What is the electric field at the end of a thin rod with a distributed charge?

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Homework Statement



Thin rod AB has length l=100 cm and total charge q0=37 nC that is distributed in such a way that its line density λ is proportional to the square of the distance from the end A, i.e. λ(x) =kx2. Determine electric field E at the end A of the rod.

Homework Equations



E = kq/r^2, λ = Q/l, dq = λdl

The Attempt at a Solution



I tried doing substituting the line density into the integral of the electric field making: dE = kλdl/r^2 and then I get stuck because if I try to make r = x then the line density will cancel with the x^2 and leaving just the K and I don't know what K is or how to solve for it. If someone can help me with this problem it would be greatly appreciated. Thanks!
 
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You are told explicitly, [tex]\lambda = kx^2[/tex]
So stating [tex]\lambda =\frac{Q}{\ell}[/tex] is a mistake!

The total charge is a quantity you need to determine the numerical value of [tex]k[/tex]
 
RoyalCat said:
You are told explicitly, [tex]\lambda = kx^2[/tex]
So stating [tex]\lambda =\frac{Q}{\ell}[/tex] is a mistake!

The total charge is a quantity you need to determine the numerical value of [tex]k[/tex]

Oh okay. Then if dE = kλdx/r^2, is it valid to say that r = l/2 to have dE = kλdx/(l^2/4) the integral from 0 to l? How are we suppose to find the value of k if we can't relate the λ with Q and l?
 
You know that [tex]\frac{dq}{dx}=\lambda[/tex]
Therefore, [tex]Q_{total}=\int^{\ell}_0 \lambda dx[/tex]
 
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So therefore Q total = l which is 100cm so then k in λ=kx^2 equals 100? Also how does that help me simply the integral from 0 to l of dE = kλdx/r^2?
 
Due to a typo in the TeX, the lambda didn't show up in the integral. Don't take everything you're told by fiat. Think things through. What I posted earlier was nonsense. I said that the total charge is the length of the rod. That's gibberish.

Please think the problem through, I've given you all the help you should require. Just on a final note, I suggest you use [tex]\frac{1}{4\pi\epsilon_0}[/tex] for the constant in Coloumb's Law, since k is already taken, and has a different meaning in this exercise.
Best of luck. :)