EMT: 2nd week part 3 Electric Field Intensity of a finite line of charges

Hey there~ Here we go!

Part 3 : Electric Field Intensity of a line of charges

Just like point charges, Coulomb's law applied but with a little bit complicated way~

Let's assume that we have a line of charges and they are all acting force upon a single point. It is very hard to use superposition in this situation. So, we need another method.

The best way is to take a small portion of line charges as sample. (We assume the charges are all the same and uniform along the line) 

Since the sample is small, we treated it as dE (small change).

Then, we just integral it with the limits which are the each end of the line. 

Let's say we have a line from A to B:

So the limits are the A and B, simple!

Generally, the equation will be: E = integrate dE~

Wonder what the dE is? 
dE then equal to the modified version of Coulomb's law, with dQ replaced Q1Q2.  

Then, what replaces dQ? Well, it depends on what type of line you going to handle, generally is pl dl. :)


Here you go:
  

Alright, so far so good! Let's move to the interesting part of this section.


Let use an example to explain for better understanding:

Find the electric field intensity about the finite line charge of uniform r_l distributing along the z axis.

Solution:

The question stated that it is a finite line charge uniform distributing along z-axis, which means that the line charge is along the z-axis itself. And it seems like its electric field is spread out like a cylindrical-like field. 


Just like an antenna, my lecturer said. So we draw first to identify what it actually is:

So, we assume that there is a point nearby the line, we take a small portion of sample from the line as dQ and draw a line towards the point as dE. The distance between the line and the point is considered to be r (radius). The height of the point is z while the height of the sample is z'.

#BTW, the triangle at the right. I draw it so it is easier for you guys to see. R is actually the sum of vector of (z'-z) and r.

(Try to understand the math parts....LOL. Read carefully, this can be easier understood. Hehe.)



#We choose to use cylindrical coordinates since it is easier to solve the problem, but if you pro, I think other coordinates system also can...Haha.

OK, now to integrate the equation, there are 2 formulas that is ready for us and all we need to do is just to implement them. 

For formula no.1: x = z'-z, a = r  (This will used for integration for equation vector z)
For formula no.2: x = z'-z, a = r  (This will used for integration for equation vector r)

Now, after we complete integrate the equation, we need to further reduce it. To do so, we use a triangle approve. 

Please look at the triangle in the photo, we draw a line from b to the point and another line from a to the point. Then, we identify the angles between them as alpha 1 and alpha 2. 

Then we notice that we can substitute formulas into the equations to further simplify the solution into sin and cos of the angle.

#The equation is slightly modify to suit the triangle. It is just the trick of positive sign to negative sign. :)



Happy understanding!

Should you have anythings to ask, please comment below.

Regards,

Daniel.