Field due to an infinitely long uniformly charged wire

Run a finger along a long, straight wire strung with static charge and something strange happens to the electric field around it: it does not fade the way a point charge's field does. A point charge's field collapses as $1/r^2$ — move twice as far and the field is four times weaker. A charged wire's field only falls as $1/r$ — move twice as far and it merely halves. The reason is pure geometry: a wire is a line, not a point, and the electric field lines pouring out of it have nowhere to go but straight outward, radially, like spokes of an endless wheel. This single result — $E = \lambda/2\pi\varepsilon_0 r$ — is one of the three headline applications of Gauss's law that the CBSE Class 12 syllabus names explicitly, sitting alongside the infinite plane sheet and the charged spherical shell. It is also a favourite board and competitive-exam question because it rewards exactly the kind of symmetry reasoning Gauss's law was built for: instead of grinding through Coulomb's law and integrating over every element of the wire, you choose a Gaussian surface that mirrors the wire's own symmetry — a coaxial cylinder — and the algebra collapses to three lines. This article rebuilds that derivation from the symmetry argument up, works two numerical problems (a straightforward field calculation and a reverse problem recovering the charge density from a measured field), catalogues the marks-losing traps examiners plant around end-cap flux and the wrong exponent, and closes with three practice questions and an exam-ready formula card. Everything here uses only the algebra and geometry the chapter itself supplies — no calculus beyond what Gauss's law already assumes, and nothing from outside the prescribed CBSE 2025–26 scope.

Key takeaways

An infinitely long, uniformly charged straight wire has a linear charge density $\lambda$ (charge per unit length, C/m), and by symmetry its field must point radially, with magnitude depending only on the perpendicular distance $r$ from the wire.
Gauss's law applied to a coaxial cylindrical Gaussian surface of radius $r$ and length $l$ gives the field in three algebraic steps, with the flat end-caps contributing exactly zero flux.
The result is $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ — the field falls off as $1/r$ , not $1/r^2$ , a direct fingerprint of line (cylindrical) symmetry rather than point (spherical) symmetry.
Field direction is radially outward for $\lambda > 0$ and radially inward for $\lambda < 0$ ; field lines are straight, radial, and uniformly spaced around the wire at any given $r$ .
The formula is an idealisation: it holds strictly only for an infinite wire, and in practice for points close to a long wire and far from its ends, where edge effects are negligible.
Common numerical use: given any two of $E$ , $\lambda$ , $r$ , find the third — this is exactly the pattern examiners test, both forward (find $E$ ) and reverse (find $\lambda$ from a measured $E$ ).
Key idea: because a long charged wire has line symmetry, Gauss's law over a coaxial cylinder collapses to $E(2\pi r l) = \lambda l/\varepsilon_0$ , giving the signature $1/r$ field $E = \lambda/2\pi\varepsilon_0 r$ that distinguishes line charges from point charges.

The core insight

Choosing a coaxial cylindrical Gaussian surface around an infinitely long uniformly charged wire turns Gauss's law into $E(2\pi r l) = \lambda l/\varepsilon_0$ , giving the field $E = \lambda/2\pi\varepsilon_0 r$ , which falls off as $1/r$ because the wire's symmetry is cylindrical rather than spherical.

Where it fits in the syllabus

This topic sits inside the Electric Charges and Fields chapter, in the section on electric flux and Gauss's theorem. The board syllabus states Gauss's theorem by statement only — its formal derivation from Coulomb's law is explicitly out of scope — but then requires three named applications: the infinite line charge (this article), the infinite plane sheet, and the uniformly charged thin spherical shell. So while you never have to prove why Gauss's law is true, you are fully expected to reproduce this specific application from scratch: choosing the right Gaussian surface, computing the flux, and solving for $E = \lambda/2\pi\varepsilon_0 r$ . This makes it one of the highest-yield short-derivation questions in the chapter, regularly worth 2–3 marks in board papers and a staple of JEE/NEET single-answer numericals.

The physical picture

Consider an infinitely long straight wire carrying a uniform linear charge density $\lambda$ (charge per unit length, measured in C/m; positive if the wire is positively charged, negative otherwise). Because the wire is infinite, it looks identical from every point along its length — shift your viewpoint along the wire and nothing changes. This translational symmetry, combined with rotational symmetry about the wire's axis, forces two conclusions before any calculation is done: the field $\vec E$ at any point must point directly along the radial direction (either straight away from the wire or straight toward it), and its magnitude can depend only on the perpendicular distance $r$ from the wire — never on position along the wire, and never on the angle around it. To turn this symmetry into a number, choose a cylindrical Gaussian surface: an imaginary closed cylinder of radius $r$ and length $l$ , coaxial with the wire (sharing the same axis), with the wire passing through its centre.

