Is the Calculation of Linear Density k for a Uniformly Charged Bar Correct?

[ToucanFodder]

Homework Statement

A charge Q is distributed on a insulating bar of lenght L with linear density λ, expressed in C/m. λ=kx where k is a constant and x the distance of the generic point P expressed in meters from the origin of the bar O.
1)Calculate k
2)Calculate the electric potential at the point A positioned perpendicularly from O at a distance R

Relevant Equations

λ=kx

I attached a drawing of the problem for a better understanding and my attempted solutions.

The first point is fairly simple but there's something that I can't figure out.

dq=λdx=kxdx

Q=∫ k x dx from 0 to L -> Q=k[x^2/2]0-L -> Q=(L^2/2)k -> k=2Q/L^2

This is what I came up with. I integrated on the entire bar and calculated k but I'm not quite sure that's correct honestly. I feel like it makes sense mathematically but not physically? Linear density in general is C/m and here I have something that will look like this C/m=(C/m^2)x. Is that fair? I don't understand but I'd really love to.

Point 2 wasn't too bad I just repeated a similar reasoning this time using the electric potential formula. I attached my calculations since writing them down in this format doesn't provide a great result. I think I got it right but I'd love for some feedback if I missed the point.

Also sorry for some mistakes, English is not my first language and scientific terms and expressions can be hard.

 

[PeroK]

The way that $k$ is defined it will have units of $C/m^2$. In which case $\lambda$ has units of $C/m$.

The potential looks correct.

 

[ToucanFodder]

Thanks a lot!

 

[haruspex]

"linear density λ, expressed in C/m. λ=kx where k is a constant and x the distance of the generic point P expressed in meters"

I dislike questions that prescribe units for unknowns. It should be enough to say that x is a distance, independently of any units one might choose to express its value in.
The difficulty, of course, is that if λ=kx then k has dimension of a surface density, but that feels awkward because there is no surface here. That could have been solved merely by stating that its dimension is QL^-2.