Physics · Electrostatics

Electric field of an infinite line charge

Q: When should I combine the electric field of an infinite line charge with another relation?

Combine $\mathbf F=q_{\rm t}\mathbf E$ with the field to find force on a sufficiently small test charge that does not significantly alter the original distribution. Integrating radially gives potential differences between two finite positive radii. The logarithm must contain their dimensionless ratio; zero potential at infinity is unavailable for an infinite line. To assess a wire approximation, compare the Gauss result with a Coulomb integral for a finite uniform rod. Charge distribution and observation point must match. Adding another equation does not justify treating a real conducting surface as a uniform line without checking the model.

An ideal infinite line charge produces a radially symmetric field whose magnitude falls as 1/r.

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Formula

Calculate

E (r) = \frac{∣ λ ∣}{2 π ε _{0} r}

E in V/m; λ in C/m; ε0 in F/m; r in m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$E$	Electric-field magnitude	V/m
$λ$	Charge per unit length	C/m
$ε0$	Vacuum permittivity	F/m
$r$	Perpendicular distance to the line	m

LaTeX

E(r)=\frac{|\lambda|}{2\pi\varepsilon_0 r}

QUANTA STUDYquanta-study.de

Positive line in cross-section with outward radial arrows and radius r₀. Below, the normalised curve decreases from one to a quarter as distance ratio increases from one to four. Points at one and two show halving. — Original cross-section for λ>0; the line is normal to the plane. With r₀=0.10 m and λ=20 nC/m, E₀≈3595 V/m; at 2r₀ it is E₀/2. The graph shows E/E₀=1/(r/r₀); no field value is assigned on the singular axis.

Line charge electric field calculator

Find the field strength, line charge or potential difference around a very long, uniformly charged line in vacuum, such as a thin charged wire.

Calculation model

Find

Line charge density λ

Charge per metre of length; the sign does not matter here, e.g. 10 nC/m.

Distance from the line r

Perpendicular distance from the line to the point, greater than zero, e.g. 10 cm.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for |E|, substitute, evaluate

1. Rearrange for |E|
$|E|=\frac{|\lambda|}{2\pi\varepsilon_0r}$
2. Substitute values (in SI units)
- λ = 10 nC/m = 1 · 10⁻⁸ C/m
- r = 10 cm = 0.1 m
$\frac{1\times10^{-8}}{2\cdot 3.14159\cdot 8.85419\times10^{-12}\cdot 0.1}=1797.51$
3. Result
|E| = 1798 V/m

ε₀≈8.8541878188·10⁻¹² F/m: rounded measured CODATA2022 value, not an exact SI definition.

Result: Electric field strength |E|

|E| = 1798 V/m

With λ = 10 nC/m and r = 10 cm, |E| = 1798 V/m.

Quick tips

Twice the distance, half the field strength. The field of a line falls off as 1/r, more slowly than the 1/r² of a point charge.
Estimate: 1/(2πε₀) ≈ 1.8 · 10¹⁰ V·m/C. So 10 nC/m at a distance of 0.1 m gives about 1.8 · 10¹⁰ · 10⁻⁸ ÷ 0.1 ≈ 1800 V/m.

Assumptions and limits

Uniform infinite electrostatic line in vacuum with r>0. λ is signed charge per length; Er is positive outward from the axis and field magnitude nonnegative. Not wire interior, finite rod or altered material/conductor boundary conditions. Charge-density magnitude alone is determined from |E|; its sign remains unknown.

Derivation & background

Electrostatic vacuum model of an infinite uniform line at r>0. Negative λ reverses radial direction. Thin long rods approximate it only away from their ends; finite uniform rods have a quantifiable midplane correction. Do not extend it inside conducting wire or to altered boundary conditions.

Exam blueprint

Validity range

Derivation steps

Flux through a coaxial cylinder explains inverse distance; symmetry connects flux and local field.

1Place the line on the z-axis; λ is constant signed charge per length and r>0 perpendicular distance.
2Rotation and translation symmetry imply a radial field with equal radial component on a coaxial side surface.
3Choose a closed coaxial cylinder of radius r and length ℓ. Radial field is tangent to its caps: cap fluxes are zero, but field there is not generally zero.
4Side flux is $2\pi r\ell E_r$ and enclosed charge $\lambda\ell$ .
5Gauss’s law gives $2\pi r\ell E_r=\frac{\lambda\ell}{\varepsilon_0}$ ; ℓ cancels.
6Thus $E_r=\frac{\lambda}{2\pi\varepsilon_0r}$ . Magnitude uses |λ|; the sign determines inward or outward direction.

Rearrangements

Linear charge density

|\lambda|=2\pi\varepsilon_0rE

E is the nonnegative field magnitude. Rearrangement gives only |λ|; charge sign needs radial direction. Ideal uniform infinite line in vacuum.

Task variant

An ideal infinite uniform line in vacuum has $\lambda=+2.0\,\mathrm{\mu C\,m^{-1}}$ . Find field magnitude and direction at perpendicular axis distance $r=1.0\,\mathrm{cm}$ .

Approach: $E=\frac{|\lambda|}{2\pi\varepsilon_0r}$ .
With $r=0.010\,\mathrm m$ , $E\approx3.5950\cdot10^6\,\mathrm{V\,m^{-1}}$ .
Positive λ gives radially outward direction.
This uses the uniform infinite vacuum model; doubling distance halves magnitude.