\oint \vec E\cdot d\vec A = \dfrac{q_{\text{enc}}}{\varepsilon_0}

where: the left side is the total electric flux through the closed Gaussian surface, and $q_{\text{enc}}$ is the net charge enclosed inside that surface. The physical reading is that the total number of field lines crossing out of any closed surface is fixed entirely by how much charge sits inside it — the shape of the surface is our choice, and a smart choice makes the flux integral trivial. This cylindrical surface has three distinct parts, and each behaves differently:

Figure

A cylindrical Gaussian surface of radius r and length l drawn coaxially around a long straight charged wire, with radial field arrows striking the curved side perpendicular to it and lying flat along the two end caps so they contribute no flux there

The curved lateral surface: here $\vec E$ is radial, constant in magnitude (since every point on this surface is the same distance $r$ from the wire), and parallel to the outward area-vector $d\vec A$ , so $\vec E\cdot d\vec A = E\,dA$ everywhere on it, and the flux is simply $E$ times the curved area.
The two flat end-caps: here $\vec E$ is still radial — pointing perpendicular to the wire's axis — while the area-vector $d\vec A$ on an end-cap points along the axis. Two mutually perpendicular vectors have zero dot product, so $\vec E\cdot d\vec A = 0$ at every point on both caps, and they contribute nothing to the flux.

The core derivation

Derivation AField of an infinite line charge from Gauss's law

The goal of this derivation is to turn the single stated quantity — the linear charge density $\lambda$ of an infinitely long uniformly charged wire — into an explicit formula for the electric field $E$ at any perpendicular distance $r$ from the wire, using only the symmetry argument already established and Gauss's law in its stated (not derived) form. No calculus beyond recognising that a cylinder's curved surface area is $2\pi r l$ is required; the entire result rests on choosing a Gaussian surface that matches the wire's own cylindrical symmetry so the flux integral reduces to simple multiplication rather than an integral over an unknown field.

Step 1. Set up the Gaussian surface. Choose a right circular cylinder of radius $r$ and length $l$ , coaxial with the wire, so the wire's axis coincides with the cylinder's axis and $r$ is measured perpendicular to the wire. This choice is deliberate: it is the one surface on which the symmetry argument (radial field, magnitude depending only on $r$ ) lets us pull $E$ straight out of the flux integral.

Step 2. Compute the flux through each part. On the curved surface, $\vec E$ is radial and of constant magnitude $E(r)$ , parallel everywhere to $d\vec A$ , so the flux there is $E \times (\text{curved area}) = E(2\pi r l)$ . On both end-caps, $\vec E \perp d\vec A$ , so their flux is zero, as argued from the geometry above. The total flux through the closed cylinder is therefore just the curved-surface term:

\oint \vec E\cdot d\vec A = E(2\pi r l)

Step 3. Compute the enclosed charge. The only charge inside this Gaussian cylinder is the portion of the wire that lies within its length $l$ . Since the wire carries a uniform linear charge density $\lambda$ , the enclosed charge is simply $q_{\text{enc}} = \lambda l$ — charge per unit length times the length enclosed.

Step 4. Apply Gauss's law and solve for $E$ . Substituting both results into $\oint \vec E\cdot d\vec A = q_{\text{enc}}/\varepsilon_0$ :

E(2\pi r l) = \dfrac{\lambda l}{\varepsilon_0}

The length $l$ appears on both sides and cancels — a useful check, since the field at a given $r$ must not depend on how long a Gaussian cylinder we happened to draw. Dividing through by $2\pi r l$ :

E = \dfrac{\lambda}{2\pi\varepsilon_0 r}

Step 5. Read off the physical content. The field magnitude depends only on $r$ , exactly as symmetry demanded, and falls off as $1/r$ rather than the $1/r^2$ of a point charge — a direct consequence of the wire's one-dimensional (line) symmetry versus a point's zero-dimensional (spherical) symmetry. Direction is radially outward from the wire when $\lambda>0$ and radially inward when $\lambda<0$ ; in vector form, $\vec E = \dfrac{\lambda}{2\pi\varepsilon_0 r}\hat r$ , where $\hat r$ is the unit vector pointing away from the wire, perpendicular to it.

where: $E$ is the electric field magnitude (N/C) at perpendicular distance $r$ (m) from the wire, $\lambda$ is the linear charge density (C/m), and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space. The quantity $\dfrac{1}{2\pi\varepsilon_0} \approx 1.8\times10^{10}\ \text{N}\,\cdot\,\text{m/C}$ is a convenient constant to memorise for quick numerical work, just as $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is memorised for Coulomb's law.

