6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
A closed surface encloses two small beads with charges +2.0 nC and −6.0 nC. A third bead, with charge −3.0 nC, is outside the surface. What is the net electric flux through the closed surface? Use ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
Answer and reasoning
A−7.9 × 10² N·m²/C A student who includes the bead outside the surface picks this: (2.0 − 6.0 − 3.0) nC/ε₀. The field of the outside bead enters the surface on one side and leaves on the other, so it adds nothing to the net flux.
B−4.5 × 10² N·m²/CCorrect Only the two enclosed beads count, added with their signs: qenc = +2.0 nC − 6.0 nC = −4.0 nC. ΦE = qenc/ε₀ = (−4.0 × 10⁻⁹ C)/(8.85 × 10⁻¹² C²/(N·m²)) = −4.5 × 10² N·m²/C. The flux is negative because the net enclosed charge is negative: more field enters the surface than leaves it.
C+9.0 × 10² N·m²/C A student who adds the enclosed charges as magnitudes picks this: (2.0 + 6.0) nC/ε₀. The −6.0 nC bead’s flux is negative and cancels part of the +2.0 nC bead’s, so qenc = −4.0 nC.
D+4.5 × 10² N·m²/C A student who thinks flux has no sign picks this: the right size, |−4.0 nC|/ε₀, but positive. With outward area vectors, a negative net enclosed charge gives a negative net flux, because more field enters than leaves.
Working qenc = +2.0 nC + (−6.0 nC) = −4.0 × 10⁻⁹ C; the outside bead does not count. ΦE = qenc/ε₀ = (−4.0 × 10⁻⁹ C)/(8.85 × 10⁻¹² C²/(N·m²)) = −4.5 × 10² N·m²/C. The negative sign means that more field enters the surface than leaves it.
A student is choosing a Gaussian surface in order to apply Gauss’s law. Which statement about Gaussian surfaces is correct?
Answer and reasoning
AA Gaussian surface is the physical surface of the object that carries the charge. A student who thinks a Gaussian surface must be a real boundary picks this. The surface is imaginary and is drawn through the point where the field is wanted, which is often in empty space away from the object.
BA Gaussian surface can be open, like a flat disk, as long as field lines cross it. A student who applies Gauss’s law to any surface the field crosses picks this. Gauss’s law relates flux to enclosed charge, and only a closed surface encloses anything; a disk encloses nothing.
CA Gaussian surface can be any closed surface, even one that passes through empty space.Correct A Gaussian surface is an imaginary, closed, three-dimensional surface. It can have any shape and can pass through empty space or through material, and Gauss’s law holds for every such surface. A shape that matches the symmetry is chosen only because it makes the flux integral easy to evaluate.
DA Gaussian surface must match the charges’ symmetry, or Gauss’s law fails. A student who confuses when Gauss’s law is useful with when it is true picks this. The law holds for every closed surface; a surface that matches the symmetry is needed only to calculate E from it.
A point charge is at the center of a spherical Gaussian surface of radius 0.10 m. The net electric flux through this surface is 6.8 × 10² N·m²/C. What is the magnitude of the electric field at a distance of 0.30 m from the charge?
Answer and reasoning
A5.4 × 10³ N/C A student who thinks the larger sphere has nine times the flux, because its area is nine times larger, picks this: 9 × (6.8 × 10²)/(4π(0.30 m)²). The enclosed charge is unchanged, so the flux is unchanged.
B7.6 × 10³ N/C A student who writes Gauss’s law as flux = kqenc picks this: q = Φ/k, and then E = kq/r² = Φ/r² = (6.8 × 10² N·m²/C)/(0.30 m)². The flux is qenc/ε₀ = 4πk qenc, so E = Φ/(4πr²).
C6.0 × 10² N/CCorrect A Gaussian sphere of radius 0.30 m around the same charge has the same flux, 6.8 × 10² N·m²/C. The field is radial with one magnitude over that sphere, so E = Φ/(4πr²) = (6.8 × 10² N·m²/C)/(4π × 0.090 m²) = 6.0 × 10² N/C.
D6.8 × 10² N/C A student who treats flux and field as the same quantity picks this, reading the flux value as the field. The flux, in N·m²/C, must be divided by the area of the sphere through the point, 4π(0.30 m)², to give the field in N/C.
Working A concentric sphere of radius 0.30 m encloses the same charge, so the flux through it is also 6.8 × 10² N·m²/C. By symmetry E is radial with one magnitude over that sphere: E = Φ/(4πr²) = (6.8 × 10² N·m²/C)/(4π(0.30 m)²) = 6.0 × 10² N/C.
For which of the following can Gauss’s law, with a suitably chosen Gaussian surface, be used to calculate the magnitude of the electric field at the stated point?
Answer and reasoning
AA very long, uniformly charged solid cylinder, at a point 3 cm from its axisCorrect A very long cylinder has cylindrical symmetry. On a coaxial Gaussian cylinder through the point, E is radial with one magnitude over the curved side and parallel to the end caps, so the flux is E(2πrℓ), whether the point is inside or outside the charged cylinder, and E follows from qenc.
BA uniformly charged solid cube, at a point 1 cm from the center of one of the cube’s faces A student who thinks any Gaussian surface shaped like the charged object will work picks this. Over a cube-shaped surface around a charged cube, the field changes in size and direction from point to point, so E cannot be taken out of the flux integral.