Common mistakes

Set the potential at infinity to zero for an ideal infinite line charge.

Choose a finite reference radius. For finite sources, also check distance relative to half-length.

Exam context

Radial field direction, potential difference and finite line length: check sign and approximation error separately.

Formula cluster

Electrostatics

Gauss law gives the line field with cylindrical symmetry; comparison with Coulomb clarifies the difference from point sources.

∮ E·dA = Q_innen / ε₀

Gauss's Law for Electrostatics

F = ke·|q₁q₂|/r²

Coulomb's Law

E⃗ = F⃗el/q; Ehom = |U|/d

Electric Field Strength

Worked example

For $\lambda=2.0\,\mathrm{\mu C\,m^{-1}}$ and $r=1.0\,\mathrm{cm}$ , $E=\frac{2.0\cdot10^{-6}}{2\pi(8.8541878188\cdot10^{-12})(0.010)}\,\mathrm{V\,m^{-1}}\approx3.60\cdot10^6\,\mathrm{V\,m^{-1}}$ , radially outward.

Applications

Connect Gauss’s law, exterior fields of long cylindrical charge distributions, finite potential differences and justified approximation checks.

How do you calculate with Electric field of an infinite line charge?

Here is how to work through a typical Electric field of an infinite line charge (E = |λ|/(2πε₀r)) task step by step:

1
Task
An ideal infinite uniform line in vacuum has $\lambda=+2.0\,\mathrm{\mu C\,m^{-1}}$ . Find field magnitude and direction at perpendicular axis distance $r=1.0\,\mathrm{cm}$ .
Solution path
1. Approach: $E=\frac{|\lambda|}{2\pi\varepsilon_0r}$ .
2. With $r=0.010\,\mathrm m$ , $E\approx3.5950\cdot10^6\,\mathrm{V\,m^{-1}}$ .
3. Positive λ gives radially outward direction.
4. This uses the uniform infinite vacuum model; doubling distance halves magnitude.

Understand and apply the equation

A line charge differs from a point charge through its extended geometry. A uniformly charged infinite line has a field that falls inversely with distance. Symmetry explains this behaviour; a finite rod reveals when the approximation fails.

Why a long line differs from a point

For a point charge, spherical area grows with distance squared. Around a uniform infinite line, flux spreads over a cylindrical side surface whose area, at fixed length, grows only proportionally to radius. Field therefore decreases with distance to the first power. The infinite line models the middle region of a sufficiently long, thin charge distribution; it is not a real object with finite total charge.[1][2]

Prior knowledge: linear density λ is charge per length. Its sign denotes positive or negative charge.
r is perpendicular distance from the line. Radial direction points from the nearest point on the line to the observer.
The ideal vacuum model excludes extra conductors or material boundaries that break cylindrical symmetry.

[1][2]

How do line-charge field magnitude and signed radial component differ?

Place the line on the z-axis. Its constant signed density $\lambda$ describes charge per length, not the total charge of an infinite line. Measure $r>0$ perpendicular to the axis. The unit vector $\hat{\mathbf r}$ points outward from the axis. Field magnitude and radial component differ: a negative line charge has a positive magnitude but a negative radial component.[1]

\begin{aligned}\mathbf E(r)&=\frac{\lambda}{2\pi\varepsilon_0r}\hat{\mathbf r}\\E_r(r)&=\frac{\lambda}{2\pi\varepsilon_0r}\\|\mathbf E(r)|&=\frac{|\lambda|}{2\pi\varepsilon_0r}\end{aligned}

In vacuum for a uniform infinite line; r=0 is outside the definition.[1]

Moving the observation point along the line or rotating it around the line leaves the charge distribution unchanged. The field therefore depends on neither axial position nor azimuth. Reflection in a plane normal to the line eliminates an axial component. These symmetry arguments allow a flux law to determine a local field strength.[1]

Why the denominator contains one distance

Choose a cylinder around the line. On its side, field is normal to the surface with the same radial component everywhere. At the caps, field instead lies within their planes. Its normal component is therefore zero: the cap flux vanishes, not the field itself. Cylinder length cancels because both enclosed charge and side area are proportional to it.[1]

Choose a closed imaginary cylinder with the same axis, radius r and arbitrary positive length ℓ. It is an integration surface, not an extra conducting body. The radial component is constant on its side; the field is tangential to both end caps, so their fluxes individually vanish. For negative charge the side flux is negative while the field magnitude remains positive.[1]

\begin{aligned}\oint\mathbf E\cdot\mathrm d\mathbf A&=E_r(2\pi r\ell)\\Q_{\rm enc}&=\lambda\ell\\E_r(2\pi r\ell)&=\frac{\lambda\ell}{\varepsilon_0}\\E_r&=\frac{\lambda}{2\pi\varepsilon_0r}\end{aligned}

Gauss’s law plus symmetry; the arbitrary cylinder length cancels.[1]

A point charge distributes flux over a sphere whose area grows quadratically with radius. The chosen cylinder’s side grows only linearly with radius at fixed length, giving an inverse dependence on r. Gauss’s law also holds for a finite rod, but without the same symmetry the field cannot be taken outside the side integral as a constant. Gauss’s law of electrostatics supplies the flux; only the appropriate symmetry turns it into the local field component here.[1]

