$E$ field of a charged rod of length $L$. If $Q=λL$ why is $dQ=λdx$? In reference to this page here explaining the $\mathbf{E}$ of a charged rod,  it explicitly states that $Q = λ\cdot L$. Why then does it go on to explain that $\mathrm dQ = λ \cdot \mathrm dx\;$? 
I assumed it would be $\mathrm dQ = λ \cdot \mathrm dL\;$ instead since $\mathrm dL$ would be considered an infinitesimally small portion of the rod. Can someone please help me understand ?
 A: The problem here is that you are mixing the limit of integration with the integrand. Keep the infinitesimally small portion to be $dx$ such that $ \int_{0}^{L} dx = L $. Then, $ dQ = \lambda dx $, and integrating both sides gives $ Q = \lambda L $. 
A: How you define your variables is subjective. As long as you know what you mean, then you are good to go. Of course, there are options that make your work easier to read and understand, but it is not necessary. There is no rule that says you have to name a finite length $L$ and an infinitesimal distance in that length $dL$. As long as you know what $dx$ means (which it seems like you do), then everything is fine.
With that being said, this is a common convention. $L$ is the set length of the rod. $x$ is the labeling of the coordinates of the line the rod is on. So, for example, the rod could have ends at $x=0$ and $x=L$, or $x=-\frac L2$ and $x=\frac L2$. Since we are working with $x$ as the coordinate, it makes sense that $dx$ would be used for the infinitesimal for integration. In other words, in the work you quote, $L$ is just a set constant length, not a reference to a coordinate system. 
However, like I said above, there would be nothing wrong with saying $Q=\lambda L$ and $dQ=\lambda dL$. You would just need to recognize what $L$ and $dL$ actually mean and how they relate, and this is somewhat clunky notation. And as R.G.B points out, you would need to modify any integrals where you would want to refer to $L$ in the integration limits if you are trying to also use $L$ as a reference of the coordinate system.
