Physics · Electric Charges And Fields · NEET
It comes from the shape of the Gaussian surface. For a point charge you use a sphere, whose area is 4πr² (an r² term), so E ∝ 1/r². For a line charge you use a cylinder, whose curved area is 2πrl (only one r term), so E ∝ 1/r. The geometry decides the power of r, not the charge itself.
The field near an infinite wire is radial — it points straight out, perpendicular to the wire. The flat end caps of the cylinder are parallel to the field lines (their outward normal points along the wire's axis). Since E is perpendicular to that normal, E·dA = 0 on the caps. All the flux passes through the curved side only.
A coaxial cylinder — a cylinder whose axis lies exactly on the wire, of radius r (the point where you want E) and any length l. This matches the symmetry: E is constant in magnitude and normal to the curved surface everywhere, so the flux integral becomes simply E × (2πrl).
No. The charge enclosed is q = λl and the curved area is 2πrl. When you write E(2πrl) = λl/ε₀, the l cancels on both sides, giving E = λ/(2πε₀r). This is why you may pick any convenient length — the result does not depend on it.
Radially — straight away from the wire if λ is positive, and straight toward the wire if λ is negative. It never has a component along the length of the wire, because for an infinite wire every axial contribution cancels by symmetry.
Two parallel infinite line charges with linear charge densities +λ C/m and −λ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E = λ/(2πε₀r), where λ is the linear charge density (C/m), r is the perpendicular distance from the wire, and ε₀ is the permittivity of free space. The field is directed radially.
Take a coaxial cylindrical Gaussian surface of radius r and length l. Flux through the flat ends is zero (field is radial), and through the curved side it is E×2πrl. By Gauss's law E×2πrl = q_enc/ε₀ = λl/ε₀. Cancelling l gives E = λ/(2πε₀r).
It varies as 1/r. If you double the distance, the field becomes half. On an E-versus-r graph this is a smooth curve falling off more slowly than the 1/r² curve of a point charge.
λ is the linear charge density — the charge per unit length of the wire, measured in coulombs per metre (C/m). For a length l carrying charge q, λ = q/l.
Infinite length guarantees perfect symmetry so the field is exactly radial and depends only on r. For a real finite wire this formula is a good approximation only for points close to the wire and far from its ends.
