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According to Wikipedia :

"An electric potential (also called the electric field potential, potential drop or the electrostatic potential) is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any acceleration"

But how can any object be moved without producing even the slightest acceleration upon impact ?

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    $\begingroup$ Imagine moving the charge very very slowly. The acceleration would be negligible. And this is the case, that this definition is made for. $\endgroup$ – Steeven Jul 26 '18 at 18:07
  • $\begingroup$ @Steeven but that would mean their kinetic energy is also negligible, no ? i mean, for example if we need to calculate collisions that would generate heat in a resistor $\endgroup$ – physicsnewbie Jul 26 '18 at 18:13
  • $\begingroup$ @physicsnewbie This is not saying the only way to find potential is to move a charge slowly from the reference point to the point in question. This is just one way to look at what the potential is. Electric potential exists in all cases with charged particles and electric fields. $\endgroup$ – Aaron Stevens Jul 26 '18 at 18:36
  • $\begingroup$ @AaronStevens i just think of an electron as an object falling inside a gravitational field, and can't imagine it not accelerating $\endgroup$ – physicsnewbie Jul 26 '18 at 19:04
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    $\begingroup$ You can move an object very quickly and very slowly to the same high speed. When done slowly, the acceleration is small. And it also takes a lot of time. In thermodynamics you also often assume negligible accelerations, to avoid extra terms with forces that would cause the accelerations. You must then remember that in real life that particular definition is not the full picture. $\endgroup$ – Steeven Jul 26 '18 at 19:40
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First, it is important to understand why the movement of the change should be done without any acceleration.

This is because, in order to accelerate the charge, the force applied to it has to exceed the force of the field acting on the charge and, as a result, the total work done on the charge, while moving it from the reference point A to some point B, will exceed the potential difference between A and B, by the amount of the extra work spent to accelerate the charge.

Once that understood, you should see that the suggestions in the comment - to move the charge very slow - will lead to as accurate measurement as you wish, given that, in theory, you can move the charge infinitely slow.

If, however, you want to complete the measurements quickly, there are a couple of methods which would still give you accurate results.

1) You can accelerate the charge from the initial zero velocity at point A to any desired velocity and slow it down to zero velocity, when you arrive to point B, in which case the positive work spent to accelerate the charge will be canceled by the negative work spent to slow it down and, therefore, the total work will be exactly equal to the potential difference between A and B.

2) You can accelerate the charge from the initial zero velocity at point A to any desired velocity, move the charge at that velocity all the way to B and subtract the kinetic energy of the charge from the total work. The result will be exactly equal to the potential difference between A and B.

3) You can accelerate the charge to a desired velocity before you arrive to point A and then continue moving the charge all the way to point B at that velocity. In this case the total work, performed on the charge while moving it, at a constant velocity, from A to B, will be exactly equal to the potential difference between A and B, without any adjustments.

The advantage of method 3 over 1 and 2 is that you can accurately measure the potential relative to A at any point on the way from A to B and end up with an accurate potential map of the whole path. You'll have the same ability, if you move at infinitely low speed, but it would take more time.

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  • $\begingroup$ "the force applied to it has to exceed the force of the field acting on the charge and" , the force field strength is the orbital the electron is in ? so to pull it out of that field , a bigger force would be needed just like lifting up a box would take a force to overcome the gravity force field $\endgroup$ – physicsnewbie Jul 28 '18 at 10:28
  • $\begingroup$ @physicsnewbie In the definition of potential, we are talking about a free positive unit charge, e.g., a free positron or, if the definition was for a negative unit charge, it would be a free electron, i.e., the electron that was separated from its atom earlier. $\endgroup$ – V.F. Jul 28 '18 at 10:49
  • $\begingroup$ if it's free , which field attracts it ? i mean which field does it need to be "released" from ? other charges in it's vicinity ? $\endgroup$ – physicsnewbie Jul 28 '18 at 10:55
  • $\begingroup$ also, in option 1) that you mentioned, what is the initial state of the charge, is the charge "standing still" ? what if it collides with another object before it reaches it's "destination" what would be it's momentum when they collide ? $\endgroup$ – physicsnewbie Jul 28 '18 at 10:58
  • $\begingroup$ @physicsnewbie It could be any electrostatic field, created by a charged body, like a point charge (could be another electron) or a charged sphere, located anywhere (one foot or one mile away). There is potential associated with that field and this potential changes from point to point. Here, we are defining a potential at a random point, B, of this field relative to a chosen reference point A, which is typically chosen at infinity. So, if we say a potential at B is x, we mean potential relative to infinity. If we say pot at B rel. to A, we mean potential difference between A and B. $\endgroup$ – V.F. Jul 28 '18 at 11:07

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