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When talking about bringing a charge from a reference point to another point, we consider the work done by an "operator" force opposite to the electric one, but wouldn't that just cancel the acceleration given by the electric force and not make the charge move the other way?
And what if the charges are opposite? The work would be as if I am trying to get it from its position to the symmetric of the reference point by its position.

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2 Answers 2

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The "bringing a charge from infinity" is an idealized situation to remove some ambiguity from the potential, where we explicitely take advantage of the $\tfrac 1 r$-dependence. There are some potentials that do not go to zero at infinity (the line charge is one exemple) and thus we need to arbitrarely choose a point in space as the zero of the potential.

but wouldn't that just cancel the acceleration given by the electric force and not make the charge move the other way?

The situation is even more problematic if we remember that at infinity the force is zero! What would happen in real life is we need a small net force at the beginning to make the charge move towards the other, and then for the rest of the motion we keep the net force zero.

And what if the charges are opposite?

When the charges have the same sign the operator force must pump energy into the system in order to bring them closer together. That's why the potential is positive (remember, potential = potential energy / charge). When the charges are opposite the operator force must take energy out of the system, otherwise one charge would "fall" into the other, making the potential negative.

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  • $\begingroup$ 1st quote) Then why define the concept of electric potential energy as such? The equal and opposite force I mean. 2nd quote) What did physicist originally think of when the expression of electric potential energy was defined? The work needed to oppose that of the electric field? $\endgroup$
    – GDGDJKJ
    Commented Apr 12, 2019 at 14:14
  • $\begingroup$ Yes, change in potential energy is equal to the work required to oppose the electric field. If it was equal to the work done by the field the potential would flip sign. The first was chosen to make the electric field positive when the potential is a decreasing function of $r$. $\endgroup$
    – ErickShock
    Commented Apr 12, 2019 at 17:54
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The force on a charge $q$ in a uniform electric field, $E$, is $F=qE$ and is constant. The work required to move the charge a distance $d$ in the field is

$$W=qEd$$

If the charge is positive, and the charge moves in the direction of the electric field (+ to – by convention) solely under the influence of the field, the field does positive work on the charge. It accelerates the charge gaining kinetic energy equal to the work done by the field.

On the other hand, if as you indicated an external force is applied to the charge against the electric field (- to +) and the force is equal to the force of the electric field, there will be no acceleration but the charge does move at constant velocity. The external force does positive work on the charge, but the electric field does an equal amount of negative work, so the net work done on the charge is zero. The end result is the work done by the external force on the charge is stored as electrical potential energy of the charge.

Addendum:

This will respond to your follow up comment.

You start by applying a slightly greater force to get the charge moving then reduce it to equal the opposing field force. Now you are moving at constant velocity. Then before reaching the second point reduce the force to less than the field to bring it to a stop. Then change in kinetic energy is zero and no net work done. But charge has gained PE. Similar to gravitational PE when raising a msss

Hope this helps.

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  • $\begingroup$ Yes I am aware of that and it is my reason for asking, why define an energy that contributes to "nothing"? I am targeting the claim when teaching the concept of potential of bringing a charge from a reference point to another which can't happen by applying an equal opposite force. $\endgroup$
    – GDGDJKJ
    Commented Apr 12, 2019 at 14:09
  • $\begingroup$ See addendum to my answer $\endgroup$
    – Bob D
    Commented Apr 12, 2019 at 14:32
  • $\begingroup$ Thank you for your answer, that is not the way it is being taught though. $\endgroup$
    – GDGDJKJ
    Commented Apr 12, 2019 at 15:05
  • $\begingroup$ @Luyw How is it being taught? $\endgroup$
    – Bob D
    Commented Apr 12, 2019 at 15:20
  • $\begingroup$ No one mentions a force bigger than the electric one to make it move the opposite way and then that force becomes equal to the electric one (in magnitude). $\endgroup$
    – GDGDJKJ
    Commented Apr 12, 2019 at 18:56

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