# Sign of scalar product in electric potential integral?

## Main Question or Discussion Point

the potential difference between b and a is defined as follows:

V(b) - V(a) = -∫E $\bullet$dl

the integral is taken from a to b.

so the potential of a positive charge, with infinity as reference, is

V(r) - V(infinity) = V(r) = -∫E $\bullet$dl

the integral is from infinity to r.

My question is this. E points outward, and since integral is from infinity to r, we would be moving inward over the integral. So dl, and hence the scalar product in the integral, should be negative. But since E is inverse square in r, integrating gives another negative sign, giving us 3 in all.This would lead to a negative potential for the positive charge.

Clearly, the potential of the positive charge is positive. But that means the scalar product in this integral is also positive, even though E and dl seem to be pointing is opposite directions.

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don't you see the minus sign before it.

don't you see the minus sign before it.
true. taking E$\bullet$dl to be negative, the negative signs combine to give a positive.

But E is inverse square in r. When you compute the integral in terms of r, you get an extra negative sign, which makes the potential negative. Integral of 1/r^2 is negative.

$\vec{dl}$ is a vector whose direction is opposite the direction of E and produces a minus sign. The scalar under integral should be integrated over $dl$ ( scaler) and this happens to be $-dr$. The last minus sign solves your problem.

vanhees71
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Let's do the integral. That might clarify this issue. As the electric field, let's take the Coulomb field of a point charge at rest in the origin,

$$\vec{E}(\vec{x})=\frac{Q}{4 \pi r^3} \vec{x}, \quad \text{with} \quad r=|\vec{x}|.$$

Since $\vec{\nabla} \times \vec{E}=0$ for all $\vec{x} \in \mathbb{R}^3 \setminus\{0\}$ and since this domain is simply connected, there exists a scalar potential, i.e.,

$$\vec{E}=-\vec{\nabla} \Phi.$$

It's most easy to calculate the potential in spherical coordinates, i.e.,

$$\vec{E}=\frac{Q}{4 \pi r^2} \vec{e}_r.$$

Obviously the potential is a function of $r$ only, and from that we get

$$-\vec{\nabla} \Phi=-\vec{e}_r \partial_r \Phi.$$

Comparison with the electric field gives

$$\Phi'(r)=-\frac{Q}{4 \pi r^2} \; \Rightarrow \Phi(r)=+\frac{Q}{4 \pi r} + \text{const}.$$

The arbitrary constant can be set to 0, which is the usual convention for the boundary condition of the potential at infinity.

Of course, you can get the same result by a direct evaluation of the line integral. You can take any path that connects a fixed point $\vec{x}_0 \neq 0$ with the point $\vec{x}$ and take

$$\tilde{\Phi(x)}=-\int_{C(\vec{x})} \mathrm{d} \vec{x} \cdot \vec{E}.$$

The only constraint is that the path may not run through the origin. You can alwahys use a circle with the center at the origin and then a radial piece to connect the two points $\vec{x}_0$ and $\vec{x}$. The circle doesn't contribute since the electric field is perpendicular to it and thus your result is

$$\tilde{\Phi}(\vec{x})=\Phi(\vec{x})-\Phi(\vec{x}_0)=\frac{Q}{4 \pi} \left (\frac{1}{r}-\frac{1}{r_0} \right)$$

in accordance with the result obtained above using the more direct method of solving the differential equation.

I'll do it too, just to see it the same way, differently... (with no animosity towards the above whatsoever),

with all the numbers that can get in the way we set equal to 1, we have for the electric field of a point charge
$$\vec{E}=\frac{q}{r^2}\hat{r}$$
we are going to preform
$$\Phi=-\int_{a}^{b}\vec{E}\cdot d\vec{l}=-\int_{\infty}^{r}\frac{q}{r'^2}\hat{r}'\cdot (\hat{r}'dr')=-\int_{\infty}^{r}\frac{q}{r'^2}dr'$$
Above I put in the electric field, Now I'll switch the limits of integration at the cost of a sign change
$$\Phi=\int_{r}^{\infty}\frac{q}{r'^2}dr'=\left(0-\left(-\frac{q}{r}\right)\right)=\frac{q}{r}$$
Now
$$-\partial_{r}\left(\frac{q}{r}\right)\hat{r}=\frac{q}{r^2}\hat{r}=\vec{E}$$
which goes back also. Now you have it two ways from two different people :)

$\vec{dl}$ is a vector whose direction is opposite the direction of E and produces a minus sign. The scalar under integral should be integrated over $dl$ ( scaler) and this happens to be $-dr$. The last minus sign solves your problem.
I have to disagree. This is a common mistake. When we write $\vec{dl}=-\vec{dr}$ and keep the limits from infinity to r, we obtain the wrong sign.
Even though we "see" we've chosen the "right" $\vec{dl}$, the choice is wrong. $\vec{dl}$ points in the positive direction of the coordinates ($\vec{dl}=\vec{dr}$) and the limits take care of the proper sign. Check it.

I have to disagree. This is a common mistake. When we write $\vec{dl}=-\vec{dr}$ and keep the limits from infinity to r, we obtain the wrong sign.
Even though we "see" we've chosen the "right" $\vec{dl}$, the choice is wrong. $\vec{dl}$ points in the positive direction of the coordinates ($\vec{dl}=\vec{dr}$) and the limits take care of the proper sign. Check it.
I didn't wrote $\vec{dl}=-\vec{dr}$. I wrote the relation for the scalar differential lengrh: $dl=-dr$ and I have another negative sign before, due the opposite direction of the vectors $\vec{dr}$ and $\vec{E}$

thanks to everyone that replied. my mistake was thinking that dl = - dr when in fact they are the same quantity, dl = dr .