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As of 28/05/2019, Electric Potential is defined in Wikipedia in the article with the same name as:

The amount of work needed to move a unit of positive charge from a reference point to a specific point inside the field without producing an acceleration

What is the meaning of "without producing an acceleration"?

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    $\begingroup$ Any better sources for defining Electric Potential than Wikpedia? $\endgroup$ – user207455 May 28 at 16:24
  • $\begingroup$ It certainly is poorly worded, but perhaps the way to think about it is the test charge is at rest at both the reference point and the point inside the field. So, no net acceleration in the field. $\endgroup$ – Jon Custer May 28 at 16:45
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    $\begingroup$ Another way of looking at it: If an acceleration is produced, the final KE might not be the same as the initial. In that case, the work done by the applied force is not equal to the change in PE. As @Unique points out below, this definition of potential (energy) has difficulties. $\endgroup$ – garyp May 30 at 20:52
  • $\begingroup$ If you move a charge at constant velocity, it becomes a current and generates a magnetic field which is perpendicular to the motion of the charge. If you accelerate the charge, then the charge produces a squeezed electric field - shaped like an hour glass - it's no longer spherical.However, implicit in the definition of work is the assumption you're a moving test charge - which is a small charge small enough not to perturb the electric field or potential - and you're moving the test charge infinitesimally slow, i.e., $W=-q\int_{a}^{b} E\cdot dl$. $\endgroup$ – Cinaed Simson Jun 1 at 3:48
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    $\begingroup$ That Wiki article needs a good edit! $\endgroup$ – Andrew Steane Jun 3 at 17:36
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The total energy of a body is the sum of kinetic energy and potential energy.
Kinetic energy depends on speed and potential energy depends on position.

The wiki text is a special case of the fact that if we move a body from A to B without altering the kinetic energy then its energy difference is only potential.

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  • $\begingroup$ Hi, what is "position" in this context (some wiki link will be helpful)? Also, what is the "special case of the fact"? Thanks, $\endgroup$ – JohnDoea Jul 2 at 18:36
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An accelerating charge radiates EM waves that carry energy,
the static electric potential difference does not account for the radiative exchange.
To measure therefore the work done by the static potential difference while moving a charge from one place to another, must be performed so that the charge does not radiate.

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  • $\begingroup$ How do you take a charge that is at rest and move it to another place without accelerating it? Maybe the potential should include the radiation too. $\endgroup$ – J Thomas May 28 at 17:15
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    $\begingroup$ @J_Thomas you move it very very slowly. Of course, what is meant by "slow" depends on context, if the radiated energy is much smaller than the potential energy difference then the motion is "slow" enough; on the other hand if it is not negligible then the motion is not slow enough... This is very similar to the concept of reversible process in thermodynamics: It has to be quasi-static but how do we know to be so? $\endgroup$ – hyportnex May 28 at 17:28
  • $\begingroup$ I misunderstand the answer as a whole. Maybe its because I lack background knowledge... $\endgroup$ – JohnDoea May 28 at 19:44
  • $\begingroup$ @JThomas "How do you take a charge that is at rest and move it to another place without accelerating it?" You do it by decelerating it prior to bringing it to rest at the other place. In that way the net work done on the charge is zero and the net change in kinetic of the charge is zero. $\endgroup$ – Bob D May 28 at 20:36
  • $\begingroup$ @hyportnex Regarding your answer, the acceleration needed to get the charge moving and to decelerate it to bring it to rest can both be infinitely small. In between the net force can be zero and the charge move at very low constant velocity. If the accelerations can be made infinitely small, then so will the radiated energy be infinitely small. So I don't think radiant energy loss is the key here, but that's just my opinion. $\endgroup$ – Bob D May 28 at 20:41
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I'm going to try a different approach to answer your question than my original answer.

The amount of work needed to move a unit of positive charge from a reference point to a specific point inside the field without producing an acceleration

What is the meaning of "without producing an acceleration"?

This is not a good definition of electric potential, which is supposed to be synonymous with voltage. A better definition is "The potential difference $V$ between two points is the work per unit charge required to move the charge between the points" (re: NCEE FE exam reference manual for Electrical and Computer Engineering).

