Applying Gauss’s Law: Electric Field of an Infinite Line Charge

Gauss’s law tells us that the total electric flux through any closed surface equals the enclosed charge divided by $\varepsilon_0$. That is a powerful statement, but turning it into an actual field calculation requires something more: symmetry. Without symmetry, the integral on the left side of Gauss’s law cannot be simplified, and we are no better off than doing a brute-force Coulomb’s-law summation. In this topic, we see how a specific symmetric charge configuration, an infinitely long uniformly charged wire, lets us extract the electric field cleanly using Gauss’s law.

Why Symmetry Matters

The general expression for the electric field of any charge distribution involves a summation (or integration) over every small charge element (Eq. 1.27). For an arbitrary arrangement of charges, carrying out this summation at every point in space is extremely difficult. There is no shortcut.

However, certain charge configurations possess enough geometric symmetry that we can determine the direction of $\mathbf{E}$ and identify surfaces where its magnitude is constant. On such surfaces, the flux integral simplifies to a simple product, and Gauss’s law gives us the field in one clean step.

The infinite line charge is one of the most important of these symmetric cases.

Setting Up the Problem: An Infinite Uniformly Charged Wire

Picture a thin, straight wire that stretches to infinity in both directions. It carries a uniform linear charge density $\lambda$ (charge per unit length, in $\text{C/m}$). We want to find the electric field at a perpendicular distance $r$ from the wire.

Identifying the Symmetry

The wire serves as an axis of symmetry. Think about what happens when you pick a point P at distance $r$ from the wire and then rotate that point around the wire. Every rotated image, P, P’, P”, sits at the same distance from the same wire, seeing the same charge distribution. Nothing has changed. This rotational equivalence tells us two things:

• The magnitude of $\mathbf{E}$ can depend only on $r$, not on the angle around the wire or the position along it.
• The direction of $\mathbf{E}$ must be radial: straight outward from the wire if $\lambda > 0$, straight inward if $\lambda < 0$.

Why Only the Radial Component Survives

There is a neat argument that confirms the field direction. Take any point P near the wire and consider two small charge elements on the wire that are placed symmetrically above and below the horizontal plane through P. Each element produces its own electric field at P. The radial components of these two fields add up, but the components along the wire point in opposite directions and cancel exactly. Since every charge element on the wire can be paired with a partner on the other side, the total field at P has no component along the wire. It is purely radial.

Fig 1.26: (a) The electric field of an infinite line charge is radial; paired charge elements cancel the axial component. (b) A cylindrical Gaussian surface of radius $r$ and length $l$ placed coaxially around the wire

Furthermore, because the wire is infinitely long, sliding along its length changes nothing. The field at a given radial distance is the same everywhere along the wire. Combining all of this: $\mathbf{E}$ depends only on $r$ and points radially.

Choosing the Gaussian Surface

To exploit this symmetry with Gauss’s law, we need a closed surface on which the flux integral becomes simple. The natural choice is a cylinder of radius $r$ and length $l$, placed coaxially with the wire. This cylinder has three parts: two flat circular end caps and one curved side surface.

Let us compute the flux through each part separately.

Flux Through the Flat Ends

On the two circular end caps, the outward normal points along the wire (parallel to the axis of the cylinder). But the electric field is radial, pointing straight outward from the wire, which means $\mathbf{E}$ is perpendicular to the outward normal on each flat face. The dot product $\mathbf{E} \cdot \Delta\mathbf{S}$ is therefore zero at every point on either end cap.

$\text{Flux through flat ends} = 0$

Flux Through the Curved Surface

On the curved surface, things are different. The outward normal at every point on this surface points radially away from the axis, exactly the same direction as $\mathbf{E}$. So $\mathbf{E}$ and $\hat{\mathbf{n}}$ are parallel everywhere on the curved surface, and the dot product simply gives $E$.

