Infinite Line of Charge

A line of charge is a Continuous Charge Distribution that extends in only 1 dimension. It uses $\lambda$ for its charge density.

Properties §

• Length (of the line) = $l$
• Charge density per meter = $\lambda =\frac{q}{l}=\frac{dq}{dl}$ (in $C/m$)

Electric Field §

Given a line of infinite length, with charge density $\lambda$, and a point charge $A$ placed at a distance of $x$ away from the line, perpendicular to it. Then the Electric Field of the line can be represented as:

$$\vec{E}=\frac{2k_e\lambda }{x}$$

Derivation (Without Gauss’ Law) §

Let’s start by assuming we have an infinite line of charge, $L$, with a uniform charge density, $\lambda$ and point charge that we use as a reference, $A$.

In order to use Coulomb’s Law, we need point charges. But a line of charge is continuous, so how can we get a specific point of charge? Well, we can use infinitesimals#tosee. Let the charge of the line at the point $(0,\text{d}y)$ equal $\text{d}q$

$$\text{d}{\vec{F}}_A=k_e\frac{q_A\text{d}{q}_B}{r^2}\hat{u}$$

And because we are using the y-axis as a reference, we can let the length of the line, $l$ equal $y$.

$$\lambda =\frac{\text{d}q}{\text{d}y}\to \text{d}q=\text{d}y\lambda$$

$$\text{d}{\vec{F}}_A=k_e\frac{q_A\times \text{d}y\times \lambda }{r^2}\hat{u}$$

Now, as we go through different sections of the line (i.e. as $y$ varies), the angle, $\theta$ between the line and $A$ also changes. We can use trigonometry to express $\hat{u}$ as a unit vector that can be split up into it’s x and y components, which also means the force, $F$ can be divided up:

$$\hat{u} = [\cos (\theta), \sin (\theta)]$$$$\text{d}\vec{F}{A} = k{e} \dfrac{q_{A}\times \text{d}{y} \times\lambda}{r^2}[\cos(\theta), \sin(\theta)]

Now, using the properties of symmetry and [vector superposition](vector%20superposition), we can eliminate the Y-components of the force.  As such, the total force at $A$ is simply the x-component:

\text{d}\vec{F}{A} = k{e} \dfrac{q_{A}\times \text{d}{y} \times\lambda}{r^2}\cos(\theta)

$$W{e}^{\prime }lluseintegrationtoeliminatethe{d}^{\prime }s:$$

\int \text{d}\vec{F}{A} = \int k{e} \dfrac{q_{A} \times\lambda \times \cos(\theta)}{r^2} \ \text{d}{y}

And pull out the constants, $k_e, q_{A}$. $\lambda$ is also a constant because we specified the line has a **uniform** charge density.

\vec{F}{A} = k{e}\times q_{A}\times \lambda \int \frac{\cos(\theta)}{r^2} \ \text{d}{y}

We can use trigonometry to express both $y$ and $r$ in terms of $\theta$

\tan(\theta) = \frac{y}{x} \to y = x \tan (\theta)

\cos(\theta) = \frac{x}{r} \to r = \frac{x}{\cos (\theta)}

Use *change* of variable to change $dy$ to $d\theta$

\frac{\text{d}{y}}{\text{d}{\theta}} = x \sec^2(\theta) \to \text{d}y = x \sec^2(\theta)\ \text{d}\theta

And finally sub it back into the integral. Remember that because it's vertical line of charge, the $x$ displacement is constant, so x can be pulled out.

\vec{F}{A} = k{e}\ q_{A} \lambda \int \frac{\cos(\theta)}{\left( \frac{x}{\cos(\theta)} \right)^2} \times x\sec ^2(\theta) \ \text{d}\theta

= \frac{k_{e}q_{A}\lambda}{x} \int \cos(\theta) \ \text{d}\theta

The infinite nature of the line means that $\theta$ is bound between $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, so that's what we integrate over:

\vec{F}{A} = \frac{k{e}q_{A}\lambda}{x} \int_{-\frac{\pi}{2}}^{\pi/2} \cos(\theta) \ d\theta = \frac{k_{e}q_{A}\lambda}{x} [\sin \theta]{\frac{-\pi}{2}}^{\frac{\pi}{2}} = 2\frac{k{e}q_{A}\lambda}{x}

$$Sofinally,weget:$$

\vec{F}{A} = 2\frac{k{e}q_{A}\lambda}{x} \propto \frac{1}{x}

Which shows that the force is inversely proportional to $x$, the distance between $A$ and the line. And to generalise this we can identify that the [Electric Field](electric%20field.md) of $L$ is simply:

\vec{E}{L} = \frac{2k{e}\lambda}{x}

But [Gauss' Law](Gauss'%20Law.md) makes this way, way easier. ### Derivation With Gauss' Law Firstly, we will use knowledge of symmetry to identify the general shape of the electric field of the line. We know that the line is symmetrical around it's axis, and as such can be rotated, reflected or translated along that axis. As such the [symmetry of that electric field](Electric%20Field%20Symmetry.md) must match:  We need a [Gaussian Surface](Gaussian%20Surface.md), that is, a closed 3D object that matches the symmetry of the electric field in order to calculate the total electric field. A finite cylinder is a good choice. Why?  Well, the electric flux on the caps of the cylinder is $0$, since the surface is parallel to the field (i.e. $\vec{A} \perp \vec{E}$). And since curved surface of the cylinder is always perpendicular to the electric field. Gauss' Law:

\Phi = \oint \vec{E} \cdot \text{d}\vec{A} = \dfrac{q}{\epsilon_{0}}

$$Becausethecapsareparallel,andthecurvedsurfaceisperpendicular,thedotproductisjusttheproductofthemagnitude.$$

\Phi = EA = \dfrac{q}{\epsilon_{0}}

\lambda = \dfrac{q}{L} = \dfrac{q}{h} \to q = \lambda h

A = 2\pi rh \ \text{(Surface Area of curved part of cylinder)}

E = \dfrac{q}{\epsilon_{0}A} = \dfrac{\lambda h}{\epsilon_{0}2\pi rh} = \dfrac{\lambda}{r}2k_{e}

And that's it! $r$ is equivalent to $x$ in the section above. ### Resources: * https://spinningnumbers.org/a/line-of-charge2.html