Secondly, the statement "without producing an acceleration" is misleading. Electrons are accelerated by an external electric field, but at the same time they are decelerated due to constant collisions with atoms and molecules in the material in which they flow losing kinetic energy in the form of heat. After each collision they once again accelerate. The end result is a constant average or "drift" velocity.

Since the potential difference between two points is intended to only refer to a difference in potential energy between the points, I'm guessing the statement "without producing an acceleration" may allude to the fact acceleration would result in a change in kinetic energy of the charge between the points, which is not part of the change in potential energy. But it's only a guess.

Hope this helps.

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  • $\begingroup$ This is actually the only correct answer among them all. The definition that you cite is the actual correct definition of potential. $\endgroup$ – gented Jun 4 at 14:05
  • $\begingroup$ @gented This definition does account for any difference of the initial and final velocity, so it is incomplete. This is the very point of the OP. Of course in electronics any kinetic energy difference is negligible and you can get away with this. If not, I'd be interested in a counterexample. $\endgroup$ – my2cts Jun 5 at 20:39
  • $\begingroup$ @my2cts I think you meant "does NOT account". Correct? In my opinion the definition from the NCEE manual is complete. The key word in the definition is "required" which can be interpreted to mean the minimum work needed to move the charge between the two points. A net change in kinetic energy is not "required" to move the charge between the two points. Without the word "required" the definition would be incomplete. $\endgroup$ – Bob D Jun 5 at 20:58
  • $\begingroup$ @BobD Thanks for spotting that. I meant "not". You are also right about "minimum required". Note that that is more like an operational definition. But then, how would you measure the minimum work? $\endgroup$ – my2cts Jun 5 at 21:55
  • $\begingroup$ @my2cts You may be right about it being more like an operational definition. After all, its from a PE FE exam reference guide meaning it's probably more an engineering definition than a physics definition. Not sure how to "measure" the minimum work. But per the work energy principle the net work done on an object equals its change in kinetic energy. So per the definition, the minimum work done between two points would be when there is no change change in kinetic energy, or zero. $\endgroup$ – Bob D Jun 5 at 22:50
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Let us imagine that we want to find the electric potential at a point $P$. To do so, we first select a reference point $R$. We then find the work done by us in moving a unit charge from $R$ to $P$. Let us understand the calculation of work in greater details. We apply a force $\vec{F}$ on the charge such that it just compensates the electric force on it. (The electric force comes from the electric field in the region. If there was no field then one would not find the potential.) Thus, $F$ has to be $-q\vec{E}$ and the work done will be \begin{equation} W = -\int_R^P q\vec{E}\cdot d\vec{x}. \end{equation} Since $\vec{F}$ just balances $q\vec{E}$ there is no acceleration of the charge. We say that the charge is moved `quasi-statically' from $R$ to $P$. Why do we need $\vec{F}$ to just balance $q\vec{E}$? It is because we want to apply just enough force to overcome the electric field and reach $P$ from $R$.

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  • $\begingroup$ If you consider the definition of potential energy of the system then it is defined as follows:-The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system So work done by us or taking the particle in equilibrium is not meaningful. $\endgroup$ – Unique May 28 at 18:22
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The work energy theorem states that the net work done on an object equals the change in its kinetic energy.
Therefore, in order for the work done on an object to only change the potential energy of object, the net work done on the object must equal zero;
"equal zero" means that no change in kinetic energy and therefore no acceleration causing that change.

Charge

In the case of charge, when work is done to move a positive charge between two points in opposition to the force the electric field exerts on the charge, that work only will result in an increase in potential energy if no net work is done on the charge which would cause an increase in kinetic energy.

The process is as follows:

The force exerted by an external agent moving a positive charge is in the same direction as the movement of the charge. Therefore that work done by the agent is considered positive and transfers energy to the charge.

Simultaneously the force on the electric charge due to the field is opposite to the movement of the charge. The work done by the field is therefore negative. In order for the charge to only acquire electrical potential energy the negative work done by the field has to equal the positive work done by the external agent in order that the net work done on the charge is zero. If it were less, the charge would have both potential and kinetic energy.

Bottom line: For there to only be a change in electrical potential energy the field has to take all the energy given the charge by the external agent so that there is no net work done on the charge, and store it as electrical potential energy of the charge electric field system.

