Calculation of the Electric Field of an Infinitely Long Uniformly Charged Straight Line

In the study of electrostatics, mastering the analysis of highly symmetric charge distributions is a fundamental skill. The model of an infinitely long, uniformly charged straight line serves as a classic paradigm in physics. It is not only a cornerstone for understanding how electric fields propagate in space but also an ideal case study for comparing two primary analytical techniques: direct integration via Coulomb’s Law and the application of Gauss’s Law.
To begin, let us define the physical model. Consider an infinitely long line positioned along the $z$-axis in a three-dimensional Cartesian coordinate system. This line carries a uniform linear charge density, denoted by $\lambda$ (measured in $\text{C/m}$). Our objective is to determine the electric field intensity $\mathbf{E}$ at a point $P$ located at a perpendicular distance $r$ from the line.

Before performing any mathematical derivations, we can significantly simplify the problem by leveraging symmetry arguments:

  • Radial Symmetry: Because the line is infinite, the charge distribution remains invariant under any rotation around the $z$-axis. Consequently, the magnitude of the electric field $E$ must depend only on the radial distance $r$ and cannot depend on the angular position.
  • Axial Symmetry: For any charge element at a position $+z$, there is a corresponding element at $-z$. The vertical ($z$-direction) components of the electric fields produced by these two elements are equal in magnitude but opposite in direction, causing them to cancel out. Therefore, the net electric field at point $P$ has no component along the $z$-axis.

Based on these observations, we can conclude that the electric field is purely radial, directed away from the line (if $\lambda > 0$) or toward it (if $\lambda < 0$). We can represent the field vector as:
$$\mathbf{E} = E(r) \mathbf{e}_r$$
where $\mathbf{e}_r$ is the unit vector in the radial direction.

Method 1: The Integral Approach via Coulomb’s Law

The most fundamental way to solve this problem is through the principle of superposition, using Coulomb's Law to sum the contributions from every infinitesimal charge element along the line.

1. Defining the Charge Element

We consider an infinitesimal segment of the line of length $dz$ located at a position $z$ on the $z$-axis. The charge contained within this segment is:
$$dq = \lambda dz$$

2. Calculating the Differential Field

Let point $P$ be located at $(r, 0, 0)$ in cylindrical coordinates. The distance $s$ from the charge element $dq$ to point $P$ is given by the Pythagorean theorem:
$$s = \sqrt{r^2 + z^2}$$
According to Coulomb's Law, the magnitude of the differential electric field $dE$ produced by $dq$ at point $P$ is:
$$dE = \frac{1}{4\pi\epsilon_0} \frac{\lambda dz}{s^2} = \frac{1}{4\pi\epsilon_0} \frac{\lambda dz}{r^2 + z^2}$$

3. Projection and Integration

As established by our symmetry analysis, we only need to integrate the radial component, $dE_r$. If $\theta$ is the angle between the vector $d\mathbf{E}$ and the radial vector, then:
$$dE_r = dE \cos\theta$$
From the geometry of the system, we see that $\cos\theta = \frac{r}{s} = \frac{r}{\sqrt{r^2 + z^2}}$. Substituting this into our expression:
$$E_r = \int_{-\infty}^{+\infty} \frac{1}{4\pi\epsilon_0} \frac{\lambda dz}{r^2 + z^2} \cdot \frac{r}{\sqrt{r^2 + z^2}} = \frac{\lambda r}{4\pi\epsilon_0} \int_{-\infty}^{+\infty} \frac{dz}{(r^2 + z^2)^{3/2}}$$

4. Solving the Integral

To solve this, we use the trigonometric substitution $z = r\tan\theta$, which implies $dz = r\sec^2\theta d\theta$. The limits of integration change from $z \in (-\infty, +\infty)$ to $\theta \in (-\pi/2, \pi/2)$.
$$E_r = \frac{\lambda r}{4\pi\epsilon_0} \int_{-\pi/2}^{\pi/2} \frac{r\sec^2\theta}{(r^2\sec^2\theta)^{3/2}} d\theta = \frac{\lambda r}{4\pi\epsilon_0} \int_{-\pi/2}^{\pi/2} \frac{r\sec^2\theta}{r^3\sec^3\theta} d\theta$$
Simplifying the integrand:
$$E_r = \frac{\lambda}{4\pi\epsilon_0 r} \int_{-\pi/2}^{\pi/2} \cos\theta d\theta = \frac{\lambda}{4\pi\epsilon_0 r} [\sin\theta]_{-\pi/2}^{\pi/2}$$
$$E_r = \frac{\lambda}{4\pi\epsilon_0 r} (1 - (-1)) = \frac{2\lambda}{4\pi\epsilon_0 r} = \frac{\lambda}{2\pi\epsilon_0 r}$$

Method 2: The Elegant Solution via Gauss’s Law

While integration is robust, Gauss's Law provides a much more efficient route for problems with high degrees of symmetry.

1. Selecting the Gaussian Surface

Given the cylindrical symmetry of the charge distribution, we choose a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the charged line.

2. Applying Gauss’s Law

Gauss's Law states that the total electric flux through a closed surface is proportional to the enclosed charge:
$$\oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{encl}}{\epsilon_0}$$

3. Calculating Electric Flux ($\Phi_E$)

The total flux through the cylinder is the sum of the flux through the two end caps and the curved side surface:

  • The End Caps: The electric field $\mathbf{E}$ is radial, while the area vector $d\mathbf{A}$ of the end caps points along the $z$-axis. Since $\mathbf{E} \perp d\mathbf{A}$, the dot product $\mathbf{E} \cdot d\mathbf{A} = 0$. Thus, there is no flux through the ends.
  • The Curved Side Surface: On this surface, $\mathbf{E}$ is parallel to $d\mathbf{A}$ and its magnitude $E$ is constant due to symmetry. The surface area of the side is $2\pi r L$.
    $$\Phi_E = E \cdot (2\pi r L)$$

4. Determining Enclosed Charge

The total charge enclosed within our Gaussian cylinder of length $L$ is:
$$Q_{encl} = \lambda L$$

5. Final Derivation

Substituting the flux and the enclosed charge back into Gauss's Law:
$$E(2\pi r L) = \frac{\lambda L}{\epsilon_0}$$
Dividing both sides by $2\pi r L$, we obtain:
$$E = \frac{\lambda}{2\pi\epsilon_0 r}$$

Conclusion and Physical Insights

Both methods yield the identical result, confirming the validity of our derivation. The electric field of an infinitely long uniformly charged line is:
$$E = \frac{\lambda}{2\pi\epsilon_0 r}$$

Key Takeaways:

  1. Decay Rate: Unlike a point charge, where the electric field strength follows an inverse-square law ($1/r^2$), the field of an infinite line decays more slowly, following an inverse law ($1/r$).
  2. The "Infinite" Approximation: In practical applications, such as analyzing the field around a long power line, the "infinite" model is a highly accurate approximation as long as the distance $r$ is much smaller than the actual length of the wire.
  3. Methodological Comparison:
    • Coulomb's Law (Integration) is a "brute force" method. It is universally applicable to any charge distribution but requires significant mathematical effort.
    • Gauss's Law is an "elegant" method. It is incredibly fast and efficient, but its utility is strictly limited to distributions that possess high levels of symmetry (spherical, cylindrical, or planar).