Worked examples

Board level - field at a fixed distance from a charged wire

A long straight wire carries a uniform linear charge density $\lambda = 3.5\times10^{-8}\ \text{C/m}$ . Calculate the magnitude of the electric field at a point $r = 10\ \text{cm}$ from the wire, and state its direction.

Step 1. Write down the formula and the known values: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ , with $\lambda = 3.5\times10^{-8}\ \text{C/m}$ , $r = 0.10\ \text{m}$ , and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ .

Step 2. Compute the denominator first: $2\pi\varepsilon_0 r = 2\pi(8.85\times10^{-12})(0.10) = 5.56\times10^{-12}$ .

Step 3. Divide: $E = \dfrac{3.5\times10^{-8}}{5.56\times10^{-12}} \approx 6.3\times10^{3}\ \text{N/C}$ .

Step 4. Fix the direction: since $\lambda>0$ , the wire is positively charged, so the field points radially outward, straight away from the wire at the point in question. Answer: $E \approx 6.3\times10^{3}\ \text{N/C}$ , directed radially away from the wire.

Competitive level - recovering the linear charge density from a measured field

At a perpendicular distance of $r = 2\ \text{cm}$ from a uniformly charged long wire, the electric field is measured to be $E = 4.5\times10^{5}\ \text{N/C}$ , directed toward the wire. Find the linear charge density of the wire, including its sign, and then find the field at twice this distance without recomputing $\lambda$ from scratch.

Step 1. Fix the sign from the direction. The field points toward the wire, and a field points toward negative source charges, so the wire must carry $\lambda < 0$ .

Step 2. Solve $E = \dfrac{|\lambda|}{2\pi\varepsilon_0 r}$ for the magnitude: $|\lambda| = 2\pi\varepsilon_0 r E$ , with $r = 0.02\ \text{m}$ and $E = 4.5\times10^{5}\ \text{N/C}$ .

Step 3. Substitute: $2\pi\varepsilon_0 r = 2\pi(8.85\times10^{-12})(0.02) = 1.11\times10^{-12}$ , so $|\lambda| = (1.11\times10^{-12})(4.5\times10^{5}) \approx 5.0\times10^{-7}\ \text{C/m}$ , giving $\lambda \approx -5.0\times10^{-7}\ \text{C/m}$ .

Step 4. Use the $1/r$ scaling shortcut for the second part, without touching $\varepsilon_0$ or $\lambda$ again: since $E \propto 1/r$ at fixed $\lambda$ , doubling $r$ to $4\ \text{cm}$ simply halves the field, $E' = E/2 = 2.25\times10^{5}\ \text{N/C}$ , still directed toward the wire. Answer: $\lambda \approx -5.0\times10^{-7}\ \text{C/m}$ , and at $r = 4\ \text{cm}$ the field is $2.25\times10^{5}\ \text{N/C}$ toward the wire. Timing note: under ninety seconds if the $1/r$ shortcut in Step 4 is recognised — recomputing the full formula for the second distance wastes time and is a common self-inflicted delay in competitive papers. Whenever a line-charge or plane-sheet question asks for the field at a second distance, check first whether simple proportional scaling ( $1/r$ for a wire, constant for a sheet, $1/r^2$ for a point charge or outside a shell) avoids redoing the algebra.

Common traps

Using the point-charge exponent by mistake: writing $E \propto 1/r^2$ for a line charge instead of the correct $1/r$ . The exponent is a direct signature of the symmetry — line (cylindrical) symmetry gives $1/r$ , point (spherical) symmetry gives $1/r^2$ — and mixing them up is the single most common error on this topic.
Measuring $r$ wrong: the formula uses the perpendicular distance from the wire's axis, not the straight-line distance from some arbitrary point marked on the wire, and not the distance from one end of the wire.
Including the end-cap flux: since $\vec E \perp d\vec A$ on the flat caps, their contribution is always exactly zero and must not be added into $q_{\text{enc}}/\varepsilon_0$ — a stray extra term here throws off the whole algebra even if the setup is otherwise correct.
Forgetting the cylinder's length $l$ must cancel: if $l$ survives in the final answer for $E$ , an algebra slip has occurred, because the field at a fixed $r$ cannot depend on the arbitrary length of Gaussian cylinder chosen.
Applying the infinite-wire result to a short or finite wire, especially near its ends, where the field is neither purely radial nor well approximated by $\lambda/2\pi\varepsilon_0 r$ — the formula is strictly an idealisation for length much greater than $r$ .
Dropping the sign of $\lambda$ when stating direction: examiners routinely test whether a student can read the field's direction (outward or inward) directly from the sign of the charge density, not just compute a magnitude.

