Path orientation for calculating electric potential

[OmegaKV]

For line integrals in vector calculus,

\int^a_b F \cdot dl

I almost always see the path oriented from a to b.

But my textbook has the following (look at the first equation for V(r):

Since the integral's limits are from O to r, I would have expected dl to also be pointing in the direction from O to r (i.e. pointing in the radially inward (minus r hat) direction), but the math in the textbook seems to imply that dl points radially outward (positive r hat direction, from r to O). I say this because the result of E dot dl has no minus sign in front of it.

How do you know which direction to orient dl?

 

[vanhees71]

It doesn't matter at which point you start your integration path as long as it is not at the singularity at $\vec{r}=0$ of the Coulomb potential (I don't know what's discussed in your book, because you didn't tell us; maybe it's a spherical surface carrying some charge?). Changing the starting point of your path just adds a constant to the potential, but the only thing interesting is the field $\vec{E}=-\vec{\nabla} V$.

Here they used a path starting from $r=|\vec{r}| \rightarrow \infty$, making the potential $0$ at infinity, which is a convenient standard choice. Since the field is radial always, the only contribution is from the part going from infinity radially in, and you get the integral solved in your book.

 

[Gordianus]

For line integrals in vector calculus,

\int^a_b F \cdot dl

I almost always see the path oriented from a to b.

But my textbook has the following (look at the first equation for V(r):

Since the integral's limits are from O to r, I would have expected dl to also be pointing in the direction from O to r (i.e. pointing in the radially inward (minus r hat) direction), but the math in the textbook seems to imply that dl points radially outward (positive r hat direction, from r to O). I say this because the result of E dot dl has no minus sign in front of it.

How do you know which direction to orient dl?

The path goes from b to a. If you reverse the order the modulus stay the same but the sign changes.