Calculating Electric Field from Charge Distributions

Here we will calculate the electric field about a point $p$, a certain horizontal distance away from a rod of uniform charge $Q$. To help better visualize the scenario, simply imagine a bar of length $L$ horizontally on the x-axis, and a point $p$ a certain distance away from it to the left, along the same axis as the bar. Let us call this distance that the point is from the nearest end of the bar $a$. What we will attempt to do here, is that we will divide the bar into infinitely small components, and treat each of them as a point charge, we will then use the formula for the electric field about point $p$ due to each point charge, and then use an integral to add up all the electric fields for the infinite number of "points." This will give us the total electric field about our point $p$. First, however, we will need to define some variables:

$\lambda$ will be defined as the linear charge density, or the total charge divided by the length of the rod, $L$. We will define $dq$ as the charge of one of these infinitely small sections of the rod we will be treating as a point charge. We will say that $dl$ will be the infinitely small length of each one of these regions. Lastly, $k$ will be our coulombic constant, which is also represented as $\dfrac{1}{4\pi\epsilon_0}$, where $\epsilon_0$ is a constant called the permittivity of free space. Let us quickly represent $\lambda$ on a more mathematical manner.

$$\lambda = \frac{Q}{L}$$

or $$\lambda = \frac{dq}{dl} \Rightarrow dq=\lambda(dl)$$

Let us set up our basic integral now using the formula for the electric field due to a point charge, $\dfrac{kQ}{l^2}$. (Here $l$ is the general distance from any $dq$ to our point $p$).

$$\int_{a}^{a+L} \dfrac{k(dq)}{l^2}$$

In this integral, however, we are integrating on bounds that are in terms of length, so we need to convert our $dq$ to something compatible with our bounds. Fortunately, we have our equation $dq=\lambda(dl)$. Thus, substituting into our integral we have:

$$\int_{a}^{a+L} \dfrac{k\lambda(dl)}{l^2}$$

Now, removing the constant terms from our integral, we are left with a manageable integral:

$$k\lambda\int_{a}^{a+L} \dfrac{dl}{l^2}$$

Integrating, we receive:

$$k\lambda [\frac{-1}{l}]_{a}^{a+L} = k\lambda[\dfrac{-1}{a+L}+\dfrac{1}{a}]$$

Using some algebra and multiplying both parts inside the brackets to get a common denominator, we can represent this as:

$$k\lambda[\dfrac{-a+a+L}{(a)(a+L)}]$$

Simplifying and substituting $\lambda$ with $\dfrac{Q}{L}$, we receive:

$$\dfrac{kQ}{L}[\dfrac{L}{(a)(a+L)}]= \dfrac{kQ}{(a)(a+L)}$$

Thus, the electric field about a point a horizontal distance $a$ from the end of a bar with uniformly distributed charge $Q$ is equal to:

$$\dfrac{kQ}{(a)(a+L)}$$

or equivalently,

$$\dfrac{Q}{4\pi\epsilon_0(a)(a+L)}$$