CA uniformly charged rod 20 cm long, at a point 3 cm from the midpoint of the rod A student who treats a rod of finite length like an infinite line picks this. Along a coaxial cylinder around a 20 cm rod the field has components along the rod and changes in size, so the curved side’s flux is not E(2πrℓ) and the end caps carry flux.
DA sphere with +Q on its upper half and −Q on its lower half, at a point far from it A student who takes zero net flux to mean zero field picks this: a concentric Gaussian sphere encloses zero net charge, which seems to give E = 0. Zero net flux means only that inward and outward flux cancel; this sphere’s field is not zero, and it varies in size and direction over the Gaussian sphere.
Working Only the very long cylinder has one of the three symmetries (cylindrical): a coaxial Gaussian cylinder through the point 3 cm from the axis has E radial and uniform over its curved side and zero flux through its ends, so E(2πrℓ) = qenc/ε₀. The cube, the 20 cm rod and the sphere with oppositely charged halves give a field that varies in size and direction over any closed surface, so Gauss’s law gives the flux but not E.
A thin, flat, insulating disk of radius R carries a surface charge density that varies with distance r from its center as σ(r) = σ₀(1 − r²/R²), where σ₀ is a positive constant. What is the total charge on the disk?
Answer and reasoning
Aπσ₀R² A student who treats the density as uniform at its central value, σ₀, picks this: σ₀ × πR². The density falls to zero at the rim, so the charge must be less than πσ₀R².
B2πσ₀R²/3 A student who averages the density over the radius, getting (2/3)σ₀, and multiplies by the area picks this. The outer rings have the most area but the smallest densities, so the mean density over the area is σ₀/2, less than the average over r.
C2σ₀R/3 A student who integrates the density over r alone, ∫0R σ₀(1 − r²/R²) dr, picks this. The units show the error: σ dr is in C/m, not C. Each ring’s area is 2πr dr.
Dπσ₀R²/2Correct Split the disk into thin rings of radius r and width dr, each of area 2πr dr and density σ₀(1 − r²/R²). Adding them: Q = ∫0R σ₀(1 − r²/R²) 2πr dr = 2πσ₀(R²/2 − R²/4) = πσ₀R²/2.
Working Thin ring of radius r and width dr: dA = 2πr dr, dq = σ(r) dA. Q = ∫0R σ₀(1 − r²/R²) 2πr dr = 2πσ₀(R²/2 − R²/4) = πσ₀R²/2. Distractors (sympy-checked): central value σ₀ over the whole area → πσ₀R²; average of σ over r, (2/3)σ₀, times πR² → 2πσ₀R²/3; ∫σ dr → 2σ₀R/3 (units C/m).
Gauss’s law is the first of Maxwell’s equations, the set of equations that together fully describe electromagnetism. Which statement about Gauss’s law is correct?
Answer and reasoning
AIt holds for highly symmetric charge distributions and fails for others. A student who confuses when Gauss’s law is true with when it is useful picks this. The law holds for every distribution; only calculating E from it directly needs spherical, cylindrical or planar symmetry.
BIt gives the field of any charged object when the Gaussian surface matches its shape. A student who thinks Gauss’s law yields E whenever the Gaussian surface copies the object’s shape picks this. The law always gives the net flux, but E can be taken out of the flux integral only with spherical, cylindrical or planar symmetry; over a cube-shaped surface around a charged cube, E varies in size and direction.
CIt is Coulomb’s law restated, so it holds for point charges but no others. A student who sees Gauss’s law as Coulomb’s law rewritten for a point charge picks this. For a point charge the two give the same field, but Gauss’s law holds for any charge distribution; it is one of the equations that fully describe electromagnetism.
DIt holds for every closed surface, whatever charges lie inside or outside it.Correct As one of Maxwell’s equations, Gauss’s law is a general law: the net flux through any closed surface equals qenc/ε₀, for any arrangement of charges. Symmetry decides only whether the law can be used on its own to calculate E, and charges outside the surface add zero net flux.
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.6.A.1 Gauss’s law Fix
Gauss’s law
The net electric flux through any closed surface equals the net charge enclosed by the surface divided by ε₀: ΦE = ∮ E⃗ · dA⃗ = qenc/ε₀. The same flux is obtained whatever the shape of the surface, as long as it encloses the same charge.
Enclosed charge, qenc
The algebraic (signed) sum of all the charge inside a closed surface, in C. Charges outside the surface add nothing to the net flux through it, although they do change the electric field at points on the surface.
Net electric flux through a closed surface
ΦE = ∮ E⃗ · dA⃗, with each area element dA⃗ pointing outward. Flux is positive where the field leaves the surface and negative where it enters. SI unit: N·m²/C (equivalently V·m).
Vacuum permittivity, ε₀
The constant in Gauss’s law, ε₀ = 8.85 × 10⁻¹² C²/(N·m²). It is related to the Coulomb constant by k = 1/(4πε₀), so qenc/ε₀ = 4πk qenc.
Students often think Every charge nearby adds to the net flux through a closed surface, so the net flux depends on charges outside the surface as well as inside it. In fact No. The field of a charge outside a closed surface enters the surface and leaves it again, so its inward and outward flux cancel. The net flux depends only on the enclosed charge, although outside charges do change the field at points on the surface.
