Electron Force and Electric Field

AI Thread Summary
The discussion revolves around calculating the electric field at the origin due to a thin rod with a variable linear charge density defined by λ = λ_0(x/l)^2 sin (x/lπ). Participants suggest visualizing the rod as a series of thin discs to facilitate the calculation. The approach involves determining the charge in a small section of the rod and using the point charge formula to find the electric field contribution from each disc. Integration of these contributions from the origin to the end of the rod is necessary to derive the total electric field at the origin. The final goal is to show that E(0) = -λ_0/(2π²€_0l).
Paul2011
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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 couldn't provide any start to this problem, kind of hit a mental block. A jump start would be much appreciated.
 
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Paul2011 said:

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 couldn't 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
 
are you familiar with calculus physics?
 
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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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
 
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. I am sure I'm over complicating this problem. I definitely understand the concept but somethings not clicking.
 
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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