Units and useful rearrangements

\begin{aligned}\left[\lambda\right]&=\mathrm{C\,m^{-1}}\\\left[\varepsilon_0\right]&=\mathrm{F\,m^{-1}}\\\left[\frac{\lambda}{\varepsilon_0r}\right]&=\mathrm{V\,m^{-1}}\\|\lambda|&=2\pi\varepsilon_0r|\mathbf E|\\r&=\frac{|\lambda|}{2\pi\varepsilon_0|\mathbf E|}\end{aligned}

The last rearrangement requires a positive field magnitude.[1][3]

The calculations use ε₀≈8.8541878188·10⁻¹² farad per metre. Convert centimetres to metres and nanocoulombs to coulombs before substitution. A doubling check tests the structure: doubling charge density doubles the magnitude; doubling distance halves it. Reversing the charge sign reverses direction only, so magnitude alone cannot recover the charge sign.[3]

Example: negative line and potential difference

Move a measurement position outward

Given: Ideal line in vacuum; $\lambda=-30\,\mathrm{nC\,m^{-1}}$ , $r_1=6\,\mathrm{cm}$ , $r_2=12\,\mathrm{cm}$ .
Find: Radial field component at r₁ and potential change from r₁ to r₂.

Model: retain the sign in the field component. Obtain potential by integrating minus the field along the radial path.
$\begin{aligned}E_r(r_1)&=\frac{\lambda}{2\pi\varepsilon_0r_1}\\V(r_2)-V(r_1)&=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_2}{r_1}\end{aligned}$
Substitute radii of 0.06 m and 0.12 m; the logarithm’s ratio is dimensionless.
$\begin{aligned}E_r(r_1)&=\frac{-30\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})(0.06)}\,\mathrm{V\,m^{-1}}\\&\approx-8987.55\,\mathrm{V\,m^{-1}}\\\Delta V&=\frac{30\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})}\ln2\,\mathrm V\\&\approx373.78\,\mathrm V\end{aligned}$

Result: The field magnitude is about 8.99 kV per metre and points toward the line. The outer potential is about 374 V higher.

Check: Check: at twice the distance the radial component is −4493.78 V per metre. A positive outward potential change agrees with the inward field; magnitude and potential difference are different quantities.

A finite rod gives an approximation error

Consider a prescribed uniformly charged thin rod from z=−a to z=+a. Axial contributions cancel pairwise only in its midplane. A real finite conductor generally does not have a prescribed uniform line density; its charge distribution must be determined separately. Here we integrate radial Coulomb components. The parameter a is half the rod length.[2]

\begin{aligned}E_{r,\rm finite}&=\frac{\lambda}{4\pi\varepsilon_0}\int_{-a}^{a}\frac{r\,\mathrm dz}{(r^2+z^2)^{\frac32}}\\&=\frac{\lambda}{2\pi\varepsilon_0r}\frac{a}{\sqrt{a^2+r^2}}\\\frac{|E_{\infty}|}{|E_{\rm finite}|}-1&=\sqrt{1+\left(\frac ra\right)^2}-1\end{aligned}

Last line: relative overestimate by the infinite model, referenced to the finite field; λ≠0.[2]

For r much smaller than a the correction factor approaches one. Far from a finite rod the field returns to that of the total point charge $Q=2a\lambda$ . Both limits check the integral result. Away from the midplane this scalar correction is insufficient; an axial component must also be included.[2]

View diagram: How much does finite line length reduce the field?

How much does finite line length reduce the field?

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For a uniformly charged rod from −a to +a, correction depends only on r/a. At

\frac ra=0.4

, as in the finite example, it is about 0.9285. The point at zero shows only the limiting ratio; the ideal line field itself is undefined on the line.[2][1]

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Example: is a metre-long rod sufficient?

Assess the midpoint approximation

Given: Uniform rod in vacuum: total length $2a=1.00\,\mathrm m$ , $\lambda=20\,\mathrm{nC\,m^{-1}}$ ; midplane point at $r=0.20\,\mathrm m$ .
Find: Finite field, infinite approximation and relative overestimate.

Model: first find the infinite field, then the geometric correction. Reference the error to the finite result.
$\begin{aligned}E_\infty&=\frac{\lambda}{2\pi\varepsilon_0r}\\E_{\rm finite}&=E_\infty\frac a{\sqrt{a^2+r^2}}\\\delta&=\frac{E_\infty-E_{\rm finite}}{E_{\rm finite}}\end{aligned}$
Substitute a half-length of 0.50 m.
$\begin{aligned}E_\infty&\approx1797.51\,\mathrm{V\,m^{-1}}\\\frac a{\sqrt{a^2+r^2}}&=\frac{0.50}{\sqrt{0.50^2+0.20^2}}\approx0.928477\\E_{\rm finite}&\approx1668.95\,\mathrm{V\,m^{-1}}\\\delta&=\frac{1797.51-1668.95}{1668.95}\approx0.07703\end{aligned}$

Result: The finite field is about 1.669 kV per metre. The infinite approximation overestimates it by 7.70 percent.

Check: Check: relative to the infinite field the deficit is instead 7.15 percent. These are consistent because their reference values differ. The correction factor lies between zero and one.

Two oppositely charged lines

Given: Two uniform infinite lines run parallel to z at x=−5.00 cm and x=+5.00 cm. The left density is +20.0 nC/m and the right density −20.0 nC/m. Observe at the origin in vacuum.
Find: Find the total field vector at the midpoint.

