Understanding the Electric Field of a Rod: Differences in Integrals Explained

In summary: In summary, the conversation discusses a confusion regarding a differential equation in a textbook and how it differs from the solution the individual came up with. The textbook's derivation uses a different integral and has a d in the denominator instead of a y. The individual realizes that their integrand is the same as the textbook's but they took the wrong integral. They should be taking the first integral instead of the second because their numerator is dy and the constant next to it is d, not y.
  • #1
livewire5
7
0
I am not sure whether this is the correct section to post in but here it goes...

I am really confused as to why the textbook has a "d" in denominator and why their integral is different from mine...I used the integral in 2nd column 2nd row but the textbook is different :S I am not sure why difference in y (I used difference in L to represent the L's segment..) approaches Dy as the textbook states...Anyhow let's break the question into one main problem...I don't understand in the last few textbook lines why my integral is different from theirs even though I used the common integrals from appendix and also, how they got a d in denominator..


I solved question on my own then looked at textbook's answer.

http://img7.imageshack.us/img7/2696/integralsd.png
http://img834.imageshack.us/img834/5516/qsanspg1.jpg
http://img560.imageshack.us/img560/8483/qsanspg2.jpg
http://img195.imageshack.us/img195/525/qspg1.png
 
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  • #2
The textbook's derivation is using a different integral. Their integrand does not have a y in the numerator but instead has d. They use the first integral in the second column of the list you give and from there we see where the extra fact of d comes from. So what you need to do is look to see why your integrand differs from the textbook's. This seems to come from the beginning of the derivation as you can see that the d in the numerator comes from the expansion of \cos \theta_i in terms of y_i and d.
 
  • #3
Im so confused why is the integral the first one rather than the 2nd one? If I turned my difference in L to difference in y which becomes dy then d*dy is what numerator would be and 2nd integral looks closer match for it than the 1st integral.
 
  • #4
Read through the textbook's explanation. It's really just the first step. They write out the electric field due to an infinitesimal length of charged rod and then convert that equation into Cartesian coordinates. Then they convert the charge element on the infinitesimal length into a charge density (assuming that the rod is uniformly charged). Finally, they just integrate along the length of the rod.

But looking through your notes, it appears that you have an equivalent set of steps. Your integrand is the same, but you took the wrong integral. You should be taking the first integral.
 
  • #5
Born2bwire said:
Read through the textbook's explanation. It's really just the first step. They write out the electric field due to an infinitesimal length of charged rod and then convert that equation into Cartesian coordinates. Then they convert the charge element on the infinitesimal length into a charge density (assuming that the rod is uniformly charged). Finally, they just integrate along the length of the rod.

But looking through your notes, it appears that you have an equivalent set of steps. Your integrand is the same, but you took the wrong integral. You should be taking the first integral.


thats what I am wondering, how do you know which integral to take because 1 and 2 are both very similar and 2 matches more closely to my equation so any ideas on why I should be taking integral 1?
 
  • #6
got it! its because my numerator is dy and the constant next to it is d not "y" so I can't use the 2nd integral.
 
  • #7
Your integrand is the same, but you took the wrong integral. You should be taking the first integral.http://www.bosin.info/g.gif
 
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FAQ: Understanding the Electric Field of a Rod: Differences in Integrals Explained

What is the electric field of a rod?

The electric field of a rod refers to the force that a charged rod exerts on other charged particles or objects in its surrounding space. It is a vector quantity that describes the direction and magnitude of the force at any given point.

How is the electric field of a rod calculated?

The electric field of a rod is calculated using the integral of the Coulomb's Law equation, which takes into account the distance, charge, and direction of the rod. This integral can be solved using mathematical techniques such as substitution or integration by parts.

What are the differences between a point charge and a charged rod?

A point charge is a theoretical concept that assumes a charged particle has no physical size, while a charged rod has a finite length and can be divided into smaller charged sections. This leads to different mathematical representations of the electric field, as the rod must be treated as a continuous distribution of charge.

How does the electric field vary along the length of a charged rod?

The electric field along the length of a charged rod is not constant, as it depends on the distance from the rod and the distribution of charge along its length. The field is strongest at the ends of the rod, where the charge is concentrated, and decreases as the distance from the rod increases.

What are some real-world applications of understanding the electric field of a rod?

Understanding the electric field of a rod is important in various fields such as physics, engineering, and chemistry. It can be used to design electrical circuits, analyze the behavior of charged particles in particle accelerators, and study the properties of materials such as semiconductors. It also plays a crucial role in understanding the behavior of lightning and other atmospheric electrical phenomena.

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