Students often think Zero net flux through a closed surface means that the electric field is zero at every point on the surface. In fact No. Zero net flux means that the inward and outward flux cancel. The field at points on the surface can be large, for example on a surface around a positive charge and an equal negative charge.
8.6.A.2 Gaussian surface Fix
Gaussian surface
An imaginary, closed, three-dimensional surface to which Gauss’s law is applied. It has an inside and an outside and no edge; it may pass through empty space or through charged material, and it need not coincide with any object. An open surface, such as a disk or a hemisphere, is not a Gaussian surface.
Students often think Gauss’s law applies to any surface that surrounds or faces a charge, open or closed, so the flux through such a surface is q/ε₀. In fact No. Gauss’s law is a statement about closed surfaces. The flux through an open surface depends on how much of the field crosses it; for a hemisphere centered on a point charge q it is q/(2ε₀), half of the flux through the whole sphere.
Students often think A Gaussian surface is the physical surface of the charged object, so the field found with Gauss’s law is the field at that surface. In fact No. A Gaussian surface is imaginary. It is drawn through the point where the field is wanted, which may be in empty space or inside the charged material.
8.6.A.3 Independence of flux from the size of the Gaussian surface Fix
Independence of flux from the size of the Gaussian surface
If the enclosed charge stays the same, the net flux is the same for a large or a small Gaussian surface. For a point charge, E falls as 1/r² while the area of a concentric sphere grows as r², so E(4πr²) = q/ε₀ for every r. If a different surface encloses a different charge, the flux changes with the enclosed charge.
Students often think A larger closed surface around the same charge has more flux through it, because more area is exposed to the field. In fact No. A larger surface has more area, but the field at the surface is weaker in the same proportion (for a point charge, E ∝ 1/r² while the area of a concentric sphere ∝ r²), so the net flux stays qenc/ε₀.
Students often think The flux through a closed surface follows the strength of the field at the surface, so a surface where the field is stronger has more flux and one where it is weaker has less. In fact No. Close to the charge the field is stronger, but the surface’s area is smaller, and the two changes cancel: while the enclosed charge is unchanged, so is the flux.
8.6.A.4 Choosing a Gaussian surface using symmetry Fix
Choosing a Gaussian surface using symmetry
A Gaussian surface is built so that on each part of it the field is either perpendicular to the surface with the same magnitude everywhere on that part, or parallel to the surface (zero flux). The flux integral then becomes E times the area where E is perpendicular. This works for spherical, cylindrical and planar symmetry.
Spherical symmetry
For a charge distribution that depends only on the distance r from a center, use a concentric sphere of radius r: E(4πr²) = qenc/ε₀. Outside the distribution the field is that of a point charge at the center; inside a uniformly charged solid sphere, E = kQr/R³.
Cylindrical symmetry
For a very long (infinite) distribution that depends only on the distance r from an axis, use a coaxial cylinder of radius r and length ℓ. The field is radial, so the end caps carry no flux: E(2πrℓ) = qenc/ε₀. For a line of charge, E = λ/(2πε₀r).
Planar symmetry
For a large (infinite) plane sheet or slab, use a pillbox with end faces of area A parallel to the plane; its sides carry no flux. For a sheet with charge density σ, flux leaves through both faces: 2EA = σA/ε₀, so E = σ/(2ε₀), independent of the distance from the sheet.
Students often think Gauss’s law gives the field of any charged object if the Gaussian surface is drawn with the same shape as the object, such as a cube around a charged cube. In fact No. Gauss’s law gives the total flux, not the field at each point. E can be taken out of the flux integral only where symmetry makes it constant in magnitude and perpendicular (or parallel) on each part of the surface: spherical, cylindrical or planar symmetry. On a cube-shaped surface around a charged cube, the field varies in size and direction.
Students often think The enclosed charge in Gauss’s law is the total charge of the object, even when the Gaussian surface lies inside the object. In fact No. Only the charge inside the Gaussian surface through the point counts. Inside a uniformly charged solid sphere, a concentric Gaussian sphere of radius r < R encloses Qr³/R³.
8.6.A.5 Charge densities λ, σ and ρ Fix
Charge densities λ, σ and ρ
Charge per unit length λ (C/m), per unit area σ (C/m²) and per unit volume ρ (C/m³). Each may be uniform or may vary with position.
Total charge from a varying density
Q = ∫ λ dℓ, ∫ σ dA or ∫ ρ dV over the length, area or volume of the distribution. With symmetry the element is a thin ring or shell: dA = 2πr dr for a disk, dV = 4πr² dr for a sphere, dV = 2πrℓ dr for a cylinder of length ℓ. The charge enclosed within radius r uses the same integral with upper limit r.
Students often think A varying charge density can be treated as uniform: the charge is one value of the density, such as its value at the center, at the edge or at the point of interest, times the volume (or area or length). In fact No. With a varying density a single value gives the wrong charge; the charge must be found by integrating ρ dV (or σ dA, or λ dℓ) over the region, each thin piece with its own density.
Students often think The total charge is the integral of the density with respect to r alone, ∫ρ dr, which is the area under a graph of ρ against r. In fact No. dr is a length, not a volume. For spherical symmetry the thin-shell volume is dV = 4πr² dr; for a cylinder of length ℓ, dV = 2πrℓ dr; for a disk, dA = 2πr dr.
