Understanding Electric potential I am reading about Electrostatics in Griffiths "Introduction to Electrodynamics". Referring to the integral $\int \vec{E}. d\vec{l}$ The book says: "Because the line integral is independent of path, we can define a function
$$V(\vec{r})=-\int_\mathcal{O}^\vec{r} \vec{E}\cdot d\vec{l},$$
Where $\mathcal{O}$ is some standard reference point..."
I have a few questions about the intuition behind the above
I understand that the Electric field is a conservative field. So I think, if we were to consider V about a closed loop would always be zero. This would mean that the above V can only be defined between two points on the loop. This is fairly logical (if I am correct in thinking that). 
My issue comes with the minus sign, I don't understand its significance intuitively, and in fact I don't even understand what the integral means intuitively.
If someone could offer me some help in understanding this, that'd be great.
 A: I'll address your questions out of order:
This integral tells you that the potential from a source is a product of that source's field and how far away you are from it ($d\vec l$). However, you cannot know how far away you are from the source without some reference point $\mathcal{O}$. Typically we take this reference point to infinitely-far away (i.e., the potential from a source charge is zero at infinity). 
Now Griffiths writes the formula as 
$$ V = - \int_\mathcal{O}^\vec{r} \vec{E} \cdot d\vec{l}, $$
as you mention. The minus sign is strictly a pedagogical choice. It encourages students to think of integrating from infinity down to their location of interest, because it forces the reference point to be the beginning of the bounds of integration (in other words, you integrate from $\mathcal{O}$ to $\vec r$). This cultivates conceptual understanding because it makes you consider your reference point as the start of the problem.
Let's do a simple example. Let's suppose we are distance $x$ from a point charge $q$. We know from Coulomb's law that the field from this charge is
$$ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}.$$
Now let's plug this into our integral. We have
$$ V = - \int_\mathcal{O}^\vec{r} \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \cdot d\vec{l}.$$
We have now choose to our reference point. Let's choose $\mathcal{O} \rightarrow \infty$, and let's plug in ${r} = x$ and $d\vec{l} = d\vec{r}$. Then
$$ V = - \int_\infty^x \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \cdot d\vec{r} = -\frac{q}{4\pi\epsilon_0} \int_\infty^x \frac{dr}{r^2}$$
$$ =  -\frac{q}{4\pi\epsilon_0} \left(-\frac{1}{x}\right) = \frac{1}{4\pi\epsilon_0}\frac{q}{x}.$$
And that's our potential at $x$! Easy!
Now to answer your first question. The electric field is conservative, but its closed loop integral is not necessarily zero. It is in electrostatics. Time-variant magnetic fields create electric fields. See Faraday-Lenz law:
$$ \oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}$$
or
$$ \nabla \times \vec{E} = -\frac{\partial B}{\partial t}. $$
A: You are no doubt aware that the work done by a force is $\int \vec{F}\cdot d\vec{l}$. Well the force on a charge due to an electric field is $q\vec{E}$, so the integral in your question is minus the work done per unit charge in moving it from O to $\vec{r}$.  If the electric field does positive work on the charge then it gains kinetic energy, so you must lose potential energy in order that the total energy is conserved - hence the minus sign. $V(\vec{r})$ is the potential energy per unit charge (with respect to its starting position).
A: First, a mathematical intuition: 
The fact that the line integral between any two points in space is independent of the path implies that you can assign each point in space a number ("potential") and calculate the line integral from A to B by subtracting the "potential" assigned to B from the "potential" assigned to A. 
This pre-assignment is very convenient for problem solving. 
Proof follows these lines:
$$
\int_A^B{Edl}=\int_A^O{Edl}+\int_O^B{Edl}=\int_O^B{Edl}-\int_O^A{Edl}
$$
The proof also suggests a procedure for assigning the potential: pick an arbitrary reference point O and calculate $\int_O^A{Edl}$ for each point A in space. 
From physics point of view, the work you apply for moving a particle from A to B is
$$
-\int_A^B{Edl}
$$
So by defining potential as $$-\int_O^A{Edl}$$
you can calculate the work that you apply when moving a particle from A to B as the potential in B minus the potential in A. 
With the minus sign, everything "falls in place". For example, a particle at a high potential point tends to fall to a low potential location. 
A: Sure, glad to help because you've overcome the most difficult part. Yes, you're right in your reasoning.
The "minus sign issue" cannot be easier: it's just a convention. We can choose the sign like that. In other words, we want to work with that minus sign in the middle. 
If it's not satifying enough, do the following: let's call that integral just "$B$".
$B=\int_a^b \vec{E}\cdot d\vec{r} $.
Okay, now we just say "we want to define another quantity $V$ as $V=-B$; and we say we don't wanna see B's anymore, they're too many variables, let's replace $V$ everywhere. That's it.
It means that
$\vec{E}=\vec{\nabla}B=\vec{\nabla}(-V)=-\vec{\nabla}V$
