Explain $-\int^{a}_{b}\frac{Q}{4 \pi \epsilon \bar{r}^{2}} \cdot d \bar{r}= \left[\frac{Q}{4\pi \epsilon \bar{r}}\right]^{a}_{b}$ Explain $-\int_{b}^{a}\frac{Q}{4 \pi \epsilon \bar{r}^{2}} \cdot d \bar{r}= \left[\frac{Q}{4\pi \epsilon \bar{r}}\right]_{b}^{a}$. What is the term $\frac{1}{\bar{r}^{2}} \cdot d\bar{r}$? I find it hard to think about it.
$r$ is a radius of a ball.
 A: In rigorous mathematics, one does not usually write this as you did with a multiplication dot, as the symbol $\mathrm{d}\bar{r}$ is considered as a formal part of the notation of the integral (indicating the variable of integration) and not as a factor to be multiplied. However, this integral comes from physics, and in physics it's quite usual to treat expressions like $\mathrm{d}\bar{r}$ as if they were "infinitesimal quantities" which can be manipulated algebraically much like ordinary quantities (though not exactly like them). In fact, this was also the viewpoint taken by many in mathematics before analysis was put on a rigorous basis with definitions that don't use such "infinitesimal quantities". However, there's a branch of mathematics called non-standard analysis that does deal with basing analysis on a formal definition of what such "infinitesimal quantities" mean.
Having said all that, in the field where this integral occurs, in physics, $\mathrm{d}\bar{r}$ is often considered as an infinitesimal change in the radius $\bar{r}$, and then $\frac{1}{\bar{r}^2}\mathrm{d}\bar{r}$ means the product of that change with the inverse square of the radius. That is, the integral is considered as "summing" over a large/infinite number of small/infinitesimal radial intervals, and for each interval the product of the change in radius on that interval, $\bar{r}$, with (some proportionality constant times) the inverse square $\frac{1}{\bar{r}^2}$ of the radius is added to the value of the integral. Does that answer your question?
A: Maybe, I write up your equation in more conventional (physics) notation. You have the electric field $$\vec{E}(\vec{r}) = \frac{Q}{4 \pi \epsilon_0 r^2} \frac{\vec{r}}{r} \in \mathbb{R}^3$$ at the position $\vec{r}\in \mathbb{R}^3$ of a particle with charge $Q$ at the origin ($r=|\vec{r}|$). Integrating this field over a line gives the potential difference between the endpoints of the line. In your case you want to integrate the electric field from somewhere on the surface with $r=a$ to the surface with $r=b$ (along which path exactly you integrate the field does not matter as the system is radially symmetric and we are in a static situation  [mathematically $\nabla \wedge \vec{E}(\vec{r}) =0$]. So we want to integrate (this is a line integral)$$\int_{r=a}^{r=b} \vec{E}(\vec{r}) \cdot d\vec{r} = [-U(r)]_{r=a}^{b} $$
where we introduced the potential $U(r) = \frac{Q}{4 \pi \epsilon_0 r}$ with the property that $E= - \nabla U$ and the gradient theorem.
