Electric Flux and Gauss's Law: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
An infinitely long straight wire lies along the z-axis and carries a uniform linear charge density λ>0; nothing else is present (ε0 is the permittivity of free space). A flat rectangle lies in the plane x=d (d>0) and is defined by −d≤y≤d, 0≤z≤L. The electric flux through this rectangle, with its normal taken along +x, is ______. Exactly one option is correct.
A2πε0λL
B2πε0λLln2
C4ε0λL
D8ε0λL
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Answer: C
Symmetry: the field is radial from the wire, E=2πε0sλ, with no z-component, so the flux through the rectangle is L times the flux per unit length through the segment x=d, −d≤y≤d of the cross-section.
In a cross-section, a line element whose ends subtend angle dϕ at the wire receives flux (per unit length) Esdϕ=2πε0λdϕ. The segment subtends 2tan−1(d/d)=π/2.
Hence Φ=2πε0λL⋅2π=4ε0λL.
Check with Gauss's law: the rectangle is one face of a square tube of side 2d and length L around the wire; the four faces share λL/ε0 equally and the end squares get no flux. Answer (C).
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A cube occupies 0≤x≤a, 0≤y≤a, 0≤z≤a. A point charge q is fixed at the midpoint (a/2,0,0) of the cube edge that lies on the x-axis; nothing else is present. Fluxes are taken with the outward normals of the cube. Let Φ1 be the electric flux through the face x=a and Φ2 the flux through the face y=a. The value of Φ1+Φ2 is ______. Exactly one option is correct.
A4ε0q
B8ε0q
C12ε0q
D16ε0q
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Answer: B
Faces y=0 and z=0 contain the charge: the field of q at their points lies in their planes, so their flux is zero.
Place four identical cubes around the edge so that the edge is their common line. Together they form a a×2a×2a box with q at its centre, so the total outward flux of the box is q/ε0, and by the fourfold symmetry about the edge each cube's four charge-free faces carry q/(4ε0).
In one cube, the reflection x→a−x (through the plane x=a/2, which contains q) maps face x=a to x=0, and the reflection y↔z maps face y=a to z=a. So 2Φ1+2Φ2=4ε0q, giving Φ1+Φ2=8ε0q.
(Individually Φ1=4πε0qtan−1(4/3) and Φ2=4πε0qtan−1(3/4), which do add to q/(8ε0); symmetry alone fixes only the sum.) Answer (B).
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
In each case below, the stated charge distribution is the only charge present. For which one can the magnitude of the electric field at every stated point be obtained from Gauss's law together with symmetry alone, without integrating Coulomb's law? Exactly one option is correct.
AA thin square plate of side L with uniform surface charge density: the field at points on its central perpendicular axis.
BA solid cylinder of finite length L and radius R with uniform volume charge density: the field at all points.
CA thin hemispherical shell with uniform surface charge density: the field at all points of the flat circular region bounded by its rim.
DAn infinite slab −d≤x≤d (unbounded in y and z) with charge density ρ=ρ0x2/d2, ρ0>0: the field at all points.
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Answer: D
Gauss's law gives a field only when symmetry makes E constant in magnitude and fixed in direction over a whole closed surface.
(A) The plate has fourfold symmetry about its axis but no surface exists on which the field is constant; the axial field needs integration.
(B) A finite cylinder has neither translational symmetry along its axis nor spherical symmetry.
(C) Mirror symmetry shows the field on the base is perpendicular to it, but its magnitude is not fixed by any Gauss surface.
(D) The slab is invariant under translations in y,z and under x→−x, so E=E(x)x^ with E(−x)=−E(x), hence E(0)=0. A pillbox from −x to x gives 2EA=ε0A∫−xxρdx′, so E=3ε0d2ρ0x3 for ∣x∣≤d and E=3ε0ρ0d outside. Answer (D).
Difficulty: Very hard
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