# What is the difference between Electrostatic potential (not “Potential Energy”!) And Electric potential

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

• What you just stated is correct. Although both terms you used mean the same thing. I believe what you are trying to say is "What is the difference between electrostatic potential and electrodynamic potential?" For electrostatic cases (which is often taught for introduction to E&M courses), the potential is somewhat straight forward, but for moving charges (i.e. electrodynamics), you need to start consider other variables such as retarded time, for example. Look up Lienard-Wiechert potential. I think you will find your answer there. – Kane Billiot Jul 7 '18 at 21:35
• I belive potential energy and potential are two diffrent things – Anonymous Jul 7 '18 at 21:36
• @Kane Billiot Excuse me,I'm sorry but by moving charges I mean electricity not electrodynamics – Anonymous Jul 7 '18 at 21:48

"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.

• Sorry @Jan for being a Little bit implicit in my question I can understand your answer to my problem to a much extent, I've not yet progressed to wave equation until now so, could you please put it in terms of work done, cheers! – Anonymous Jul 8 '18 at 10:41
• I can't, retarded electric potential has no direct connection to work. – Ján Lalinský Jul 8 '18 at 20:51
• Oh Ok, I didn't know that before 👍 – Anonymous Jul 10 '18 at 10:34

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 $\phi$.

• Frankly speaking @Phoenix87 All this is fancy for me, No offense your answer is indeed a very professional one but Its at a level much much higher than I can understand ,I'm like literally a noob I feel in this amazing huge community. I'm acquainted till Biot-Savart law ! – Anonymous Jul 8 '18 at 10:50
• You can safely ignore the first half of the answer that shows how to derive the expressions for the electric and magnetic field in terms of the potentials. – Phoenix87 Jul 9 '18 at 15:20