Case II: P along the central axis of a finite line of charge Instead of looking for the magnitude of the electric field adjacent to one of the ends of the finite line of charge, let's now examine the electric field along a perpendicular bisector at point P as shown in the diagram below. The charge present on a small segment of the rod, Δx_{i,} can be expressed as Δq_{i} which equals
We do not have to be concerned with the horizontal, or x-components, of the electric field at point P since symmetry reveals that they will cancel. Consequently, all we have to do is add together all of the vertical components of the electric field at P and we will be done. Considering this small section to represent one of hundreds or thousands of point changes comprising the rod, each Dq_{i} has a vertical contribution to the electric field at point P can be represented by
Adding together all of the contributions gives us the following expressions. Taking the limit as Dx approaches 0, we get the expression However, there is a problem with this integral. We cannot integrate **r** and ** cos q**with respect to **dx**. We must standardize our variables. To achieve this, we are going to use two trigonometric substitutions: one for cos q and the other for tan q.
Referencing our initial charge diagram, we see that Substituting this for 1/r in our previous equation gives us Our expression is now slightly better since it no longer contains **r **and the constant **a** is easily removed from the integral whenever it is convenient. Our next step is to replace **dx** with **dq**. To do this will now use the trig function tan q. Isolating **dx**, gives us the expression Now we can write our final integral and evaluate. To finish we now need to return to our charge diagram and pay attention to assigning the limits to our integral. Looking from left to right, q sweeps from q_{1} to q_{2}. Since we know that q_{1}and q_{2} have the same magnitude (remember that P is on the perpendicular bisector of our finite line of charge) we can write the relationship that q_{2} = -q_{1 } Since q_{1} and q_{2 }are equal, we can renamed our angle as just q without the need of any further subscripts. But what about the length of the rod and its overall charge. How can we write an expression containing Q and L? To do this, we will use the following expression for sin q. Substituting gives us and we are finished. |