Approach: field points rightwards away from the positive left line and towards the negative right line. The components add.
$\begin{aligned}E_{x,L}&=+\frac{|\lambda|}{2\pi\varepsilon_0a}\\E_{x,R}&=+\frac{|\lambda|}{2\pi\varepsilon_0a}\\E_{x,\mathrm{net}}&=2E_{x,L}\end{aligned}$
Substitute $a=0.0500\,\mathrm m$ .
$\begin{aligned}E_{x,L}&=\frac{20.0\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})(0.0500)}\,\mathrm{V\,m^{-1}}\\&\approx7190.04\,\mathrm{V\,m^{-1}}\\E_{x,\mathrm{net}}&\approx14380.1\,\mathrm{V\,m^{-1}}\end{aligned}$

Result: The total field is about 14.38 kV/m along +x, from the positive towards the negative line location. These are prescribed line charges, not conductor charges allowed to redistribute.

Check: Making the right line positive reverses its contribution alone and gives zero midpoint field. A negative probe experiences force along −x in the field calculated here.

Explore further: Logarithmic potential difference

V(r)-V(r_0)=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac r{r_0}

Both r and r₀ must be positive; only their ratio belongs inside the logarithm.[2]

The line field decays too slowly to set an infinite line’s potential to zero at infinity: the logarithm diverges as distance grows without bound. A finite reference radius fixes the arbitrary potential constant; it does not remove a physical singularity. A test charge experiences force $q_{\mathrm t}\mathbf E$ and potential-energy change $q_{\mathrm t}\Delta V$ . It should not significantly alter the original charge distribution.[2]

What wire diameter and surroundings change

The ideal line has no radius. A long conducting cylinder has the same exterior radial field at equal charge per length, but its interior is field-free at electrostatic equilibrium. A uniformly volume-charged insulator has a different interior profile. Do not extend the line formula to the axis of a real wire. Other conductors, grounded surfaces and spatially varying dielectrics change boundary conditions and can destroy radial symmetry.[1]

In a measurement, geometric errors often exceed the uncertainty of physical constants. Record distance from the axis, position relative to the ends and surrounding objects. Agreement between two numbers does not prove ideal symmetry. A full Coulomb integral or boundary-value solution is often the appropriate check for a finite object. Coulomb’s law permits this source integration without inferring a spatially constant field from flux alone.[2]

Five exercises using different kinds of reasoning

Calculate · Foundation

An ideal line has $\lambda=20\,\mathrm{nC\,m^{-1}}$ . Find the field at $r=10\,\mathrm{cm}$ .

Show hint

Convert distance to metres; the positive sign determines direction.

Learning goal: Determine magnitude and direction together.

Show solution and calculation

Model and approach: Ideal uniform infinite line in vacuum; λ carries its sign.
$E_r=\frac{\lambda}{2\pi\varepsilon_0r}$ Use the radial Gauss result.
Substitution gives 3595.02 V per metre, radially outward.
Twice the distance would halve the magnitude; the unit is volts per metre.

E_r=\frac{20\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})(0.10)}\,\mathrm{V\,m^{-1}}\approx3595.02\,\mathrm{V\,m^{-1}}

[1]

Rearrange · Application

At $r=4.0\,\mathrm{cm}$ the measured magnitude is $|\mathbf E|=1000\,\mathrm{V\,m^{-1}}$ . Which line charge densities are compatible?

Show hint

Rearrange for charge-density magnitude; direction is unspecified.

Learning goal: Distinguish magnitude and sign measurements.

Show solution and calculation

Model and approach: Measured magnitude carries no direction sign.
$|\lambda|=2\pi\varepsilon_0r|\mathbf E|$ Multiply the positive field magnitude by the side geometry.
This gives $|\lambda|=2.2253\,\mathrm{nC}$ per metre.
Without a direction measurement both signs are possible.
Substituting either sign gives the same magnitude.

|\lambda|=2\pi\varepsilon_0r|\mathbf E|\approx2.2253\cdot10^{-9}\,\mathrm{C\,m^{-1}}

[1]

Interpret · Foundation

A closed Gaussian cylinder encloses a negative uniform line. Are the cap fluxes positive, negative or zero, and what sign has the side flux?

Show hint

Surface normals point outward; the field points radially inward.

Learning goal: Read flux sign and local field direction.

Show solution and calculation

Model and approach: Radial field is perpendicular to cap normals; radial component determines side flux.
$\Phi=\int\mathbf E\cdot\hat{\mathbf n}\,\mathrm dA$ On both caps the field is perpendicular to the normal, giving zero flux.
On the side the field opposes the outward normal, giving negative flux.
Field magnitude remains positive; the signed component carries the sign in Gauss’s law.

\Phi_{\rm caps}=0,\qquad\Phi_{\rm side}=2\pi r\ell E_r=\frac{\lambda\ell}{\varepsilon_0}<0

[1]

Error analysis · Application

A solution integrates the infinite line field to infinity and sets V=0 there, then claims a finite absolute value. Where is the error?

Show hint

Examine the integral of an inverse-distance function.

Learning goal: Identify an invalid potential reference.

Show solution and calculation

The potential-difference model is correct, but integration to infinity diverges for λ≠0.
That reference is invalid.
Choose a finite positive radius r₀ instead.
The logarithm of the radius ratio is finite and dimensionless; measurable differences are independent of the additive potential constant.

\Delta V=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac r{r_0}

[2]

Transfer · Challenge

A uniform rod is observed in its midplane at $r=0.10\,\mathrm m$ . How large must half-length a be for the infinite model to overestimate by at most 1% of the finite field?

