Charged Rod

This page provides an explanation of the electric field caused by a uniformly charged thin rod.

The explanation for a uniformly charged rod is the simplest example for a uniformly charged object. The algorithm applied here can be applied to any object.

The Algorithm

Consider a uniformly charged thin rod of Length [math]\displaystyle{ L }[/math] and positive Charge [math]\displaystyle{ Q }[/math].

First Step

Imagine dividing the rod into a series of very thin slices, each with the same charge [math]\displaystyle{ \Delta Q }[/math]. This charge [math]\displaystyle{ \Delta Q }[/math] is a small part of the overall charge. Imagine it as a point charge. Each slice contributes its own electric field, [math]\displaystyle{ \Delta E }[/math]. Summing all these individual slices of [math]\displaystyle{ E }[/math] gives you the Electric Field of the rod. You may recall from your calculus training that this is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

Second Step

The second step is to write a mathematical expression for the field [math]\displaystyle{ \Delta E }[/math] contributed by a single slice of the rod. Because we are imagining each slice as a point charge, we use the formula for a point charge. First, determine [math]\displaystyle{ r }[/math], the vector pointing from the source to the observation location. For our example, this is [math]\displaystyle{ r = obs - source = \lt 0,y,0\gt - \lt x,0,0\gt = \lt -x,y,0\gt }[/math]. Now use this to calculate the magnitude and direction of [math]\displaystyle{ r }[/math]. So [math]\displaystyle{ |\vec{r}| = \sqrt{x^2 + (-y)^2} = \sqrt{x^2 + y^2} }[/math] and [math]\displaystyle{ \hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{\lt x,-y,0\gt }{\sqrt{x^2 + y^2}} }[/math]. [math]\displaystyle{ \hat{r} }[/math] is the vector portion of the expression for the field. The scalar portion is [math]\displaystyle{ \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2} }[/math]. Thus the expression for one slice of the rod is: [math]\displaystyle{ \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot \lt x,-y,0\gt }[/math].

Determining [math]\displaystyle{ \Delta Q }[/math] and the integration variable In the first step we determined that the changing variable for this rod was its x coordinate. This should signify to you that the integration variable is [math]\displaystyle{ dx }[/math]. We need to put this integration variable into our expression for the electric field. More specifically, we need to express [math]\displaystyle{ \Delta Q }[/math] in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any slice of it is: [math]\displaystyle{ \Delta Q = (\frac{\Delta x}{L})\cdot Q }[/math]. This quantity can also be expressed in terms of the charge density.

Expression for [math]\displaystyle{ \Delta \vec{E} }[/math]

Substituting what we just found into our previous expression and dividing into x and y components we get [math]\displaystyle{ \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dy }[/math] and [math]\displaystyle{ \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dy }[/math]. Note that we have replaced [math]\displaystyle{ \Delta y }[/math] with [math]\displaystyle{ dy }[/math] in preparation for integration.

Third Step

The third step is to sum all of our slices. One way is with Numerical summation. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from [math]\displaystyle{ -L/2 }[/math] to [math]\displaystyle{ +L/2 }[/math]. So the expression is [math]\displaystyle{ \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dy. }[/math] Solving this gives the final expression [math]\displaystyle{ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} }[/math] Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.

Work in progress --Spennell3 (talk) 14:47, 2 December 2015 (EST)