8.6.A.6 Maxwell’s equations Fix
Maxwell’s equations
The set of equations that together fully describe electromagnetism. Gauss’s law, ∮ E⃗ · dA⃗ = qenc/ε₀, is the first of them: it holds for every closed surface and every arrangement of charges, even when it cannot be used on its own to calculate E.
Students often think Gauss’s law is true only for charge distributions with high symmetry; for other distributions the flux through a closed surface is not qenc/ε₀. In fact No. Gauss’s law holds for every closed surface and every charge distribution. Symmetry is needed only to use it to calculate the field, because symmetry lets E be taken out of the flux integral.
Students often think Gauss’s law is Coulomb’s law rewritten for a point charge, so it holds for point charges but not for extended charge distributions. In fact No. For a point charge, Gauss’s law gives the same field as Coulomb’s law, but Gauss’s law holds for any closed surface and any charge distribution; it is the first of Maxwell’s equations, which together fully describe electromagnetism.
13 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 13
The figure shows cross sections of three closed surfaces, S₁, S₂ and S₃, and three point charges. Φ₁, Φ₂ and Φ₃ are the net electric fluxes through S₁, S₂ and S₃. Which ranking of the fluxes is correct?
Answer and reasoning
AΦ₁ > Φ₂ > Φ₃Correct Each flux is the charge enclosed by that surface, with signs, divided by ε₀. S₁ encloses +2q, S₂ encloses +2q − q = +q, and S₃ encloses −q + q = 0, so Φ₁ = 2q/ε₀ > Φ₂ = q/ε₀ > Φ₃ = 0. The size and shape of each surface do not matter, and a charge outside a surface adds nothing to its net flux.
BΦ₂ > Φ₁ = Φ₃ A student who adds the enclosed charges as magnitudes picks this: 2q for S₁, 2q + q = 3q for S₂ and q + q = 2q for S₃. Field enters a surface toward the −q charge, so its flux is negative: S₂ encloses a net +q and S₃ a net zero.
CΦ₃ > Φ₂ > Φ₁ A student who thinks a larger closed surface catches more flux picks this, ranking the surfaces by size. For a given enclosed charge the flux does not depend on the size of the surface; what differs here is the net charge each surface encloses.
DΦ₁ = Φ₂ = Φ₃ A student who thinks every charge in the figure adds to the flux through each surface picks this: each would get (2q − q + q)/ε₀. The field of a charge outside a closed surface enters and leaves it, adding zero net flux; only the enclosed charge counts.
Working Φ = qenc/ε₀ with signed charges. S₁ encloses +2q: Φ₁ = 2q/ε₀. S₂ encloses +2q and −q: Φ₂ = q/ε₀. S₃ encloses −q and +q: Φ₃ = 0. So Φ₁ > Φ₂ > Φ₃. (Magnitudes added: 2q, 3q, 2q. Ranked by size of surface: S₃ > S₂ > S₁. Every charge counted for every surface: 2q/ε₀ each.)
A point charge +q is at the center of a thin, insulating spherical shell of radius R. The shell carries charge −3q spread uniformly over its surface. What are the magnitude and direction of the electric field at a distance r > R from the center?
Answer and reasoning
Aq/(πε₀r²), away from the center A student who adds the magnitudes of the enclosed charges picks this: q + 3q = 4q, a positive total, giving 4q/(4πε₀r²) directed outward. With signs, qenc = +q − 3q = −2q, so the field is half as large and points toward the center.
B3q/(4πε₀r²), toward the center A student who thinks the charged shell blocks the field of the charge inside it picks this, using only the shell’s −3q. Outside the shell the field depends on all the enclosed charge; the shell does not cut off the point charge’s field.
Cq/(2πε₀r²), away from the center A student who thinks flux, and so qenc, has no sign picks this: the magnitude is right, but the field is taken to point outward as for a positive charge. The net enclosed charge is −2q, so more field enters the Gaussian sphere than leaves it: the field points toward the center.
Dq/(2πε₀r²), toward the centerCorrect A concentric Gaussian sphere of radius r encloses both the point charge and the shell: qenc = +q − 3q = −2q. By symmetry E is radial and has one magnitude over the sphere, so E(4πr²) = 2q/ε₀ and E = q/(2πε₀r²). The net enclosed charge is negative, so the net flux is inward and the field points toward the center.
Working Spherical symmetry: use a concentric Gaussian sphere of radius r > R. The field is radial with one magnitude everywhere on it, so |∮E⃗·dA⃗| = E(4πr²). qenc = +q + (−3q) = −2q. E(4πr²) = 2q/ε₀, so E = 2q/(4πε₀r²) = q/(2πε₀r²); the net enclosed charge is negative, so the flux and the field point inward, toward the center. Distractors: magnitudes added (+4q) → q/(πε₀r²), outward; shell charge only (−3q) → 3q/(4πε₀r²), inward; sign of qenc ignored → q/(2πε₀r²), outward.
A point charge +q is at the center of an imaginary sphere of radius R. Consider only the upper half of the sphere’s surface: a hemisphere, open at its rim. Which claim about the magnitude of the electric flux through the hemisphere, with its justification, is correct? (k = 1/(4πε₀).)
Answer and reasoning
Aq/ε₀: the hemisphere surrounds the point charge, so Gauss’s law gives the flux through it. A student who applies Gauss’s law to an open surface picks this. The hemisphere is not closed: half of the field lines leave through the missing lower half, so only half of q/ε₀ passes through the hemisphere.
