact Not in general. If the field changes in magnitude or direction over the surface, the flux must be found by adding E⃗ · dA⃗ over every area element, ΦE = ∫E⃗ · dA⃗.
Students often think The flux through a surface is the integral of the field’s magnitude over the surface, ∫E dA, whatever the angle at which the field meets it. In fact No. At each area element only the component of the field along dA⃗ passes through: the integrand is E cosθ dA, where θ is the angle between the field and the normal at that element.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
The figure shows an edge view of a flat rectangular surface in a uniform electric field. What is the magnitude of the electric flux through the surface?
Answer and reasoning
A12 N·m²/C A student who uses the angle between the field and the surface itself picks this: (50)(0.40)(cos 53°) = 12 N·m²/C. The angle in E⃗ · A⃗ is measured from the area vector, the normal to the surface, and here it is 37°.
B20 N·m²/C A student who ignores the tilt picks this: (50 N/C)(0.40 m²) = 20 N·m²/C. That is the flux only when the field is perpendicular to the surface; the tilted surface lets less of the field through.
C16 N·m²/CCorrect The flux is E⃗ · A⃗ = EA cosθ, where θ is the angle between the field and the area vector, which is perpendicular to the surface. The field makes 53° with the surface, so θ = 37°: ΦE = (50 N/C)(0.50 m × 0.80 m)(cos 37°) = (50)(0.40)(0.80) = 16 N·m²/C.
D50 N·m²/C A student who takes the flux to be the field strength picks this. Flux combines the field with the surface’s area and orientation: (50 N/C)(0.40 m²)(cos 37°), in N·m²/C.
Working A = (0.50 m)(0.80 m) = 0.40 m². The field makes 53° with the surface, so it makes 90° − 53° = 37° with the area vector (the normal to the surface). ΦE = EA cos 37° = (50 N/C)(0.40 m²)(0.80) = 16 N·m²/C. Distractors: angle to the surface used, EA cos 53° = (50)(0.40)(0.60) = 12 N·m²/C; orientation ignored, EA = 20 N·m²/C; flux read as the field value, 50.
The figure shows a point charge on the central axis of a flat disk. Taking the disk’s area vector to point away from the charge, what is the electric flux through the disk? Use k = 9.0 × 10⁹ N·m²/C².
Answer and reasoning
A25 N·m²/C A student who treats the field as uniform at its value at the disk’s center, kq/d² = 50 N/C, and multiplies by the area πR² picks this. Toward the rim the field is weaker and meets the disk at a slant, so the flux is much less.
B14 N·m²/C A student who integrates the field’s magnitude over the disk without the factor cosθ picks this: πkq ln(1 + R²/d²) = 14 N·m²/C. Only the component of the field along the area vector passes through the disk, and toward the rim that component is small.
C11 N·m²/CCorrect The field varies over the disk in magnitude and direction, so integrate over rings of area 2πr dr. At radius r, E⃗ · dA⃗ = [kq/(r² + d²)][d/√(r² + d²)](2πr dr). From r = 0 to R this gives ΦE = 2πkq[1 − d/√(d² + R²)] = 2π(9.0 × 10⁹)(0.50 × 10⁻⁹)(1 − 0.30/0.50) = 11 N·m²/C.
D50 N·m²/C A student who takes the flux to be the field at the disk’s center, kq/d² = 50 N/C, picks this. Flux is E⃗ · dA⃗ added over the whole disk, in N·m²/C, not the field at one point.
Working Divide the disk into rings of radius r and width dr: dA = 2πr dr. At a ring, the field has magnitude kq/(r² + d²) and makes angle θ with the axis (the area vector), where cosθ = d/√(r² + d²). ΦE = ∫0R kqd(2πr dr)/(r² + d²)3/2 = 2πkq[1 − d/√(d² + R²)] = 2π(9.0 × 10⁹)(0.50 × 10⁻⁹ C)(1 − 0.30/0.50) = 11 N·m²/C. Distractors: field at the center, kq/d² = 50 N/C, times πR²: 25 N·m²/C; magnitudes integrated without cosθ, πkq ln(1 + R²/d²) = 14 N·m²/C; the field at the center read as the flux, 50.
In a region of space the electric field is E⃗ = bx î, where b is a positive constant. A closed cubical surface of edge length s has two faces perpendicular to the x-axis, at x = a and x = a + s (a > 0); its other four faces are parallel to the x-axis. What is the net electric flux through the cube’s surface?
Answer and reasoning
Abs³Correct With outward area vectors, the face at x = a + s has flux +b(a + s)s², and the face at x = a has flux −bas², because there the field points into the cube, opposite to the area vector. The other four faces are parallel to the field, so no flux passes through them. The net flux is b(a + s)s² − bas² = bs³.
