Electrostatic Potential Energy Derivation

How is the boxed step , physically as well as mathematically justified and correct ?

Source:Wiki http://en.wikipedia.org/wiki/Electric_potential_energy

As work done = $- \Delta U$. for Conservative force and it shouldn't matter whether we take $ds$ or $-dr$ ?

And when $dr$ is just a notation to specify the variable and the real thing behind it , is a limit , why is it that $dr$ is so important here .

Image : http://www.artofproblemsolving.com/Forum/download/file.php?id=43358&mode=view

What is wrong here ?

$$\newcommand{\newln}{\\&\quad\quad{}} \begin{align}&\int^{r_b}_{r_a}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_a-U_b) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_r-U_\infty) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}} =-U_r ~~~~~~~ [U_\infty = 0], \cos\theta=-1 , \vec{A} \cdot \vec{B}=|A||B|\cos\theta\newln \Rightarrow -\int^{r}_{\infty}k\cdot\frac{q.q_o}{r^2}dr=-U_r ~~~~~~~ [\textrm{Coulomb's Law}]\newln \Rightarrow kq\cdot q_o\int^{r}_{\infty}\frac{1}{r^2}dr=U_r\newln \Rightarrow kq\cdot q_o\left[\frac{-1}{r} \right]^r_\infty=U_r\newln\Rightarrow \frac{-kq.q_o}{r}=U_r\newln \Rightarrow U_r=-\frac{kq.q_o}{r} \end{align}$$

Answer

When you calculate work, you do so along a given path. Here, that path has tangent vector $d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward.

Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate $r$ to parameterize the path, in that $r$ is larger at the start of the path and smaller at the end. This runs counter to what you would usually do when parameterizing such a path with an arbitrary parameter.

Let $\mathbf s_0$ and $\mathbf s_1$ be the starting and ending points of a path $\mathbf s(\lambda) = \mathbf s_0 + (\mathbf s_1 -\mathbf s_0)\lambda$. The work integral is then

$$W = \int_{\mathbf s_0}^{\mathbf s_1} \mathbf F(\mathbf s) \cdot d\mathbf s= \int_0^1 \mathbf F(\mathbf s(\lambda)) \cdot \frac{d\mathbf s}{d\lambda} \, d\lambda = \int_0^1 \mathbf F (\mathbf s(\lambda))\cdot (\mathbf s_1 - \mathbf s_0) \, d\lambda$$

For two finite points, the basic approach is sound, but it breaks down when you have a point at infinity involved. This is the reason that the problem of assembling a configuration is usually attacked with a different basic parameterization.

Instead, set $\mathbf s(\lambda) = \lambda \hat{\mathbf a}$ for some unit vector $\hat{\mathbf a}$ and set the bounds of the integral as being from $[\infty, R)$. This is the important point: even though the path is being traversed coming in from infinity, the parameterization means that $d\mathbf s/d\lambda = + \hat{\mathbf a}$, not minus as I originally thought. The path's still oriented outward; we're just traversing it backwards.

Here's how that integral looks:

$$W = \int_{\infty}^R \mathbf F(\lambda \hat{\mathbf a}) \cdot \hat{\mathbf a} \, d\lambda$$

Of course, we know the expression for the electric force:

$$\mathbf F(\mathbf r) = k\frac{qq_0 \mathbf r}{|r|^3}$$

Plug in $\mathbf r = \lambda \hat{\mathbf a}$ to get

$$\mathbf F(\lambda \hat{\mathbf a}) = k \frac{qq_0 \lambda \hat{\mathbf a}}{\lambda^3} = k \frac{q q_0 \hat{\mathbf a}}{\lambda^2}$$

We find that the integrand is then

$$W = \int_{\infty}^R k \frac{q q_0}{\lambda^2} \hat{\mathbf a} \cdot \hat{\mathbf a} \, d\lambda = \int_\infty^R k \frac{q q_0}{\lambda^2} \, d\lambda = - k \frac{q q_0}{R} < 0$$

The work is negative, so the change in potential energy $\Delta U = - W$ is positive as required.

So where is the problem then? As we've seen, there actually shouldn't be an extra negative sign coming in on line 4 (as posted in the OP's question). This is somewhat obscured because an explicit parameterization of the path is never written down in the first place--usually, you don't have to, but this problem is tricky enough that it helps immensely.