Since $\EE$ is the derivative of $V$, we should be able to recover $V$ from $\EE$ by integrating. This is in fact correct, as can be seen by recalling the Master formula: \begin{eqnarray*} dV = \grad{V}\cdot{d\rr} \end{eqnarray*} Integrating both sides yields the fundamental theorem for gradients, namely \begin{eqnarray*} V \bigg|_A^B = \Int_A^B\grad{V}\cdot{d\rr} \end{eqnarray*} This integral does not depend on the path of integration! If you move from one place to the other, the difference in potential at your initial and final positions does not depend on the path you took.

A special case of this is \begin{eqnarray*} \oint\grad{V}\cdot d\rr = 0 \end{eqnarray*} where the circle on the integral sign indicates that the path is closed. If you return to your starting point, adding up the change in potential wherever you go, the net change will of course be zero.

Recall that \begin{eqnarray*} \EE = -\grad V \end{eqnarray*} Thus, \begin{eqnarray*} \oint\EE\cdot d\rr = 0 \end{eqnarray*} Both of these are properties of conservative vector fields. We also have \begin{eqnarray*} V = -\int\EE\cdot{d\rr} \end{eqnarray*} which can be used to find the potential from the field, as we now illustrate.

Suppose you have a positive charge $q$ located at the origin. You know the electric field is \begin{eqnarray*} \EE(\rr) = \frac{1}{4\pi\epsilon_0} \frac{q\,\rhat}{r^2} \end{eqnarray*} What is the potential at the point $B=(0,b)$?

We're really looking for the potential difference between some reference point and $B$. It is customary to take the reference point to be at infinity. So we need to choose a path from infinity to $B$, and integrate $\EE$ along it. Which path? The most obvious one to choose is along the $y$-axis. We therefore have $x=0=z$, so \begin{eqnarray*} d\rr = dx\,\ii + dy\,\jj + dz\,\kk = dy\,\jj \end{eqnarray*} Furthermore, on this path clearly $\rhat = \jj$, so that \begin{eqnarray*} V \bigg|_B = - \Int_\infty^b \frac{q}{4\pi\epsilon_0} \frac{dy}{y^2} = + \frac{1}{4\pi\epsilon_0} \frac{q}{y} \Bigg|_\infty^b = \frac{1}{4\pi\epsilon_0} \frac{q}{b} \end{eqnarray*} which should come as no surprise.

Let's try another path from infinity. Suppose we come in along the line $y=b$. Now we have $d\rr=dx\,\ii$, but $r=\sqrt{x^2+b^2}$, so this integral appears to be a bit harder. Try it yourself! So let's try spherical coordinates. Recall that \begin{eqnarray*} d\rr = dr\,\rhat + r\,d\theta\,\that + r\,\sin\theta\,d\phi\,\phat \end{eqnarray*} We know that $\theta=\frac\pi2$ in the $xy$-plane, but the relationship between $r$ and $\phi$ doesn't seem obvious. However, when we compute the integral, we get \begin{eqnarray*} V \bigg|_B &=& - \Int_\infty^B \frac{q}{4\pi\epsilon_0} \frac{q\,\rhat}{r^2} \cdot d\rr\\ &=& - \Int_\infty^b \frac{q}{4\pi\epsilon_0} \frac{dr}{r^2}\\ &=& + \frac{1}{4\pi\epsilon_0} \frac{q}{r} \Bigg|_\infty^b\\ &=& \frac{1}{4\pi\epsilon_0} \frac{q}{b} \end{eqnarray*} as before. The messy $d\phi$ term disappeared from the integral! This vividly demonstrates the path-independent nature of this integral. It is remarkable that nature produces electric fields with this property.