Point charge potential (sign problem)

Question

I'm a bit embarrassed, but I'm not able to compute the electric potential at point $P$ (at a distance $R$ from the origin) generated by a positive unitary point charge in the origin with the right sign. Simply use the definition $$V(P) = -\int_\infty^P \vec E\cdot d\vec l,$$ forgetting the constant and choosing a straight line to integrate from infinity (so in the direction of $d\vec l=-\frac{\vec r}{r})$:

$$V(P) = -\int_\infty^P \frac{\vec r}{r^3}\cdot d\vec l = -\int_\infty^P \frac{\vec r}{r^3}\cdot d\left(-\frac{\vec r}{r}\right) = \int_\infty^R \frac{1}{r^2}dr = -\frac{1}{R} $$

Answer 1

You're calculating the line integral incorrectly; the direction of $\vec l$ is determined by the parametrization you use to describe the curve you want to integrate over, not by the direction of the curve itself. Because you're integrating backwards over $r$, you have to use $\vec l = \frac{\vec r}{r}$, not $\vec l = - \frac{\vec r}{r}$.

The easiest way to avoid this problem is to reverse the direction of the integration. I won't write out this calculation since I see someone already beat me to it. :-)

Another way is to use a more explicit parametrization, e.g., $\vec r(t) = (\frac{R}{t}, 0, 0)$ with $t$ running from 0 to 1.

$$V(R) = -\int_\infty^R \frac{\vec r}{r^3}\cdot d\vec l = -\int_0^1 \frac{\vec r}{r^3}\cdot \vec r' dt $$
$$ = -\int_0^1 ((\frac{R}{t})^{-2}, 0, 0) \cdot (\frac{-R}{t^2}, 0, 0) dt $$ $$ = \int_0^1 \frac{1}{R} dt = \frac{1}{R} $$

For extra credit (and to really see what's going on) try $\vec r(t) = (\frac{-R}{t}, 0, 0)$ with $t$ running from 0 to -1. The reversal of the direction of $\vec r'$ cancels out the fact that the integration is backwards.

EDIT:

If I understand correctly, you want to understand why the intuitive approach doesn't work. Here's another way of looking at it.

Conceptually, what you are trying to do is to add up the infinitesimal changes in potential $$\delta V = - \vec E \cdot d \vec l$$ over the curve from infinity to R. (I say "add up" rather than "integrate" deliberately, you'll see why in a moment.) On this curve, going in that direction, $\vec E$ and $\vec l$ are in opposite directions so the dot product is negative, making $\delta V$ positive. If all the $\delta V$s are positive, then of course so is the sum.

So if the sum is positive, why is the integral negative? Because you've silently switched from doing a line integral (now over a scalar field) to doing an ordinary integral. By convention, for an ordinary integral, if $a < b$ then $$\int_b^a = -\int_a^b$$

But for a line integral over a scalar field, $$\int_b^a = \int_a^b$$

So since in this case you are integrating from $\infty$ to $R$, mistaking a line integral for a regular integral causes the sign to switch.

The reason for the difference between a line integral and an ordinary integral is that the line integral represents (loosely speaking) the sum of the values along the curve whereas the ordinary integral is defined as the inverse of differentiation. When summing up scalars, it doesn't matter which end you start at, the result is the same; but differentiation reverses sign when you change directions.

Answer 2

Apart from a sign problem (which basically is caused by wrongly doing an integration in a direction opposite to the $\vec{E}$ field direction), there is also a problem(v3) with an apparent identification of (the change of) $\vec{\ell}$ (which has dimension of length) with (the change of) $\pm \frac{\vec r}{r}$ (which is dimensionless).

Try to compare with the following reasoning. (Let us for simplicity assume that the charge in the origin satisfies $\frac{Q}{4\pi\varepsilon_0}=1$.) The electrostatic field $\vec{E}$ is

$$-\vec{\nabla} V~=~ \vec{E}~=~\frac{\vec{r}}{r^3}. $$

Its length is

$$ E~=~ |\vec{E}|~=~\frac{1}{r^2}. $$

Then the potential is

$$ V(R)~=~ - \int_{r=\infty}^{r=R} \vec{E}\cdot {\rm d}\vec{r}~=~\int_{r=R}^{r=\infty} \vec{E}\cdot {\rm d}\vec{r}~=~\int_{r=R}^{r=\infty} E ~{\rm d}r$$ $$=~\int_{r=R}^{r=\infty} ~\frac{{\rm d}r}{r^2}~=~\left[\frac{-1}{r}\right]_{r=R}^{r=\infty} ~=~\frac{1}{R}. $$