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Electron Force and Electric Field

  • Thread starter Paul2011
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  • #1
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Homework Statement



A thin rod runs along the x axis from the origin to
x = l. Its linear charge density (C/m) is given by
λ = λ_0(x/l)^2 sin (x/lπ), where λ_0 is a constant.
Show that at the origin,

E(0)=〖-λ〗_0/(2π²€_0l)

Homework Equations





The Attempt at a Solution



Sorry I couldnt provide any start to this problem, kind of hit a mental block. A jump start would be much appreciated.
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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Homework Statement



A thin rod runs along the x axis from the origin to
x = l. Its linear charge density (C/m) is given by
λ = λ_0(x/l)^2 sin (x/lπ), where λ_0 is a constant.
Show that at the origin,

E(0)=〖-λ〗_0/(2π²€_0l)

Homework Equations





The Attempt at a Solution



Sorry I couldnt provide any start to this problem, kind of hit a mental block. A jump start would be much appreciated.
What law applies?

Think of the rod as a series of thin disks of thickness dx. Calculate the electric field of that disk as a function of x along the entire length of the rod.

AM
 
  • #3
1,137
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are you familiar with calculus physics?
 
  • #4
9
0
Ok, after reading what you have had to say I am a little embarrassed to say that I am not grasping the whole thing. I under stand that the rod can be visualized as a bunch of discs which would run from the origin to x=l, which I wrote on my coordinate plane as (l,0). the end points are from (0,0) to (l,0). So I know my bounds would run from 0 to l. And also, the thickness of the disc is dx. What I'm trying to understand is where does the given equation,λ = λ_0(x/l)^2 sin (x/lπ) come into the picture. https://www.physicsforums.com/library.php?do=view_item&itemid=2" if I use the first equation I'm not quite sure what I would replace all the variables with.
 
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  • #5
1,137
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λ is the density

suppose you are considering any dics of negligible thickness dx at distance x from origin.
how will you find the charge in that disc?

and a little hint: as rod is thin, you may take the discs to be acting like small point charges and instead of using Electric field formula of disc, use formula of point charge
 
  • #6
9
0
If I use the point charge formula F= (kq_1q_2)/r^2 then my r would be from origin to l so total distance of l. And I would have to add them up from the origin to x=l. Im sure I'm over complicating this problem. I definitely understand the concept but somethings not clicking.
 
  • #7
1,137
0
you are not over complicating it.

lets go step wise:
1. assume a section dx at distance x from origin
2. find charge in it.
3. find the electric field dE at origin due to that charge
4. Integrate it from x=0 to x=l
 

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