# Determing the direction of dl when calculating V

## Homework Statement

Hi. I'm wondering how one determines the direction of dl (vector) when calculating V, the electric potential. So, suppose we want to calculate the electric potential caused by a point charge q at a distance a from the point charge, if we know the electric field of a point charge.

## Homework Equations

The textbook I have says to calculate the potential by Va - Vb = integral from a to b of E*dl.

## The Attempt at a Solution

So, I can let b = infinity, to get Va - Vinfinity = integral from a to infinity of k*q/r2 dr, where I have said that dl = dr*rhat, since I am integrating from r = a to r = infinity. Doing the integral, and assuming Vinfinity = 0, I get the correct answer,

V = kq/a.

But suppose instead that I had integrated from infinity to a instead of the other way. So,

Vinfinity - Va = integral from infinity to a of E*dl.

Now, since I am coming from infinity toward a, I make dl = -dr*rhat. That introduces one negative sign. But the order of integration is also flipped, which introduces another. BUT, since I have Vinfinity - Va, that introduces a third negative, so I end up with

V = -kq/a

the opposite of what I got before, and the wrong answer. Where did I go wrong? I'm guessing it must have been when I said that dl = -dr*rhat - but why isn't that correct? What sets the proper direction of dl?

tiny-tim
Homework Helper
welcome to pf!

hi kc8ukw! welcome to pf!

(have an infinity: ∞ and an integral: ∫ )

the potential V is defined by …

V(R) - V(∞) = ∫R E(r) dr …

what exactly are you saying?

Hi Kc8ukw,
For your particular problem, (E in the $\hat{r}$ direction), d$\vec{l}$ is in the direction of $\hat{r}$ (you know this already). But in spherical coordinates $\hat{r}$ always points radially outward. But radially outward can point, for instance, along the $\hat{x}$ direction or along the -$\hat{x}$ direction. Both of these directions, $\hat{x}$ and -$\hat{x}$, are called $\hat{r}$ when we are in spherical coordinates. This is because in general the direction of $\hat{r}$ depends on the angles θ and $\phi$. So your d$\vec{l}$ vector here is in the direction called $\hat{r}$ (not -$\hat{r}$!) regardless of if it points in from infinity or out to infinity.

An example from Griffiths's awesome book: Introduction to Electrodynamics might help:
Imagine two vectors $\vec{A}$ = $\hat{y}$ and $\vec{B}$= -$\hat{y}$. Both of these vectors are called $\hat{r}$ is spherical coordinates and in particular $\vec{A}$+ $\vec{B}$ is 0 NOT 2$\hat{r}$!

tiny-tim
Homework Helper
welcome to pf!

hi finnsmama! welcome to pf!

but that wasn't yesterday, it was last year …

i don't think kc8ukw is coming back!

Hey Tiny tim,
yeah, I knew that...but people search for answers to questions all the time and this one was not answered...and I had just gone over this in class! So answering seemed the right thing to do...