Confusion concerning notion and origin of electrostatic potential

An elementary circulation is mathematically defined by: $$dC$$=$$\overrightarrow{OM}$$.$$\overrightarrow{X}$$. Now, we aplly this to a ponctual charge: $$dC$$=$$\overrightarrow{OM}$$.$$\overrightarrow{E}$$

And we know that $$\overrightarrow{E}$$=$$\frac {q}{4\pi \epsilon r^2}$$.$$\overrightarrow{e_r}$$ while d$$\overrightarrow{OM}$$=$$dr$$.$$\overrightarrow{e_r}$$+...$$\overrightarrow{e_\theta}$$+...

Thus: $$dC$$=$$\frac {q}{4\pi \epsilon r^2}$$.$$dr$$

Now, what I don't understand is how did we keep going to get this result:

$$\frac {q}{4\pi \epsilon r^2}$$.$$dr$$=-d($$\frac {q}{4\pi \epsilon r}$$+cte)??

And why did we pose the electrostatic potential V=$$\frac {q}{4\pi \epsilon r}$$+cte???

Is V in this case the "circulation"? If you have any other ways to demonstrate the expression of the the electrostatic potential then please let me know. According to Wikipedia : V is defined as the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. If that is the case then does anyone know the mathematical proof of that?(and by that I mean a mathematical relation between the work and the potential)

Also how did we get to the conclusion that $$\overrightarrow{E}$$=-$$\nabla$$V ???

Not sure what you mean via circulation. But, Performing a line integral for a general path, $$\int -E \cdot dl$$ shows that the closed line integral is zero. applying stokes theorem on this gives that $$\nabla × -E = 0$$ from this we can conclude that$$E = - \nabla V$$ If we want to actually find this potential function we can use multiple methods. Given we actually have the equation for the electric field we have to set each vector components of this equal to the corresponding component of $$- \nabla V$$ and solve the system of equations. determining the physical interpretation of this potential function, Given it satisfies (1) $$E = - \nabla V$$ is easy, using the fundamental theorem of gradients

(2) $$V(b)-V(a)= \int_a^b \nabla V \cdot dl$$ Substituting (1) into (2) gives

$$V(b)-V(a) = \int_a^b - E \cdot dl$$

obviously the term on the right side corresponds to the work done against the field to move a unit charge from a to B, which corresponds to the difference in potentials. going further, when you solve for V there is a +c attached to the definition. this corresponds to the boundary conditions surrounding where we define potential to be zero. V is non unique , there are infinitely many V's that satisfy (1) only inserting a boundary condition to we get a unique function

$$\int_a^b-E \cdot dl = \int_k^b-E \cdot dl - \int_k^a -E \cdot dl$$ the left integral is obviously v(b) and the right integral is obviously V(a) saying that e.g b is now a variable gives $$V(b)= \int_k^b-E \cdot dl$$

Notice I have split the right hand side into 2 integrals that are measured from some reference point K this corresponds to the +c in the definition and has the physical interpretation as the work done against the field STARTING FROM REFERENCE to some point. most cases Reference is taken as infinity in which case the +c is 0

Given (1) is true (Using spherical gradient is actually way easier but I'm going to stick with cartesian) In cartesian, the system of equations u need to solve is

$$-\partial V /\partial x = E \cdot i$$ $$-\partial V /\partial y = E \cdot j$$ $$-\partial V /\partial z = E \cdot k$$

Relation between work and potential, Given (1)

$$\int_a^bE \cdot dl = - \int_a^b -E \cdot dl$$ $$\int_a^b E \cdot dl = - (V(b)-V(a))$$ Work from A to B = -Q (difference in potential)