Common trapThe exponent is the whole point: 1/r, never 1/r squared

The single biggest mark-loser on this topic is silently reusing the point-charge formula $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ for a line charge, or reusing the line-charge exponent for a point or shell problem in the same paper. There is no shortcut around remembering why the exponent changes: a point charge's field lines spread out over an expanding sphere of area $4\pi r^2$ , so the field dilutes as $1/r^2$ ; a line charge's field lines spread out only over an expanding cylinder of area $2\pi r l$ (the length $l$ does not grow), so the field dilutes as $1/r$ alone. If you can reconstruct that area argument in ten seconds, you will never again default to the wrong power of $r$ under exam pressure.

You should be able to…

Explain, from symmetry alone, why the electric field of an infinite line charge must be radial and must depend only on the perpendicular distance $r$ .
Justify the choice of a coaxial cylindrical Gaussian surface, and explain why the flat end-caps contribute zero flux.
Reproduce the full derivation, from Gauss's law to $E = \lambda/2\pi\varepsilon_0 r$ , showing the length $l$ cancelling.
State how the field's direction (outward or inward) depends on the sign of $\lambda$ .
Solve numerically for any one of $E$ , $\lambda$ , $r$ given the other two, including reverse problems that start from a measured field.
Explain why the field falls off as $1/r$ for a line charge but as $1/r^2$ for a point charge or the exterior of a charged sphere.

Ask the tutor: Want to see why swapping the Gaussian surface to a sphere would make this derivation impossible, or try a few more forward-and-reverse numeric variants? Ask the AI tutor to work through it with you.

Why does a line charge give a 1/r field but a point charge gives 1/r squared, if both use Gauss's law the same way?

It comes down to the shape of the Gaussian surface that matches each source's symmetry. For a point charge, the natural surface is a sphere, whose area grows as $4\pi r^2$ , so the same flux gets spread over an area growing like $r^2$ , diluting the field as $1/r^2$ . For an infinite line charge, the natural surface is a cylinder, whose curved area grows as $2\pi r l$ — proportional to $r$ alone, since the length $l$ is fixed by our own choice of Gaussian surface, not by geometry. Same flux, area growing only as $r$ , so the field dilutes only as $1/r$ . The exponent is really just telling you how fast the "area available to the field lines" grows with distance.

Is a real wire, like a metal rod on a lab bench, actually infinite, and does the formula still work?

No real wire is infinite, so the formula $E = \lambda/2\pi\varepsilon_0 r$ is always an approximation. It works very well, though, for points close to the wire compared with the wire's total length and far from its two ends — in that region the wire genuinely "looks infinite" to a nearby point, because the missing far-away charge contributes very little to the field there anyway. Near the ends of a finite wire, the field picks up components along the wire's length too, and the simple radial, $1/r$ picture breaks down. Board and competitive problems always specify "infinitely long" or "very long compared to r" precisely to sidestep this complication.

How is this different from the field due to a charged conducting cylinder, like a charged pipe?

A solid or hollow charged conducting cylinder of finite radius is a different, more advanced problem — it is explicitly out of the CBSE 2025-26 scope, which restricts this application of Gauss's law to an idealised line (thread-like) charge with negligible thickness. Outside such a cylinder, far from its surface, the field does still reduce to the same $1/r$ form with $\lambda$ taken as the cylinder's total charge per unit length, because from far away the cylinder's charge looks like it is concentrated on its axis. But the field inside a charged conducting cylinder's material, and the detailed field near a cylinder of finite radius, involve reasoning about conductors and charge distribution that goes beyond what this chapter's line-charge idealisation covers.

Why does the length l of the Gaussian cylinder cancel out — doesn't that mean the answer is wrong somehow?

The opposite — the cancellation is exactly the sign that the answer is right. The Gaussian cylinder's length $l$ is not a real physical quantity; it is an imaginary construction tool you chose so that Gauss's law becomes easy to apply. The actual electric field at a point a distance $r$ from the wire cannot possibly depend on how long an imaginary surface you happened to draw around it. If $l$ had survived in the final formula for $E$ , that would be the red flag showing an error had crept into the derivation, since it would mean the field changes depending on an arbitrary choice with no physical meaning.