B0: an open surface encloses no charge, so by Gauss’s law zero net flux passes through it. A student who takes ‘no enclosed charge’ to mean ‘no flux’ for any surface picks this. Gauss’s law applies to closed surfaces only; field lines from the charge do cross the hemisphere, carrying flux q/(2ε₀).
Ckq/R²: the flux through a surface equals the field there, which is kq/R² on the hemisphere. A student who treats flux and field as the same quantity picks this. The field at the hemisphere is kq/R², in N/C, but flux is field times area, in N·m²/C: (kq/R²)(2πR²) = q/(2ε₀).
Dq/(2ε₀): the closed sphere has flux q/ε₀, and by symmetry each half carries half of it.Correct Gauss’s law applies to the whole sphere, a closed surface, whose flux is q/ε₀. The field is radial and has the same magnitude, kq/R², everywhere on the sphere, so the two halves carry equal flux: q/(2ε₀) each. As a check, field times area gives (kq/R²)(2πR²) = q/(2ε₀).
Working Whole sphere (closed): Φ = q/ε₀. By symmetry the field is radial with magnitude kq/R² everywhere on the sphere, so each half carries half the flux: q/(2ε₀). Check: (kq/R²)(2πR²) = q/(2ε₀).
An infinitely long line of charge with uniform linear charge density λ lies along the axis of a closed cylindrical Gaussian surface of radius r and length ℓ. The net electric flux through this surface is Φ₁. The surface is then replaced by a closed coaxial cylinder of radius 3r and length 2ℓ, and the net electric flux through the new surface is Φ₂. What is the ratio Φ₂/Φ₁?
Answer and reasoning
AΦ₂/Φ₁ = 1 A student who remembers that flux does not depend on the size of a Gaussian surface, but not the condition attached to it, picks this. That holds only if the enclosed charge is the same; the longer cylinder encloses twice as much of the line.
BΦ₂/Φ₁ = 2Correct The flux equals the enclosed charge divided by ε₀. The new cylinder encloses a length 2ℓ of the line, twice as much charge, so the flux doubles. Tripling the radius changes nothing: at the curved surface the field falls to one-third while the area per unit length triples.
CΦ₂/Φ₁ = 6 A student who thinks flux grows with the area the field crosses picks this: the curved surface has 3 × 2 = 6 times the area. The field at the larger radius is one-third as strong, so the flux grows only as the enclosed length does, by a factor of 2.
DΦ₂/Φ₁ = ⅓ A student who follows only the field at the surface picks this: E ∝ 1/r, so the field at 3r is one-third as strong. Flux is field times area, and the curved area is six times larger: (1/3)(6) = 2.
Working Φ = qenc/ε₀, with qenc = λ × (length of line inside). Original: λℓ/ε₀. New: λ(2ℓ)/ε₀. Ratio = 2. The radius does not matter: E = λ/(2πε₀r) falls as 1/r while the curved area 2πrℓ grows as r. (Curved area ×3 × 2 = ×6; field alone ×1/3; size-independence applied without its condition: ×1.)
The figure shows an edge view of a large, flat slab of insulating material with uniform volume charge density ρ, and a point P. In terms of the quantities labeled in the figure, what is the magnitude of the electric field at P?
Answer and reasoning
A2ρy/ε₀ A student who lets flux leave the pillbox through only one end face picks this: EA = ρ(2yA)/ε₀. The field points away from the central plane on both sides, so flux leaves through both faces.
Bρy/(2ε₀) A student who uses the result 2EA for a pillbox with one face on the central plane picks this: 2EA = ρ(yA)/ε₀. On the central plane the field is zero by symmetry, so that pillbox has flux EA, through its outer face only, which gives ρy/ε₀.
Cρy/(4πε₀) A student who writes Gauss’s law as flux = kqenc picks this: 2EA = kρ(2yA) gives E = kρy = ρy/(4πε₀). The flux is qenc/ε₀, which is 4πk times qenc.
Dρy/ε₀Correct Take a pillbox with end faces of area A at distance y on each side of the central plane. By symmetry the field is perpendicular to both faces, equally strong at each and pointing outward, so the flux is 2EA; the pillbox encloses ρ(2yA). Then 2EA = 2ρyA/ε₀ and E = ρy/ε₀. The thickness d does not matter while P is inside the slab.
Working Planar symmetry: the field is perpendicular to the slab, points away from the central plane on both sides, and has equal magnitude at equal distances from it. Pillbox with end faces of area A at distance y on each side of the central plane: flux 2EA (none through its sides); qenc = ρ(2yA). 2EA = 2ρyA/ε₀, so E = ρy/ε₀. Distractors: one face only, EA = 2ρyA/ε₀ → 2ρy/ε₀; 2EA with one face on the central plane, 2EA = ρyA/ε₀ → ρy/(2ε₀); flux written as kqenc, 2EA = k(2ρyA) → ρy/(4πε₀).
A very long, thin-walled, insulating cylindrical shell of radius R carries a uniform surface charge density σ. What is the magnitude of the electric field at a distance r > R from the axis of the shell?
Answer and reasoning
Aσ/ε₀ A student who uses the area of the shell itself, 2πRℓ, instead of the Gaussian surface through the point picks this. That gives the field just outside the shell; at distance r the same flux is spread over the larger area 2πrℓ.