Show hint

Use relative error referenced to the finite field.

Learning goal: Derive a quantitative approximation limit from the finite model.

Show solution and calculation

Use $\sqrt{1+\left(\frac ra\right)^2}-1\le0.01$ .
Squaring positive sides gives a≥0.70535 m, hence total length at least about 1.411 m.
Charge density cancels for λ≠0.
This limit applies only in the midplane of a uniform line, excluding other objects.

\begin{aligned}\frac ra&\le\sqrt{1.01^2-1}\\a&\ge\frac{0.10\,\mathrm m}{\sqrt{1.01^2-1}}\approx0.70535\,\mathrm m\end{aligned}

[2]

What to take away

Cylindrical symmetry gives radial field. Its magnitude uses |λ|; its signed radial component uses λ.
Cap fluxes vanish because the field is tangential, not because field on the caps is zero.
For a finite line, check end distances and which reference defines your relative error.

[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Developed radial signs, Gaussian cylinder, finite-rod correction and potential reference with two checked examples and five exercises.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Quanta. Electric field of an infinite line charge. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrisches-feld-unendliche-linienladung

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Frequently asked questions about Electric field of an infinite line charge

What does the electric field of an infinite line charge describe physically?+

The equation describes the radial electric field of a uniformly charged infinite line in vacuum. Distance is measured perpendicular to the line and density means charge per length. Its sign sets direction: outward for positive charge and inward for negative charge. Magnitude falls inversely with distance because flux spreads over a cylindrical side whose area grows linearly at fixed length. Doubling distance therefore halves magnitude. The signed component $E_r=\frac{\lambda}{2\pi\varepsilon_0r}$ differs from the nonnegative field magnitude. Infinite total length is a model assumption rather than a measurable dimension of a real wire.

How can I check a calculation of the electric field of an infinite line charge?+

First decide whether magnitude or signed radial component is required. For the existing inputs $\lambda=2.0\,\mathrm{\mu C\,m^{-1}}$ and $r=1.0\,\mathrm{cm}$ , substitute radius 0.010 metre and density two millionths of a coulomb per metre. The result is about 3.60 million volts per metre, outward. Check inverse distance dependence: two centimetres gives half that value. Equal negative density gives the same magnitude with reversed direction. Finally assess geometry; for a short rod or near an end, correct numerical substitution is not an accurate description of the actual field.

What assumptions and limits apply to the electric field of an infinite line charge?+

The ideal line is infinite, uniformly charged and treated electrostatically in vacuum. The observation point must lie outside the line. For a finite uniform rod, the approximation works especially in its midplane when distance is small compared with half-length. The correction factor a divided by the square root of a²+r² permits a quantitative check there. Away from the midplane additional axial components appear. Real conducting rods need not have uniform charge density. Wire radius, other conductors and dielectrics change model conditions; in particular the line equation does not describe a conductor’s field-free interior at electrostatic equilibrium.

How do I check units and scale for the electric field of an infinite line charge?+

Charge density has units coulomb per metre, ε₀ farad per metre and r metres, giving volts per metre for the field. Do not confuse nanocoulombs and microcoulombs: that changes the result by a factor of one thousand. A radius given in centimetres also needs conversion before substitution. Dimensions alone cannot identify a wrong charge sign; that needs a radial direction convention. An additional trend check compares two radii at equal density. For the finite model also verify that $\frac a{\sqrt{a^2+r^2}}$ lies between zero and one, so the infinite approximation overestimates magnitude.

When should I combine the electric field of an infinite line charge with another relation?+

Combine $\mathbf F=q_{\rm t}\mathbf E$ with the field to find force on a sufficiently small test charge that does not significantly alter the original distribution. Integrating radially gives potential differences between two finite positive radii. The logarithm must contain their dimensionless ratio; zero potential at infinity is unavailable for an infinite line. To assess a wire approximation, compare the Gauss result with a Coulomb integral for a finite uniform rod. Charge distribution and observation point must match. Adding another equation does not justify treating a real conducting surface as a uniform line without checking the model.

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[1]OpenStax, University Physics Vol. 2, §6.3: Applying Gauss’s Law

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Symbol

Meaning

Unit

E

Electric-field magnitude

V/m

λ

Charge per unit length

C/m

ε0

Vacuum permittivity

F/m

r

Perpendicular distance to the line

Exam blueprint

Validity range

Derivation steps

Flux through a coaxial cylinder explains inverse distance; symmetry connects flux and local field.

1Place the line on the z-axis; λ is constant signed charge per length and r>0 perpendicular distance.
2Rotation and translation symmetry imply a radial field with equal radial component on a coaxial side surface.
3Choose a closed coaxial cylinder of radius r and length ℓ. Radial field is tangent to its caps: cap fluxes are zero, but field there is not generally zero.
4Side flux is $2\pi r\ell E_r$ and enclosed charge $\lambda\ell$ .
5Gauss’s law gives $2\pi r\ell E_r=\frac{\lambda\ell}{\varepsilon_0}$ ; ℓ cancels.
6Thus $E_r=\frac{\lambda}{2\pi\varepsilon_0r}$ . Magnitude uses |λ|; the sign determines inward or outward direction.

How do you calculate with Electric field of an infinite line charge?

Here is how to work through a typical Electric field of an infinite line charge (E = |λ|/(2πε₀r)) task step by step:

Task

An ideal infinite uniform line in vacuum has $\lambda=+2.0\,\mathrm{\mu C\,m^{-1}}$ . Find field magnitude and direction at perpendicular axis distance $r=1.0\,\mathrm{cm}$ .

