What is the difference between Electrostatic potential (not "Potential Energy"!) And Electric potential

Question

I have spent an hour with all I had got to figure out the diffrence between them : So What I understood is please correct me if I'm wrong (I know I will be dead wrong) That electrostatic potential occurs in stationary charges and that this term means The potential or the ability of a charge to do work and for Electric potential it's the same the only diffrence is that the charges are not stationary but moving

Answer 1

"Electrostatic potential" implies that it is given by positions of all charges at the same time. Usually given by the Coulomb formula for potential $$ \phi(\mathbf x) = \sum_k K\frac{q_k}{|\mathbf x - \mathbf r_k|}. $$

"Electric potential" is a more general term, it implies only a function of position and time $\phi(\mathbf x,t)$ which is to be used in the formula for electric field

$$ \mathbf E = -\nabla \phi - \partial_t \mathbf A $$

Electrostatic potential is a special kind of electric potential, but it is the most common and most useful one. There are other kinds, like retarded electric potential, used when solving the wave equation for EM fields of accelerated charged particles.

Answer 2

Unfortunately, the vector notation is not the best one to grasp the actual nature of electromagnetism. Indeed, the best way is to dive into relativity and consider the electric and magnetic field together. In Minkowski space-time, define the 4-potential

$$\phi = (V,\mathbf A).$$

To find the associated field strength, do what you would normally do, i.e. take the differential of the potential and set

$$F = \text d\phi$$

With a slight abuse of the vector notation, the above differential expands to

$$F = \left(-\nabla V + \frac{\partial\mathbf A}{\partial t}\right)\cdot(\text dt\wedge\text d\mathbf x) + (\nabla\times\mathbf A)\cdot(\star\text d\mathbf x),$$

where $\star\text d\mathbf x$ is the vector of spatial surface elements

$$\star\text d\mathbf x = (\text dx_2\wedge\text dx_3, \text dx_3\wedge\text dx_1, \text dx_1\wedge\text dx_2).$$

From this expression you can now recognise the electric and magnetic components of the Faraday tensor field $F$:

$$\mathbf E = -\nabla V + \frac{\partial\mathbf A}{\partial t}$$ $$\mathbf B = \nabla\times\mathbf A$$

Undoubtedly, the covariant definition of the electromagnetic tensor is simpler: just $F = \text d\phi$. Thanks to this one can then produce the more general formula for the electric component and from that produce the electric potential as the path integral of $\mathbf E$, which in general is a function of the path itself:

$$V[\gamma] = \int_\gamma \mathbf E\cdot\text d\mathbf x.$$

When we talk about something static, e.g. the electrostatic potential, we can then assume that the time derivatives vanish (simply by definition), so that the electric component of the Faraday tensor reduces to the gradient of a scalar field. In this case, the electrostatic potential coincides with the scalar field $V$, i.e. the time component of the 4-potential $\phi$.