Bσ/(2ε₀) A student who uses the field of a flat sheet for any surface charge picks this. σ/(2ε₀) is the field of a large plane sheet; the curved shell’s field depends on the distance r, as Gauss’s law with a cylindrical surface shows.
CσR/(ε₀r)Correct A coaxial Gaussian cylinder of radius r and length ℓ has flux E(2πrℓ) through its curved side and none through its ends. It encloses the charge on a length ℓ of the shell, σ(2πRℓ). So E = σR/(ε₀r): the field falls off as 1/r, like that of a line of charge with λ = 2πRσ.
DσR²/(ε₀r²) A student who assumes every field falls off as 1/r², as for a point charge, picks this: the value at the shell, σ/ε₀, scaled by (R/r)². With cylindrical symmetry the Gaussian area grows as r, so the field falls as 1/r.
Working Cylindrical symmetry: coaxial Gaussian cylinder of radius r and length ℓ. E is radial and uniform over the curved side; no flux through the ends: flux = E(2πrℓ). qenc = σ(2πRℓ). E(2πrℓ) = σ(2πRℓ)/ε₀, so E = σR/(ε₀r). Distractors: area of the shell, E(2πRℓ) = σ2πRℓ/ε₀ → σ/ε₀; sheet result → σ/(2ε₀); 1/r² fall-off from σ/ε₀ at R → σR²/(ε₀r²).
A very long, solid, insulating cylinder of radius R has a positive volume charge density ρ(s) that depends only on the distance s from its axis. Which expression gives the magnitude of the electric field at a distance r < R from the axis?
Answer and reasoning
A(1/(2πε₀r)) ∫0R ρ(s) 2πs ds A student who uses the cylinder’s whole charge as the enclosed charge picks this: the upper limit R includes charge that lies outside the Gaussian surface. Only the charge within radius r is enclosed.
B(1/(2πε₀r)) ∫0r ρ(s) 2πs dsCorrect The Gaussian cylinder must pass through the point, so its radius is r and its curved area 2πrℓ. It encloses the charge between s = 0 and s = r, found by adding thin shells of volume 2πsℓ ds, each with its own density ρ(s). Dividing qenc/ε₀ by 2πrℓ gives this expression.
C(1/(2πε₀r)) ∫0r ρ(r) 2πs ds A student who treats the density as uniform, equal to its value at the point, picks this: ρ(r) is a constant in this integral, which therefore gives ρ(r)πr² per unit length. Each thin shell must carry its own density ρ(s).
D(1/(2πε₀R)) ∫0R ρ(s) 2πs ds A student who takes the cylinder’s own surface as the Gaussian surface picks this: that surface encloses the whole charge and has radius R, so the expression is the field at the surface, r = R. The Gaussian surface must pass through the point at distance r.
Working Coaxial Gaussian cylinder of radius r and length ℓ: flux E(2πrℓ). The enclosed charge is made of thin cylindrical shells of radius s, thickness ds and volume 2πsℓ ds, each with density ρ(s), from s = 0 to s = r: qenc = ℓ ∫0r ρ(s) 2πs ds. E(2πrℓ) = qenc/ε₀, so E = (1/(2πε₀r)) ∫0r ρ(s) 2πs ds. Checked with sympy for ρ = ρ₀s/R: gives ρ₀r²/(3ε₀R), the same as a direct Gauss’s-law solution.
The figure shows an edge view of two large, parallel, insulating sheets with uniform surface charge densities as labeled. Point P is between the sheets. What is the magnitude of the electric field at P? Use ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
Answer and reasoning
A2.3 × 10³ N/CCorrect Each large sheet produces a field of magnitude σ/(2ε₀), the same at any distance. Between the sheets, the positive sheet’s field points away from it and the negative sheet’s field points toward it, so both point toward the negative sheet and add: E = (3.0 × 10⁻⁸ + 1.0 × 10⁻⁸) C/m²/(2ε₀) = 2.3 × 10³ N/C.
B1.1 × 10³ N/C A student who thinks the fields of a positive and a negative sheet partly cancel picks this: (3.0 − 1.0) × 10⁻⁸ C/m²/(2ε₀). Between the sheets both fields point toward the negative sheet, so they add; they partly cancel only outside the pair.
C4.5 × 10³ N/C A student who lets flux leave a sheet’s pillbox through only one face picks this, using σ/ε₀ for each sheet: (4.0 × 10⁻⁸ C/m²)/ε₀. A sheet’s field points away from it on both sides, so 2EA = σA/ε₀ and each sheet gives σ/(2ε₀).
D1.8 × 10² N/C A student who writes Gauss’s law as flux = kqenc picks this: 2EA = kσA gives kσ/2 for each sheet, (9.0 × 10⁹)(4.0 × 10⁻⁸)/2. The flux is qenc/ε₀ = 4πk qenc, so each sheet gives σ/(2ε₀).
Working Each sheet: pillbox with 2EA = σA/ε₀, so E = σ/(2ε₀), independent of distance. Positive sheet: (3.0 × 10⁻⁸ C/m²)/(2 × 8.85 × 10⁻¹²) = 1.7 × 10³ N/C, away from it (toward the negative sheet). Negative sheet: (1.0 × 10⁻⁸)/(2 × 8.85 × 10⁻¹²) = 5.6 × 10² N/C, toward it. Same direction at P: E = (3.0 + 1.0) × 10⁻⁸/(2ε₀) = 2.3 × 10³ N/C, toward the negative sheet.