Solution path

Approach: $E=\frac{|\lambda|}{2\pi\varepsilon_0r}$ .
With $r=0.010\,\mathrm m$ , $E\approx3.5950\cdot10^6\,\mathrm{V\,m^{-1}}$ .
Positive λ gives radially outward direction.
This uses the uniform infinite vacuum model; doubling distance halves magnitude.

Why a long line differs from a point

Prior knowledge: linear density λ is charge per length. Its sign denotes positive or negative charge.
r is perpendicular distance from the line. Radial direction points from the nearest point on the line to the observer.
The ideal vacuum model excludes extra conductors or material boundaries that break cylindrical symmetry.

[1][2]

How do line-charge field magnitude and signed radial component differ?

\begin{aligned}\mathbf E(r)&=\frac{\lambda}{2\pi\varepsilon_0r}\hat{\mathbf r}\\E_r(r)&=\frac{\lambda}{2\pi\varepsilon_0r}\\|\mathbf E(r)|&=\frac{|\lambda|}{2\pi\varepsilon_0r}\end{aligned}

In vacuum for a uniform infinite line; r=0 is outside the definition.[1]

Why the denominator contains one distance

\begin{aligned}\oint\mathbf E\cdot\mathrm d\mathbf A&=E_r(2\pi r\ell)\\Q_{\rm enc}&=\lambda\ell\\E_r(2\pi r\ell)&=\frac{\lambda\ell}{\varepsilon_0}\\E_r&=\frac{\lambda}{2\pi\varepsilon_0r}\end{aligned}

Gauss’s law plus symmetry; the arbitrary cylinder length cancels.[1]

Units and useful rearrangements

\begin{aligned}\left[\lambda\right]&=\mathrm{C\,m^{-1}}\\\left[\varepsilon_0\right]&=\mathrm{F\,m^{-1}}\\\left[\frac{\lambda}{\varepsilon_0r}\right]&=\mathrm{V\,m^{-1}}\\|\lambda|&=2\pi\varepsilon_0r|\mathbf E|\\r&=\frac{|\lambda|}{2\pi\varepsilon_0|\mathbf E|}\end{aligned}

The last rearrangement requires a positive field magnitude.[1][3]

Example: negative line and potential difference

Move a measurement position outward

Given: Ideal line in vacuum; $\lambda=-30\,\mathrm{nC\,m^{-1}}$ , $r_1=6\,\mathrm{cm}$ , $r_2=12\,\mathrm{cm}$ .
Find: Radial field component at r₁ and potential change from r₁ to r₂.

Model: retain the sign in the field component. Obtain potential by integrating minus the field along the radial path.
$\begin{aligned}E_r(r_1)&=\frac{\lambda}{2\pi\varepsilon_0r_1}\\V(r_2)-V(r_1)&=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_2}{r_1}\end{aligned}$
Substitute radii of 0.06 m and 0.12 m; the logarithm’s ratio is dimensionless.
$\begin{aligned}E_r(r_1)&=\frac{-30\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})(0.06)}\,\mathrm{V\,m^{-1}}\\&\approx-8987.55\,\mathrm{V\,m^{-1}}\\\Delta V&=\frac{30\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})}\ln2\,\mathrm V\\&\approx373.78\,\mathrm V\end{aligned}$

Result: The field magnitude is about 8.99 kV per metre and points toward the line. The outer potential is about 374 V higher.

A finite rod gives an approximation error

\begin{aligned}E_{r,\rm finite}&=\frac{\lambda}{4\pi\varepsilon_0}\int_{-a}^{a}\frac{r\,\mathrm dz}{(r^2+z^2)^{\frac32}}\\&=\frac{\lambda}{2\pi\varepsilon_0r}\frac{a}{\sqrt{a^2+r^2}}\\\frac{|E_{\infty}|}{|E_{\rm finite}|}-1&=\sqrt{1+\left(\frac ra\right)^2}-1\end{aligned}

Last line: relative overestimate by the infinite model, referenced to the finite field; λ≠0.[2]

View diagram: How much does finite line length reduce the field?

How much does finite line length reduce the field?

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For a uniformly charged rod from −a to +a, correction depends only on r/a. At

\frac ra=0.4

, as in the finite example, it is about 0.9285. The point at zero shows only the limiting ratio; the ideal line field itself is undefined on the line.[2][1]

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Example: is a metre-long rod sufficient?

Assess the midpoint approximation

Given: Uniform rod in vacuum: total length $2a=1.00\,\mathrm m$ , $\lambda=20\,\mathrm{nC\,m^{-1}}$ ; midplane point at $r=0.20\,\mathrm m$ .
Find: Finite field, infinite approximation and relative overestimate.

Model: first find the infinite field, then the geometric correction. Reference the error to the finite result.
$\begin{aligned}E_\infty&=\frac{\lambda}{2\pi\varepsilon_0r}\\E_{\rm finite}&=E_\infty\frac a{\sqrt{a^2+r^2}}\\\delta&=\frac{E_\infty-E_{\rm finite}}{E_{\rm finite}}\end{aligned}$
Substitute a half-length of 0.50 m.
$\begin{aligned}E_\infty&\approx1797.51\,\mathrm{V\,m^{-1}}\\\frac a{\sqrt{a^2+r^2}}&=\frac{0.50}{\sqrt{0.50^2+0.20^2}}\approx0.928477\\E_{\rm finite}&\approx1668.95\,\mathrm{V\,m^{-1}}\\\delta&=\frac{1797.51-1668.95}{1668.95}\approx0.07703\end{aligned}$

Result: The finite field is about 1.669 kV per metre. The infinite approximation overestimates it by 7.70 percent.