B0 A student who thinks the net flux through a closed surface is zero whenever field enters on one side and leaves on the other picks this. Here the field is stronger where it leaves (x = a + s) than where it enters (x = a), so the outward flux is larger.
Cb(2a+s)s² A student who treats the flux through every face as positive picks this, adding bas² and b(a + s)s². At x = a the outward area vector points toward −x, opposite to the field, so that face’s flux is −bas².
D3b(2a+s)s² A student who ignores each face’s orientation, adding the field’s magnitude times area over all six faces, picks this. The four side faces lie along the field, so no flux passes through them, and the face where the field enters has negative flux.
Working Outward area vectors. Face at x = a + s (area vector +î): flux +b(a + s)s². Face at x = a (area vector −î): flux −bas². The four other faces have area vectors perpendicular to î: flux 0. Net: b(a + s)s² − bas² = bs³. Distractors (sympy-checked): 0 (entering and leaving taken to cancel); both end faces counted positive, b(2a + s)s²; |E| integrated over all six faces, bas² + b(a + s)s² + 4bs²(a + s/2) = 3b(2a + s)s².
A small, flat surface of area A is a distance r from a point charge and faces it: the surface’s area vector points directly away from the charge. The electric flux through the surface is Φ. The surface is replaced by a small surface of area 2A, a distance 2r from the charge and also facing it. What is the electric flux through the new surface?
Answer and reasoning
A0.25Φ A student who treats flux as the field at the surface picks this: the field falls to one-quarter at twice the distance. Flux is field times area, and the new surface has twice the area, so the flux is (1/4)(2)Φ.
B2.00Φ A student who thinks a larger surface catches proportionally more flux picks this, doubling Φ because the area doubles. The new surface is also twice as far away, where the field is one-quarter as strong.
C1.00Φ A student who thinks the flux through a surface near a charge depends only on the charge picks this. A small surface intercepts only part of the charge’s field, and how much depends on its area and distance: here (1/4)(2)Φ.
D0.50ΦCorrect The surface is small and faces the charge, so the field over it is nearly uniform and parallel to its area vector: Φ = (kq/r²)A. At twice the distance the field is one-quarter as strong, and the area is twice as large: Φnew = (1/4)(2)Φ = 0.50Φ.
Working The surfaces are small, so the field is nearly uniform over each and parallel to its area vector: Φ = (kq/r²)A. New: (kq/(2r)²)(2A) = (1/4)(2)Φ = 0.50Φ. Distractors: flux treated as the field, ×1/4 → 0.25Φ; area doubled with the field taken as unchanged → 2.00Φ; flux fixed by the charge alone → 1.00Φ.
The figure shows edge views of three flat surfaces, S₁, S₂ and S₃, in a uniform electric field; each surface extends the same distance into the page. Φ₁, Φ₂ and Φ₃ are the magnitudes of the electric flux through the three surfaces. Which ranking is correct?
Answer and reasoning
AΦ₂ > Φ₁ = Φ₃ A student who ignores each surface’s orientation, taking Φ = EA, picks this: EA, 2EA and EA. S₂ is tilted, so it presents only (2A)(cos 60°) = A across the field, and no field passes through S₃, which lies along the field.
BΦ₁ = Φ₂ > Φ₃Correct Φ = EA cosθ, with θ the angle between the field and the area vector. S₁ is perpendicular to the field (θ = 0): Φ₁ = EA. S₂ has twice the area, but its area vector is 60° from the field: Φ₂ = E(2A)(cos 60°) = EA. S₃ lies along the field (θ = 90°): Φ₃ = 0.
CΦ₂ > Φ₃ > Φ₁ A student who uses the angle between the field and each surface picks this: cos 90° = 0 for S₁, (2A)(cos 30°) for S₂ and cos 0° = 1 for S₃. The angle in E⃗ · A⃗ is measured from the area vector, so the surface perpendicular to the field has the most flux per unit area.
DΦ₁ = Φ₂ = Φ₃ A student who treats flux as the field strength picks this, since the field is the same everywhere. Flux also depends on each surface’s area and on its orientation to the field.
Working Φ = EA cosθ, with θ between E⃗ and the area vector. S₁ (area A, perpendicular to the field): θ = 0, Φ₁ = EA. S₂ (area 2A, at 30° to the field): θ = 60°, Φ₂ = E(2A)(0.50) = EA. S₃ (area A, parallel to the field): θ = 90°, Φ₃ = 0. So Φ₁ = Φ₂ > Φ₃. Distractors: orientation ignored (EA, 2EA, EA) → Φ₂ > Φ₁ = Φ₃; angle to the surface used (0, 2EA cos 30° ≈ 1.7EA, EA) → Φ₂ > Φ₃ > Φ₁; flux read as the field (the same E for each) → Φ₁ = Φ₂ = Φ₃.