A very long, solid, insulating cylinder of radius R has a uniform volume charge density. The magnitude of the electric field is E₁ at a distance R/2 from the axis and E₂ at a distance 2R from the axis. What is the ratio E₂/E₁?
Answer and reasoning
AE₂/E₁ = ¼ A student who uses the whole charge per unit length at R/2 as well picks this: E₁ = ρπR²/(2πε₀(R/2)) = ρR/ε₀, four times E₂. A Gaussian cylinder of radius R/2 encloses only the charge within R/2, one-quarter of the total.
BE₂/E₁ = 4 A student who uses the cylinder’s surface area, 2πRℓ, at both points picks this: E₁ = ρR/(8ε₀) and E₂ = ρR/(2ε₀). The area in Gauss’s law is that of the Gaussian surface through each point: 2π(R/2)ℓ and 2π(2R)ℓ.
CE₂/E₁ = 1Correct Inside, a Gaussian cylinder of radius R/2 encloses one-quarter of the charge per unit length and has half the radius: E₁ = ρR/(4ε₀). Outside, at 2R, it encloses all the charge per unit length, ρπR², spread over twice the radius: E₂ = ρR²/(2ε₀ · 2R) = ρR/(4ε₀). The two magnitudes are equal.
DE₂/E₁ = ½ A student who assumes the field outside falls off as 1/r², as for a point charge, picks this: E₂ = (ρR/(2ε₀))/4 = ρR/(8ε₀), half of E₁. Outside a long cylinder the Gaussian area grows as r, so the field falls as 1/r and E₂ = ρR/(4ε₀).
Working Gaussian cylinder of radius r, length ℓ: E(2πrℓ) = qenc/ε₀. At R/2: qenc = ρπ(R/2)²ℓ, so E₁ = ρ(R/2)/(2ε₀) = ρR/(4ε₀). At 2R: qenc = ρπR²ℓ, so E₂ = ρR²/(2ε₀ · 2R) = ρR/(4ε₀). E₂/E₁ = 1 (inside E ∝ r, outside E ∝ 1/r). Distractors: total charge at R/2 → E₁ = ρR/ε₀, ratio ¼; area 2πRℓ at both points → E₁ = ρR/(8ε₀), E₂ = ρR/(2ε₀), ratio 4; 1/r² outside → E₂ = ρR/(8ε₀), ratio ½.
The figure shows a solid, insulating sphere of radius R with charge spread uniformly through its volume, and three points P₁, P₂ and P₃; the distance of each point from the center O is given below it. E₁, E₂ and E₃ are the magnitudes of the electric field at the three points. Which ranking is correct?
Answer and reasoning
AE₁ > E₂ > E₃ A student who counts the whole charge as enclosed at P₁ picks this: kQ/(R/2)² = 4kQ/R². A Gaussian sphere through P₁ encloses only one-eighth of the charge, so E₁ = kQ/(2R²).
BE₂ > E₁ > E₃Correct Inside, a Gaussian sphere of radius r encloses Qr³/R³, so E = kQr/R³ grows in proportion to r: E₁ = kQ/(2R²). At the surface E₂ = kQ/R², the largest value. Outside, all the charge is enclosed and E = kQ/r²: E₃ = kQ/(4R²). So E₂ > E₁ > E₃.
CE₂ > E₃ > E₁ A student who thinks the field inside any charged sphere is zero picks this. That holds inside a thin charged shell, but a Gaussian sphere through P₁ inside this solid sphere encloses Q/8, so E₁ = kQ/(2R²), larger than E₃ = kQ/(4R²).
DE₃ > E₂ > E₁ A student who keeps using qenc = ρ × (volume inside the Gaussian sphere) beyond the charged sphere picks this, so that E grows with r everywhere. Outside R there is no more charge: the enclosed charge stays Q and E falls as 1/r².
Working Concentric Gaussian sphere through each point. Inside (r < R): qenc = Qr³/R³, E = kQr/R³, so E₁ = kQ/(2R²). At R: E₂ = kQ/R². Outside: E = kQ/r², so E₃ = kQ/(4R²). E₂ > E₁ > E₃. (Total charge at P₁: 4kQ/R². Zero inside: E₁ = 0. qenc = ρ × Gaussian volume outside: E ∝ r everywhere.)
A solid, insulating sphere of radius R has volume charge density ρ(r) = ρ₀(1 − r/R), where r is the distance from its center and ρ₀ is a positive constant. What is the magnitude of the electric field at a distance r < R from the center?
Answer and reasoning
Aρ₀r(4R − 3r)/(12ε₀R)Correct A Gaussian sphere of radius r encloses the charge of the thin shells from 0 to r: qenc = ∫0r ρ₀(1 − s/R) 4πs² ds = πρ₀r³(4R − 3r)/(3R). Then E(4πr²) = qenc/ε₀ gives E = ρ₀r(4R − 3r)/(12ε₀R).
Bρ₀r(1 − r/R)/(3ε₀) A student who treats the density as uniform, equal to its value at r, picks this: qenc = ρ₀(1 − r/R)(4/3)πr³. The density is larger nearer the center, so the enclosed charge must be found shell by shell.
Cρ₀(2R − r)/(8πε₀rR) A student who integrates the density over r alone, ∫0r ρ₀(1 − s/R) ds, picks this. That integral is not a charge (its units are C/m²); each thin shell has volume 4πs² ds.