Check: Check: relative to the infinite field the deficit is instead 7.15 percent. These are consistent because their reference values differ. The correction factor lies between zero and one.

Two oppositely charged lines

Given: Two uniform infinite lines run parallel to z at x=−5.00 cm and x=+5.00 cm. The left density is +20.0 nC/m and the right density −20.0 nC/m. Observe at the origin in vacuum.
Find: Find the total field vector at the midpoint.

Approach: field points rightwards away from the positive left line and towards the negative right line. The components add.
$\begin{aligned}E_{x,L}&=+\frac{|\lambda|}{2\pi\varepsilon_0a}\\E_{x,R}&=+\frac{|\lambda|}{2\pi\varepsilon_0a}\\E_{x,\mathrm{net}}&=2E_{x,L}\end{aligned}$
Substitute $a=0.0500\,\mathrm m$ .
$\begin{aligned}E_{x,L}&=\frac{20.0\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})(0.0500)}\,\mathrm{V\,m^{-1}}\\&\approx7190.04\,\mathrm{V\,m^{-1}}\\E_{x,\mathrm{net}}&\approx14380.1\,\mathrm{V\,m^{-1}}\end{aligned}$

Result: The total field is about 14.38 kV/m along +x, from the positive towards the negative line location. These are prescribed line charges, not conductor charges allowed to redistribute.

Check: Making the right line positive reverses its contribution alone and gives zero midpoint field. A negative probe experiences force along −x in the field calculated here.

Explore further: Logarithmic potential difference

V(r)-V(r_0)=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac r{r_0}

Both r and r₀ must be positive; only their ratio belongs inside the logarithm.[2]

What wire diameter and surroundings change

Five exercises using different kinds of reasoning

Calculate · Foundation

An ideal line has $\lambda=20\,\mathrm{nC\,m^{-1}}$ . Find the field at $r=10\,\mathrm{cm}$ .

Show hint

Convert distance to metres; the positive sign determines direction.

Learning goal: Determine magnitude and direction together.

Show solution and calculation

Model and approach: Ideal uniform infinite line in vacuum; λ carries its sign.
$E_r=\frac{\lambda}{2\pi\varepsilon_0r}$ Use the radial Gauss result.
Substitution gives 3595.02 V per metre, radially outward.
Twice the distance would halve the magnitude; the unit is volts per metre.

E_r=\frac{20\cdot10^{-9}}{2\pi(8.8541878188\cdot10^{-12})(0.10)}\,\mathrm{V\,m^{-1}}\approx3595.02\,\mathrm{V\,m^{-1}}

[1]

Rearrange · Application

At $r=4.0\,\mathrm{cm}$ the measured magnitude is $|\mathbf E|=1000\,\mathrm{V\,m^{-1}}$ . Which line charge densities are compatible?

Show hint

Rearrange for charge-density magnitude; direction is unspecified.

Learning goal: Distinguish magnitude and sign measurements.

Show solution and calculation

Model and approach: Measured magnitude carries no direction sign.
$|\lambda|=2\pi\varepsilon_0r|\mathbf E|$ Multiply the positive field magnitude by the side geometry.
This gives $|\lambda|=2.2253\,\mathrm{nC}$ per metre.
Without a direction measurement both signs are possible.
Substituting either sign gives the same magnitude.

|\lambda|=2\pi\varepsilon_0r|\mathbf E|\approx2.2253\cdot10^{-9}\,\mathrm{C\,m^{-1}}

[1]

Interpret · Foundation

A closed Gaussian cylinder encloses a negative uniform line. Are the cap fluxes positive, negative or zero, and what sign has the side flux?

Show hint

Surface normals point outward; the field points radially inward.

Learning goal: Read flux sign and local field direction.

Show solution and calculation

Model and approach: Radial field is perpendicular to cap normals; radial component determines side flux.
$\Phi=\int\mathbf E\cdot\hat{\mathbf n}\,\mathrm dA$ On both caps the field is perpendicular to the normal, giving zero flux.
On the side the field opposes the outward normal, giving negative flux.
Field magnitude remains positive; the signed component carries the sign in Gauss’s law.

\Phi_{\rm caps}=0,\qquad\Phi_{\rm side}=2\pi r\ell E_r=\frac{\lambda\ell}{\varepsilon_0}<0

[1]

Error analysis · Application

A solution integrates the infinite line field to infinity and sets V=0 there, then claims a finite absolute value. Where is the error?

Show hint

Examine the integral of an inverse-distance function.

Learning goal: Identify an invalid potential reference.

Show solution and calculation

The potential-difference model is correct, but integration to infinity diverges for λ≠0.
That reference is invalid.
Choose a finite positive radius r₀ instead.
The logarithm of the radius ratio is finite and dimensionless; measurable differences are independent of the additive potential constant.

\Delta V=-\frac{\lambda}{2\pi\varepsilon_0}\ln\frac r{r_0}

[2]

Transfer · Challenge

A uniform rod is observed in its midplane at $r=0.10\,\mathrm m$ . How large must half-length a be for the infinite model to overestimate by at most 1% of the finite field?