A uniform electric field of magnitude E points in the +x direction. A flat square of area A lies in a plane perpendicular to the field; it is not part of any closed surface. A student says that the electric flux through the square could be written either as +EA or as −EA. Which reasoning supports the student’s claim?
Answer and reasoning
AThe flux is +EA if the square is located at positive x, and it is −EA if the square is located at negative x. A student who takes the sign of the flux from the surface’s position picks this. The flux depends only on E⃗ and A⃗; in a uniform field the same square has the same flux wherever it is placed.
BThe flux is +EA if the square carries positive charge, and it is −EA if the square carries negative charge. A student who reads the sign of a flux as the sign of a charge on the surface picks this. The sign of E⃗ · A⃗ comes from the directions of the field and the area vector, not from whether the square is charged.
CAn open surface’s area vector may point either way along its normal; E⃗ · A⃗ changes sign with the choice.Correct The outward direction of the area vector is defined only for a closed surface. For an open surface, A⃗ may point along +x or along −x; with the field along +x, E⃗ · A⃗ is +EA for the first choice and −EA for the second. The sign has meaning once the direction of A⃗ is stated.
DFlux counts the charge that crosses the square, and charge can cross the square in either direction. A student who pictures flux as charge flowing through the square picks this. Nothing flows through the square; flux measures the field passing through it, and its sign comes from the direction chosen for the area vector.
A very long, thin rod lies along the z-axis and carries a uniform positive linear charge density λ. At a perpendicular distance r from the rod, its electric field points directly away from the rod and has magnitude λ/(2πε₀r). A flat rectangle lies in the plane x = d: it extends from y = −√3d to y = +√3d and has height h in the z-direction. Its area vector points in the +x direction. What is the electric flux through the rectangle?
Answer and reasoning
A0.55 λh/ε₀ A student who uses the field at the center of the rectangle, λ/(2πε₀d), and multiplies it by the whole area, 2√3dh, picks this: √3λh/(πε₀) ≈ 0.55 λh/ε₀. Away from the center the field is weaker and meets the rectangle at a slant, so E⃗ · dA⃗ must be integrated over the surface.
B0.42 λh/ε₀ A student who integrates the field’s magnitude over the rectangle, ∫(λ/(2πε₀r))h dy, without the factor cosθ = d/r, picks this: λh ln(2 + √3)/(πε₀) ≈ 0.42 λh/ε₀. Away from the center the field meets the rectangle at a slant, so only its x-component, E cosθ, passes through each strip.
C1.00 λh/ε₀ A student who thinks the flux through a surface near a charge depends only on the charge, not on the surface’s size or distance, picks this: it is the whole flux, λh/ε₀, that leaves a length h of rod. The rectangle intercepts only the field lines that leave within 60° of the +x direction, one third of them.
D0.33 λh/ε₀Correct Only the x-component of the field passes through the rectangle. At coordinate y the field makes an angle θ with the area vector, where cosθ = d/√(d² + y²). With strips dA = h dy, ΦE = ∫(λ/(2πε₀r))(d/r)h dy = (λh/(2πε₀))[arctan(y/d)] from −√3d to √3d = (λh/(2πε₀))(2π/3) = λh/(3ε₀) ≈ 0.33 λh/ε₀. The field lines that cross the rectangle are those leaving the rod within 60° of the +x direction: one third of them.
Working At the point of the rectangle with coordinate y, the distance from the rod is r = √(d² + y²), and the angle θ between E⃗ and the area vector (+x) has cosθ = d/r. A strip of width dy has dA = h dy, so dΦ = (λ/(2πε₀r))(d/r)h dy = (λhd/(2πε₀)) dy/(d² + y²). ΦE = (λhd/(2πε₀)) ∫−√3d√3d dy/(d² + y²) = (λh/(2πε₀))[arctan(y/d)]−√3d√3d = (λh/(2πε₀))(2π/3) = λh/(3ε₀) ≈ 0.33 λh/ε₀. Check: the rectangle’s edges lie 60° either side of the +x direction as seen from the rod, so it intercepts 120°/360° = 1/3 of the flux λh/ε₀ that leaves a length h of rod. Distractors (sympy-checked): field at the rectangle’s center, λ/(2πε₀d), times the whole area 2√3dh, gives √3λh/(πε₀) ≈ 0.55 λh/ε₀; field magnitude integrated without cosθ, (λh/(2πε₀))·2 asinh(√3) = λh ln(2 + √3)/(πε₀) ≈ 0.42 λh/ε₀; the whole flux from a length h of rod, λh/ε₀ = 1.00 λh/ε₀.
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
