Cylindrical symmetry

Before any integral, ask what the symmetry allows the field to be. An infinitely long, uniformly charged cylinder is unchanged by four operations, and each one removes a possibility. Slide it along its axis and nothing changes, so E cannot depend on the axial coordinate. Spin it about the axis and nothing changes, so the magnitude cannot depend on the angle around it. Reflect it in a plane perpendicular to the axis: an axial component would have to reverse under that reflection and also stay the same, which only zero can do, so E has no axial part. Reflect it in a plane containing the axis and the same argument kills the component that circles the axis. What survives is a single radial component depending on a single variable, E(r), pointing away from the axis for positive charge and towards it for negative.

The Gaussian surface must be one on which E ⋅ dA is trivial, and the symmetry has already named it: a cylinder coaxial with the charge, of radius r and any length L, closed by two flat discs. On the curved wall the outward normal is radial, so E is parallel to it, and E(r) is the same at every point of that wall — the flux is E(r) 2πrL with no integral left to do. On each flat cap the normal is axial while E is radial, so E ⋅ dA = 0 point by point and the caps contribute nothing. Gauss's law then reads E(r) 2πrL = λL/ε₀, and L cancels. That cancellation is the check that the work is sound: L belongs to a surface you invented, so it cannot appear in a field. If L survives into your answer, you have put the object's length where the surface's length belonged.

Dividing out gives E(r) = λ/(2πε₀r): an inverse first power, not an inverse square. The reason is geometric. Whatever flux leaves a length of wire has to cross the wall of the Gaussian cylinder, and that wall's area, 2πrL, grows in proportion to r, not as r² the way a sphere's does. Spreading fixed flux over an area that merely doubles leaves a field that merely halves. Take λ = 1.5 × 10⁻⁸ C m⁻¹: at 2.0 cm from the wire E = 1.35 × 10⁴ N C⁻¹, and at 4.0 cm it is 6.74 × 10³ N C⁻¹, exactly half. Notice also what is absent from the answer — the wire's thickness. Any cylindrically symmetric distribution with the same λ, whether a thread, a thick rod or a metal pipe, makes precisely this field everywhere outside itself.

Fill the cylinder uniformly with ρ out to radius R and the answer splits at r = R, because the enclosed charge does. For r < R the Gaussian cylinder swallows only ρπr²L, so E = ρr/(2ε₀), rising in a straight line from zero on the axis. For r > R it swallows the whole ρπR²L = λL, so E = ρR²/(2ε₀r), falling as 1/r. The two branches agree at the boundary, each giving ρR/(2ε₀), and that is where the field is largest. With ρ = 5.0 × 10⁻⁶ C m⁻³ and R = 2.0 cm the surface field is 5.6 × 10³ N C⁻¹, and the two sides of the graph balance in a memorable way: at r = 1.0 cm the field is 2.8 × 10³ N C⁻¹, and at r = 4.0 cm it is 2.8 × 10³ N C⁻¹ again.

A coaxial cable is an inner conductor of radius a carrying +λ per metre inside a tube of inner radius b and outer radius c carrying −λ per metre. Three regions, one Gaussian cylinder each. For a < r < b the enclosed charge is λL and E = λ/(2πε₀r); with a = 1.0 mm, b = 4.0 mm and λ = 3.0 × 10⁻⁹ C m⁻¹ that is 3.6 × 10⁴ N C⁻¹ at r = 1.5 mm and 1.8 × 10⁴ N C⁻¹ at r = 3.0 mm. Inside the metal of the tube, b < r < c, the field is zero in equilibrium, so the enclosed charge must be zero, which forces −λ per metre onto the tube's inner surface. Beyond the cable, r > c, the enclosed charge is λL − λL = 0, so E = 0. Equal and opposite charge inside one Gaussian cylinder is the whole of the shielding argument.

No wire is infinite, so the real question is what the idealisation costs. Integrate over a uniform rod of length L and the field on its midplane at perpendicular distance r comes out as the infinite-line value multiplied by 1/√(1 + (2r/L)²), a correction that depends only on the ratio r/L. At r = L/100 the true field is 0.02% below the idealisation; at r = L/20 it is 0.50% below; at r = L/10 it is 1.9% below; at r = L/4 it is 10.6% below; and at r = L/2 it is 29% below. Stay within about a tenth of the rod's length of the midpoint and the infinite-line field is good to 2%. Move off the midplane towards an end and a second failure appears that no correction factor repairs: the field tilts, picking up the axial component the symmetry argument had banned.