A student wants to use Gauss's law to find the electric field at distance r from an infinitely long wire carrying uniform linear charge density λ. She draws a spherical Gaussian surface of radius r centered on the wire. Why will this fail to give her the field directly?
AGauss's law only applies to spherical charge distributions, not line charges
BThe spherical surface does not enclose the correct amount of charge
COn a spherical surface, E is not uniform in magnitude and not everywhere perpendicular to the surface, so the flux integral cannot be simplified
DGauss's law requires the surface to extend to infinity for infinite charge distributions
Gauss's law is always valid — the flux through any closed surface equals Q_enc/ε₀. The problem is that for a spherical surface around a wire, the electric field is not constant in magnitude and is not everywhere perpendicular to the surface (it points radially outward from the wire, not radially outward from the sphere's center). This means the flux integral ∮ E⃗·dA⃗ cannot be replaced with E·A. The correct Gaussian surface is a coaxial cylinder, on whose curved surface E is constant in magnitude and perpendicular, allowing the integral to collapse to E·(2πrL) = λL/ε₀.
Question 2 Multiple Choice
The electric field at distance r from an infinite line charge drops off as 1/r, while the field from a point charge drops off as 1/r². What physical reason explains this difference?
ALine charges are weaker than point charges, so their fields are smaller at all distances
BFor a line charge, the flux spreads over a cylindrical surface (area ∝ r) rather than a spherical surface (area ∝ r²), so E ∝ 1/r rather than 1/r²
CThe 1/r falloff is an approximation that only holds near the wire
DPoint charges obey the inverse square law; line charges are an exception that violates it
The key insight from Gauss's law: E is determined by dividing Q_enc by the area of the Gaussian surface. For a point charge, the natural surface is a sphere with area 4πr², giving E ∝ 1/r². For an infinite line charge, the natural surface is a cylinder with curved-surface area 2πrL, giving E ∝ 1/r. The dimensionality of the charge distribution determines how the flux spreads: a point (0D) source spreads over spheres, a line (1D) source spreads over cylinders. This is not an exception to electrostatics — it follows directly from the geometry of flux spreading.
Question 3 True / False
Outside a uniformly charged spherical shell, the electric field is identical to that of a point charge with the same total charge located at the shell's center.
TTrue
FFalse
Answer: True
This is one of the most powerful results from Gauss's law applied to spherical symmetry. Drawing a concentric spherical Gaussian surface of radius r > R (the shell radius), the enclosed charge is Q regardless of how it is distributed on the shell's surface, and by symmetry E is radially outward and uniform on this surface. The result E = Q/(4πε₀r²) is identical to a point charge — the shell's internal structure is invisible to an external observer. This is the shell theorem, which also applies to gravity: Earth pulls you as if all its mass were at its center.
Question 4 True / False
Gauss's law can be used to directly calculate the electric field from any charge distribution, provided you choose the Gaussian surface carefully enough.
TTrue
FFalse
Answer: False
Gauss's law is always valid as a relation between flux and enclosed charge — but 'calculating E directly' requires being able to pull E outside the integral, which demands that E be uniform in magnitude and perpendicular to the Gaussian surface everywhere. This is only possible when the charge distribution has spherical, cylindrical, or planar symmetry. For an arbitrary charge distribution (e.g., a uniformly charged disk, or two separated point charges), no Gaussian surface has E constant and perpendicular on it, so the integral cannot be simplified. In those cases, you must use direct integration (Coulomb's law) or numerical methods.
Question 5 Short Answer
Before applying Gauss's law, you must argue from symmetry that the electric field has a certain direction and dependence on position. Why is this step essential, and what specifically must you establish?
Think about your answer, then reveal below.
Model answer: Gauss's law gives you one equation: ∮ E⃗·dA⃗ = Q_enc/ε₀. This is a vector integral with an unknown vector field E⃗. To solve for the magnitude of E, you must first use symmetry to establish two things: (1) the direction of E⃗ at every point on your chosen Gaussian surface — so that E⃗ is either parallel or perpendicular to dA⃗ — and (2) the magnitude |E| is constant on the surface (or on the parts where E⃗ is not perpendicular). Only then does the integral reduce to E times the area of the relevant surface, giving a solvable algebraic equation.
The discipline of Gauss's law applications is fundamentally the discipline of symmetry arguments. The law is always true but not always useful. What makes it useful is choosing a surface shaped by the symmetry of the source so that the direction and position-dependence of E are already determined before you write any equation. This is why the three standard geometries (spherical, cylindrical, planar) each have their canonical Gaussian surface: sphere, coaxial cylinder, pillbox — each chosen because it matches the symmetry of the corresponding charge geometry.