Show hint

Use relative error referenced to the finite field.

Learning goal: Derive a quantitative approximation limit from the finite model.

Show solution and calculation

Use $\sqrt{1+\left(\frac ra\right)^2}-1\le0.01$ .
Squaring positive sides gives a≥0.70535 m, hence total length at least about 1.411 m.
Charge density cancels for λ≠0.
This limit applies only in the midplane of a uniform line, excluding other objects.

\begin{aligned}\frac ra&\le\sqrt{1.01^2-1}\\a&\ge\frac{0.10\,\mathrm m}{\sqrt{1.01^2-1}}\approx0.70535\,\mathrm m\end{aligned}

[2]

What to take away

Cylindrical symmetry gives radial field. Its magnitude uses |λ|; its signed radial component uses λ.
Cap fluxes vanish because the field is tangential, not because field on the caps is zero.
For a finite line, check end distances and which reference defines your relative error.

[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Developed radial signs, Gaussian cylinder, finite-rod correction and potential reference with two checked examples and five exercises.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Quanta. Electric field of an infinite line charge. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrisches-feld-unendliche-linienladung

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Electric field of an infinite line charge

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatics

Worked example

Applications

How do you calculate with Electric field of an infinite line charge?

Why a long line differs from a point

How do line-charge field magnitude and signed radial component differ?

Why the denominator contains one distance

Units and useful rearrangements

Example: negative line and potential difference

Move a measurement position outward

A finite rod gives an approximation error

How much does finite line length reduce the field?

Example: is a metre-long rod sufficient?

Assess the midpoint approximation

Two oppositely charged lines

Explore further: Logarithmic potential difference

What wire diameter and surroundings change

Five exercises using different kinds of reasoning

An ideal line has λ=20 nC m−1\lambda=20\,\mathrm{nC\,m^{-1}}λ=20nCm−1. Find the field at r=10 cmr=10\,\mathrm{cm}r=10cm.

At r=4.0 cmr=4.0\,\mathrm{cm}r=4.0cm the measured magnitude is ∣E∣=1000 V m−1|\mathbf E|=1000\,\mathrm{V\,m^{-1}}∣E∣=1000Vm−1. Which line charge densities are compatible?

A closed Gaussian cylinder encloses a negative uniform line. Are the cap fluxes positive, negative or zero, and what sign has the side flux?

A solution integrates the infinite line field to infinity and sets V=0 there, then claims a finite absolute value. Where is the error?

A uniform rod is observed in its midplane at r=0.10 mr=0.10\,\mathrm mr=0.10m. How large must half-length a be for the infinite model to overestimate by at most 1% of the finite field?

What to take away

Sources for this explanation

Frequently asked questions about Electric field of an infinite line charge

Subject references and further reading

More formulas: Electricity and magnetism

Retain Electric field of an infinite line charge for exams

Electric field of an infinite line charge

Variables & units

Line charge electric field calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatics

Worked example

Applications

How do you calculate with Electric field of an infinite line charge?

Why a long line differs from a point

How do line-charge field magnitude and signed radial component differ?

Why the denominator contains one distance

Units and useful rearrangements

Example: negative line and potential difference

Move a measurement position outward

A finite rod gives an approximation error

How much does finite line length reduce the field?

Example: is a metre-long rod sufficient?

Assess the midpoint approximation

Two oppositely charged lines

Explore further: Logarithmic potential difference

What wire diameter and surroundings change

Five exercises using different kinds of reasoning

An ideal line has λ=20 nC m−1\lambda=20\,\mathrm{nC\,m^{-1}}λ=20nCm−1. Find the field at r=10 cmr=10\,\mathrm{cm}r=10cm.

At r=4.0 cmr=4.0\,\mathrm{cm}r=4.0cm the measured magnitude is ∣E∣=1000 V m−1|\mathbf E|=1000\,\mathrm{V\,m^{-1}}∣E∣=1000Vm−1. Which line charge densities are compatible?

A closed Gaussian cylinder encloses a negative uniform line. Are the cap fluxes positive, negative or zero, and what sign has the side flux?

A solution integrates the infinite line field to infinity and sets V=0 there, then claims a finite absolute value. Where is the error?

A uniform rod is observed in its midplane at r=0.10 mr=0.10\,\mathrm mr=0.10m. How large must half-length a be for the infinite model to overestimate by at most 1% of the finite field?

What to take away

Sources for this explanation

Frequently asked questions about Electric field of an infinite line charge

Subject references and further reading

An ideal line has $\lambda=20\,\mathrm{nC\,m^{-1}}$ . Find the field at $r=10\,\mathrm{cm}$ .

At $r=4.0\,\mathrm{cm}$ the measured magnitude is $|\mathbf E|=1000\,\mathrm{V\,m^{-1}}$ . Which line charge densities are compatible?

A uniform rod is observed in its midplane at $r=0.10\,\mathrm m$ . How large must half-length a be for the infinite model to overestimate by at most 1% of the finite field?

An ideal line has $\lambda=20\,\mathrm{nC\,m^{-1}}$ . Find the field at $r=10\,\mathrm{cm}$ .

At $r=4.0\,\mathrm{cm}$ the measured magnitude is $|\mathbf E|=1000\,\mathrm{V\,m^{-1}}$ . Which line charge densities are compatible?

A uniform rod is observed in its midplane at $r=0.10\,\mathrm m$ . How large must half-length a be for the infinite model to overestimate by at most 1% of the finite field?