Dρ₀r³(4R − 3r)/(12ε₀R³) A student who finds qenc correctly but divides by the sphere’s surface area, 4πR², picks this. The flux qenc/ε₀ passes through the Gaussian sphere through the point, whose area is 4πr².
Working qenc = ∫0r ρ₀(1 − s/R) 4πs² ds = 4πρ₀(r³/3 − r⁴/(4R)) = πρ₀r³(4R − 3r)/(3R). Gaussian sphere of radius r: E(4πr²) = qenc/ε₀, so E = ρ₀r(4R − 3r)/(12ε₀R). Check at r = R: E = ρ₀R/(12ε₀) = Q/(4πε₀R²) with Q = πρ₀R³/3. Distractors (sympy-checked): density taken as uniform at its value at r → ρ₀r(1 − r/R)/(3ε₀); ∫ρ ds → ρ₀(2R − r)/(8πε₀rR); area 4πR² → ρ₀r³(4R − 3r)/(12ε₀R³).
The graph shows the volume charge density ρ inside a charged, insulating sphere as a function of the distance r from its center; the graph ends at the sphere’s surface. What is the total charge of the sphere?
Answer and reasoning
A3.1 × 10⁻⁹ C A student who treats the density as uniform at its surface value picks this: (6.0 × 10⁻⁶ C/m³)(4/3)π(0.050 m)³. The density is less than 6.0 × 10⁻⁶ C/m³ everywhere inside the sphere, so the charge is less.
B2.4 × 10⁻⁹ CCorrect The graph is a straight line through the origin: ρ = ρ₀r/R, with ρ₀ = 6.0 × 10⁻⁶ C/m³ at R = 0.050 m. Adding thin shells of volume 4πr² dr: Q = ∫0R (ρ₀r/R) 4πr² dr = πρ₀R³ = π(6.0 × 10⁻⁶ C/m³)(0.050 m)³ = 2.4 × 10⁻⁹ C.
C1.6 × 10⁻⁹ C A student who takes the mean density as the graph’s midpoint value, 3.0 × 10⁻⁶ C/m³, picks this. Most of a sphere’s volume is at large r, where the density is highest, so the mean density is (3/4)(6.0 × 10⁻⁶ C/m³), not half of it.
D1.5 × 10⁻⁷ C A student who takes the area under the ρ–r graph as the charge picks this: (1/2)(0.050 m)(6.0 × 10⁻⁶ C/m³). That area is ∫ρ dr, in C/m², not a charge; each shell’s volume is 4πr² dr.
Working From the graph, ρ(r) = ρ₀r/R with ρ₀ = 6.0 × 10⁻⁶ C/m³ and R = 0.050 m. Q = ∫0R (ρ₀r/R) 4πr² dr = πρ₀R³ = π(6.0 × 10⁻⁶ C/m³)(0.050 m)³ = 2.4 × 10⁻⁹ C.
A thick spherical shell of insulating material has inner radius a and outer radius b. The material between a and b carries a uniform volume charge density ρ > 0, and the cavity inside radius a is empty. What is the magnitude of the electric field at a distance r from the center, where a < r < b?
Answer and reasoning
Aρ(b³ − a³)/(3ε₀r²) A student who takes the enclosed charge to be the shell’s whole charge, ρ(4/3)π(b³ − a³), picks this. A Gaussian sphere of radius r < b leaves out the material between r and b, so that charge is not enclosed.
Bρ(r³ − a³)/(3ε₀r²)Correct By symmetry the field is radial, so a concentric Gaussian sphere of radius r has flux E(4πr²). It encloses only the charge in the material between a and r, ρ(4/3)π(r³ − a³), since the cavity is empty. Gauss’s law gives E = ρ(r³ − a³)/(3ε₀r²), which is zero at r = a, where no charge is enclosed.
Cρ(r³ − a³)/(3ε₀b²) A student who uses the area of the shell’s outer surface, 4πb², in Gauss’s law picks this. The flux is through the Gaussian surface, a sphere of radius r through the point where the field is wanted, so the area is 4πr².
Dρ(b³ − a³)/(3ε₀b²) A student who takes the Gaussian surface to be the shell’s physical outer surface picks this: that sphere encloses the total charge and has area 4πb². This is the field at r = b, not at a point inside the material, where a < r < b.
Working Spherical symmetry: the field is radial and has the same magnitude at every point of a concentric sphere, so choose a spherical Gaussian surface of radius r. Flux: E(4πr²). Enclosed charge: only the material between a and r, qenc = ρ(4/3)π(r³ − a³); the cavity holds no charge. Gauss’s law: E(4πr²) = ρ(4/3)π(r³ − a³)/ε₀, so E = ρ(r³ − a³)/(3ε₀r²). Checks: E = 0 at r = a (no charge enclosed); E = ρ(b³ − a³)/(3ε₀b²) at r = b. Distractors (sympy-checked; with a = 1, r = 2, b = 3 and ρ/ε₀ = 1 the four options give 0.583, 2.17, 0.259 and 0.963): the shell’s total charge taken as enclosed, ρ(b³ − a³)/(3ε₀r²); the area of the outer surface, 4πb², used, ρ(r³ − a³)/(3ε₀b²); the Gaussian surface taken as the outer surface itself (total charge and area 4πb²), ρ(b³ − a³)/(3ε₀b²).
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
