Electric field and charge density

[peaceandlove]

Homework Statement

A uniform line charge of density λ lies on the x-axis between x=0 and x=L. Its total charge is 11 nC. The electric field at x=2L is (500 N/C)ˆı. Find the electric field at x=3L. Answer in units of N/C.

Homework Equations

The Attempt at a Solution

I have no clue where to even begin...I know that eventually I need to obtain a relationship between E(2L) and E(3L) but besides that I'm totally lost.

 

[LowlyPion]

Homework Statement

A uniform line charge of density λ lies on the x-axis between x=0 and x=L. Its total charge is 11 nC. The electric field at x=2L is (500 N/C)ˆı. Find the electric field at x=3L. Answer in units of N/C.

Homework Equations

The Attempt at a Solution

I have no clue where to even begin...I know that eventually I need to obtain a relationship between E(2L) and E(3L) but besides that I'm totally lost.

Develop an integral for a charged line segment along the x-axis. Won't that then tell you what the relationship between distance along x and field strength?

 

[peaceandlove]

I can't seem to get the integral right... I got (1/4(pi)e_0)*(qQ/(d(d+L)). Perhaps I calculated it wrong because I'm not sure what q is.

 

[LowlyPion]

I can't seem to get the integral right... I got (1/4(pi)e_0)*(qQ/(d(d+L)). Perhaps I calculated it wrong because I'm not sure what q is.

What you don't know is what L is.

By giving you the field intensity at 2L, then they are telling you what it is. But rather than calculate it out I'd just go to the ratio and figure the L based on the ratio of the dependency on L.

For instance you should get a result based on the charges being arranged 1L to 2L, and when measured at 3L, the field will be evaluated with the charges 2L to 3L away.

 

[peaceandlove]

Do I do that using the integral I developed? And if so does that mean the one I came up with is correct?

 

[LowlyPion]

Do I do that using the integral I developed? And if so does that mean the one I came up with is correct?

Sorry, I didn't work out your integral. I simply note that you should have an equation of the form

E = kλ∫ 1/r² = - K/r

where K is the grouped constants (which will cancel out) and r is evaluated from 1 to 2, or 2 to 3.
Then take the ratio of the equations right?

E₃/500 = (K*(1/2 - 1/3))/(K*(1-1/2)