The analogy with gravity is external positive work (say done by you) is needed to raise a mass initially at rest on the ground to a point at rest a height $h$ above the ground. Simultaneously gravity (whose force is opposite the motion of the mass) does an equal amount of negative work taking the energy you supplied the mass and storing it as gravitational potential energy in the mass earth system. Since there is no change in kinetic energy of the mass (and thus no net acceleration to cause such change), no net work is done on the mass just like no net work is done on the charge.

Hope this helps.

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  • $\begingroup$ If you consider the definition of potential energy of the system then it is defined as follows:-The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system. So work done by us or any other external agent or taking the particle in equilibrium is not meaningful. $\endgroup$ – Unique May 28 at 18:26
  • $\begingroup$ @Unique Certainly that definition describes the mechanism by which the internal electrical force gives the charge potential energy by taking energy from the charge by doing negative work. But that energy has to come from somewhere, namely the external agent that does positive work. To imply that external work is not meaningful is to imply the internal force is creating energy in violation of Conservation of Energy. It’s like saying I play no role in giving an object gravitational potential energy when I lift it, i.e. gravity “creates” potential energy by itself. $\endgroup$ – Bob D May 28 at 18:52
  • $\begingroup$ The definition of change in potential energy doesn't contain the word external agent anywhere. $\endgroup$ – Unique May 29 at 3:31
  • $\begingroup$ @Unique No external agent is needed if there is a loss of potential, that is when the field does positive work in moving positive charge in the direction of the field since that’s the natural direction that the charge would move. An external agent is needed to increase the potential of the charge, that is to move positive charge opposite the direction of the field since that’s against nature. It’s not clear if the change in potential in your definition is positive or negative. Please cite the reference for it so I can see it in context. $\endgroup$ – Bob D May 29 at 5:35
  • $\begingroup$ @Unique The wording of the definition is a little odd because it says the “negative of the work done by the internal conservative forces”. The work done by the internal conservative force in increasing potential IS negative. The negative of the negative work is positive. Do you see what I’m getting at? Please cite the reference so we can resolve this. $\endgroup$ – Bob D May 29 at 5:45
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The definition of potential energy and potential(may be electric or gravitational) has lots of misconceptions and arguments related to it.I think that the introductory textbooks are faulty in providing the exact definition of potential energy.

Please consider these 3 points for potential energy:

  1. Absolute potential energy is not defined.
  2. Change in potential energy is well defined.
  3. Potential energy is defined only for a system consisting of 2 or more than 2 particles.

Please consider the definition below

The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system.

This definition is the standard definition of potential energy of a system. This definition never says that acceleration is not allowed.
This definition gives us the whole authority to allow acceleration.
Some books provides its definition by use of external conservative forces acting on the system.
But these definitions have a fault in them because the work done by external forces on the system would be equal to change in potential energy of the system if and only if the change in kinetic energy of the system is zero.So these definitions consist of a specific case that acceleration must be zero.

So I think that you must have understood the definition of potential energy properly. Similar is the analogy of electric potential.

Consider these points for electric potential:

  1. It is defined for a system consisting of 2 or more than 2 particles.
  2. The charges due to which electric potential is calculated must be fixed.
  3. The test charge must be of small magnitude.

Electric potential for a system of charges is defined as the negative of work done by internal electrical forces of the system on the test charge per unit test charge.

Again,this definition allows acceleration.

Hope this helps!

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    $\begingroup$ Thank you. I surveyed the books on my shelf. All of them define potential energy in this way, but with varying degrees of clarity. Halliday, Serway, Chabay and Sherwood, Symon, Taylor, Cummings et. al, Kinght (prelim ed.,) all do it this way. Students take note. Part of the problem is motivating PE, almost always starting with $mgh$. It's hard to get to the general definition when starting from $mgh$. Too many presentations are unclear, incorrect, or both. $\endgroup$ – garyp Jun 1 at 12:38
  • $\begingroup$ @JohnDoea thanks for your helpful edit.Well the whole discussion about this specific topic is that acceleration is allowed for defining electric potential. $\endgroup$ – Unique Jun 2 at 7:35
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If you rather let it go, the body will gain kinetic energy in expense of potential energy (due to conservation of energy).

When you move the body at constant velocity (i.e. no acceleration) you have to do work against the force producing the potential. This work done would be equal to the gain in kinetic energy it would have, if you had let it go. Hence, for energy conservation to hold, this must be equal to the loss is potential energy.

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