Moreover, every point on the curved surface is at the same distance $r$ from the wire, so the magnitude $E$ is the same everywhere on this surface. This means we can pull $E$ out of the summation:

$\text{Flux through curved surface} = E \times (\text{area of curved surface}) = E \times 2\pi r l$

Total Flux

Since the flat ends contribute nothing:

$\phi = E \times 2\pi r l \qquad \text{(total flux through the cylinder)}$

Applying Gauss’s Law: The Derivation

The Gaussian cylinder of length $l$ encloses a length $l$ of the wire. Since the wire has $\lambda$ coulombs of charge per metre, the total enclosed charge is:

$q_{\text{enclosed}} = \lambda \, l$

Now apply Gauss’s law, $\phi = q_{\text{enclosed}} / \varepsilon_0$:

$E \times 2\pi r l = \frac{\lambda \, l}{\varepsilon_0}$

Notice that the length $l$ of the cylinder appears on both sides. Cancel it:

$E \times 2\pi r = \frac{\lambda}{\varepsilon_0}$

Finally, divide both sides by $2\pi r$ to isolate $E$:

$E = \frac{\lambda}{2\pi\varepsilon_0 \, r} \qquad \text{(1.32)}$

This is the magnitude of the electric field at a perpendicular distance $r$ from an infinitely long uniformly charged wire.

The Vector Form

To write the full vector result, we introduce $\hat{\mathbf{n}}$, the radial unit vector in the plane perpendicular to the wire at the point of interest. This unit vector points outward from the wire.

$\mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 \, r} \; \hat{\mathbf{n}} \qquad \text{(1.32)}$

Notice that $\lambda$ appears as an algebraic quantity in this formula. Its sign controls the direction of the field:

• If $\lambda > 0$ (positive charge), the scalar coefficient is positive, and $\mathbf{E}$ points along $\hat{\mathbf{n}}$, that is, radially outward.
• If $\lambda < 0$ (negative charge), the coefficient is negative, flipping the direction so that $\mathbf{E}$ points radially inward, toward the wire.

A Note on Algebraic Scalars and the Modulus

When we express any vector as $\mathbf{A} = A\,\hat{\mathbf{a}}$ (a scalar times a unit vector), the scalar $A$ can be positive or negative. If $A > 0$, $\mathbf{A}$ has the same direction as $\hat{\mathbf{a}}$. If $A < 0$, $\mathbf{A}$ points opposite to $\hat{\mathbf{a}}$. If you want a quantity that is always non-negative, use the modulus $|\mathbf{A}|$. By definition, $|\mathbf{A}| \geq 0$ always.

This distinction matters here: the formula $E = \lambda / (2\pi\varepsilon_0 r)$ can give a negative value of $E$ when $\lambda < 0$, and that negative sign is meaningful, it tells you the field points inward.

Important Observations

A few points deserve careful attention:

• The field on the surface is due to the entire wire, not just the enclosed part. Although only the charge inside the Gaussian cylinder ($\lambda l$) entered the right side of Gauss’s law, the electric field $\mathbf{E}$ at any point on that surface is produced by the charge on the whole infinite wire, including the portions stretching far beyond the cylinder in both directions. Gauss’s law works because the flux contributions from external charges cancel when integrated over the full closed surface.

• The infinite-wire assumption is essential. Without it, we cannot claim that $\mathbf{E}$ is strictly perpendicular to the wire everywhere on the curved surface. For a finite wire, the field near the ends tilts and acquires an axial component, which would prevent us from pulling $E$ out of the integral.

• Approximation for long finite wires. In practice, no real wire is infinitely long. However, Eq. (1.32) remains an excellent approximation near the central portion of a long wire, where the distance to either end is much greater than $r$. End effects become significant only close to the tips of the wire.

• The $1/r$ dependence. Unlike a point charge whose field drops as $1/r^2$, the field of an infinite line charge falls more slowly, as $1/r$. This slower fall-off arises because an infinite wire has charge spread along an entire dimension, and there is always more charge contributing from distant parts of the wire.