Magnitude of an electric field on a long rod
You place $2.7 \: \mu C$ of charge along a long nonconducting rod. The rod extends from $x=0$ farther than you can see along the positive $x$ axis. The charge distribution has the form of a decreasing exponential: $q(x)=q_0e^{-x/\ell}$, where $\ell = 28.6\: mm$. Calculate the electric field magnitude $20\:mm$ from the near end of the rod, along its long axis.
I'm not looking for an actual solution to this problem, I just need help understanding the wording of the question and making sure I know how to begin this problem. Below is my current understanding of the question and approach for tackling this problem.
Consider a two dimensional plane with a rod beginning at the origin and extending very far along the positive x-axis. We want to find the magnitude of the electric field at a point that is $20\:mm$ to the right of the origin, directly on the rod. If we imagine a positively charged test particle at this position, then the left part of the rod exerts a rightward force on the test particle while the right part exerts a leftward force.
The electric force contributed by an infinitesimally small portion of the left rod is given by $$k\frac{dq}{(20-x)^2}$$ where $k$ is <PERSON>'s constant, $dq$ is infinitesimally small charge, and $(20-x)$ is the disance between the source particle and the test particle. If we solve for $dq$ by taking the first derivative of $q(x)$ and apply this formula to the entire left rod then we obtain $$\frac{-kq_0}{\ell}\int_{0}^{20}\frac{e^{x/\ell}}{(20-x)^2}dx$$
Unfortunately, the inside function lacks an elementary antiderivative. 565
However, it just might be possible to compute the definite integral using some advanced techniques. Right now, I am in the middle of using <PERSON>'s trick to produce a Gaussian integral. (Definitely a bad sign considering this problem is from an intro level physics class) But more importantly, the integral doesn't even exist since $20$ is not an explicit boundary condition. Additionally, even if I shorten the range to make the integral converge, the value will come out to be negative, which I don't think makes sense considering the left rod should exert a rightward (positive) force.
My sources of confusion
1) Have I put the test particle in the wrong place? To me, the question's wording seems to be telling me that the test particle is directly on top of the rod. But I don't think this even makes sense in a physical context since electrical force becomes ridiculously large the closer two positively charged particles are to one another
Perhaps I should just set a boundary condition of $19.999 \: mm$. Or maybe the test particle instead is meant to be placed $20 \: mm$ to the left of the origin.
2) Am I utilizing $q(x)$ in the correct way? It seems natural to just take the first derivative to obtain $dq$, but I can't shake the feeling that $q(x)$ doesn't translate directly to $dq$. $q(0)=q_0$ and the function is constantly decreasing which leads me to believe $q(x)$ represents the amount of charge to the right of $x$. If want to find $dq$, I might to modify $q(x)$ in some way, such as $q_0-q_0e^{-x/\ell}$ for instance.
3) Why would the problem tell me the rod is longer than I can see? Does the rod extend to infinity? Or does the right part of the rod not even matter that much to the electric field at this particular point? 474