How to find the electrostatic potential due to a uniformly charged rod on its axis? I was trying to find the potential at a point (at a distance $r$) on the axis of a rod (length $L$, charge $Q$) and ran into a problem. Let's assume the farther end of the rod(relative to the point) to be $a$ and the other to be $b$. I know I can use this equation :
$$dV= \frac{Kdq}{x}$$
On integrating:
$$\int_{Va}^{Vb} dV = \frac{KQ}{L} \int_{L+r}^{L}\frac{dx}{x}$$
Every website I have referred to completely ignores putting limits on $\int dV$ and writes it as $V$. They also put opposite limits on the RHS to what I am using. Is there some convention here to choosing the upper and lower limit?
 A: The reason I can think of as to why most of the books/literature you read have $V$ instead of $\int dV$ is because they are trying to calculate the potential at the point of interest by integrating with respect to the charge distribution on the charged body at hand (i.e. the charged rod)
$$V(\vec{r})=k\int \frac{dq(\vec{r}')}{|\vec{r}-\vec{r}'|}=
\int \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}d^3\vec{r}'$$
The 1D analogue of this formula is what you should have started with in the first place. The integral is simply telling you that you sum all the contributions to the potential of the infinitesimally small charges that comprise the rod. This is like saying for a discrete contribution, that $V(\vec{r})=\sum_ik\frac{q_i}{|\vec{r}-\vec{r}_i|}$, with $k\frac{q_i}{|\vec{r}-\vec{r}_i|}$ being the contribution to the potential from a single charged particle.
What you are trying to calculate (and hence the reason you have $\int dV$ instead of $V$) is the difference in the potential of a test charge under the influence of the electric field that corresponds to $K\frac{dq}{x}$, travelling from $L$ to $L+r$.
I am not 100% sure that what I am telling you is entirely correct, but if there are any concerns, please think about it and comment back so that I can refresh my Electromagnetism and be back to you. If there are any questions, they are welcome.
To sum up, you should be using the 1 dimensional analogue of the equation I have written above, which is
$$V(x)=k\int_{x}^{x+L} \frac{\lambda dx'}{x'}$$
If you want to find the potential along the line on which the rod lies (I assume that we have constant charge density and that the one end of the rod is located at the position $x$, whereas the other end at $x+L$). If, however, you want to find the potential at some other place in space, you should use
$$V(x)=k\int_{x}^{x+L} \frac{\lambda dx'}{[(x')^2+y^2]^{1/2}}$$
if I am not mistaken. The difference here is that there exists a vertical component to the position vector (position at which we are trying to calculate the potential), as we are calculating the potential at a point which is not located at the line on which the rod lies. I hope I understood what your concern is, but if I didn't, please let me know in the comments. The reason I include both cases (potential at the points on the line of the rod and potential at some other points is because I haven't understood fully the problem you are attempting to solve, so if you could provide some more info if you require further elaboration that would be great!)
A: Let me formulate your problem in my words so that we are sure what we understand each other: we have an infinitesimally thin rod of length $L$ and a total charge $Q$ which is uniformly distributed along the rod. You are interested in the value of the electrostatic potential $V$ in a point on the line on which the rod lies, in a distance $r$ from the nearer end of the rod.
SHORT ANSWER to your question in the text:
It's not a physics convention, it's just mathematics:
if you calculate an integral over a domain of the form of an interval, the lower endpoint of the interval is the lower limit of the integral et vice versa. Hence, you should calculate:
$$V = K \frac Q L \int_r^{r+L} \frac {\mathrm d x}{x}$$
