Point charge potential (sign problem)

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 $d\vec l=-\frac{\vec r}{r})$:

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

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It's not clear to me what exactly you're asking here. What exactly do you mean by "potential of a point charge"? Are you asking about the potential caused by a point charge at some other point? Or about the potential required to move a point charge from one place to another? Are you wondering why, if it's possible to compute the potential of a sphere of charge, you can't do so for a point? I think the question could use a little more explanation. –  David Z Dec 7 '11 at 0:58
The way I understand the question, it asks about the potential that a point charge in the origin would cause at some point $P$ a distance $R$ from the origin, and apparently the sign is wrong –  Lagerbaer Dec 7 '11 at 2:15
The definition is wrong. The right definition is $V(P)-V(Q) = -\int_P^Q E dl$, that is, the change in potential is the line integral. If you integrate from P to infinity, it's the opposite sign as integrating from infinity to P. –  Ron Maimon Dec 7 '11 at 6:03
@Ron: his definition (except for the P-R typo) is the one given by Wikipedia, and it seems correct. The line integral is the amount of work done by the field on the hypothetical particle, so it is the negative of the change in potential. –  Harry Johnston Dec 7 '11 at 8:00
@Harry: it is correct if used properly, but the OP is confused because of orders of limits, things we do by physical intuition. He wants the formalism to get it right, which is reasonable. In this case you need the integration range to go from R to infinity, and to justify this, you need to explain the minus sign. This comes from the fact that one integration gives V(A)-V(B), while integrating the other direction gives V(B)-V(A). I am not saying that Wikipedia is wrong, but for OP, it is confusing. In order for a formal person to not get confused, it is better to give the difference form. –  Ron Maimon Dec 7 '11 at 9:30
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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.

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This is exactly what I did: use the definition of path integral to find a more correct way, the point is that it is not clear to my what is the direction of $d\vec l$ and why –  wiso Dec 7 '11 at 12:01
@wiso: I've added another answer. –  Harry Johnston Dec 7 '11 at 19:17

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

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

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perfect, but I don't understand what's wrong in my computation –  wiso Dec 7 '11 at 11:53
sorry, I repeat: I perfectly understand your answer, but my question is: "Where is my computation wrong?", not "Give me another way to compute the correct result" –  wiso Dec 7 '11 at 12:12
My hope was that OP by comparing the two methods step-by-step could figure out where he goes wrong. OP is basically forgetting a factor $\cos (180^\circ)$ when he evaluates a dot product. –  Qmechanic Dec 7 '11 at 12:33
@Qmechanic: no, his dot product was fine, he just had the direction of dl wrong. –  Harry Johnston Dec 7 '11 at 18:43
Well, one can argue that $\vec{r}\cdot d\vec{r}<0$ should be negative in one interpretation of OP's flawed calculation. –  Qmechanic Dec 7 '11 at 19:24
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