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

[thetest]

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!

 

[RoyalCat]

You are told explicitly, \lambda = kx^2
So stating \lambda =\frac{Q}{\ell} is a mistake!

The total charge is a quantity you need to determine the numerical value of k

 

[thetest]

You are told explicitly, \lambda = kx^2
So stating \lambda =\frac{Q}{\ell} is a mistake!

The total charge is a quantity you need to determine the numerical value of k

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?

 

[RoyalCat]

You know that \frac{dq}{dx}=\lambda
Therefore, Q_{total}=\int^{\ell}_0 \lambda dx

 

[thetest]

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?

 

[RoyalCat]

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 \frac{1}{4\pi\epsilon_0} for the constant in Coloumb's Law, since k is already taken, and has a different meaning in this exercise.
Best of luck. :)