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When a charge is moved from infinity to $r$, its electric potential energy is equal to the negative work done by the field. In this argument, is it also true that there is positive work done by the force to get from infinity to $r$. Or is it just that one exists. If both are true, is the field losing energy. I have drawn an energy drawing and am trying to capture a picture about the whole system. Note: A positive test is moved, and the distance is from a positive charge

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    $\begingroup$ What is the sign of the charge being moved? What is the direction of the electric field between infinity and $r$? $\endgroup$ – Bob D Mar 29 at 11:46
  • $\begingroup$ What Bob D said. $\endgroup$ – Gert Mar 29 at 11:51
  • $\begingroup$ A positive test is moved, and the distance is from a positive charge $\endgroup$ – Mudit Tulsianey Mar 29 at 13:33
  • $\begingroup$ Please put that information in the question itself. The purpose of the comments is to improve the question. $\endgroup$ – Dale Mar 29 at 13:49
  • $\begingroup$ Point of clarification: the potential energy "belongs" to the system of charges, NOT to any single charge. The potential energy changes by moving a charge around, but because there isn't a fixed, universal coordinate system, the change in the potential energy cannot depend on the origin, or which charge is moved. Potential energy belongs to a system, not to an individual object. $\endgroup$ – Bill N Mar 29 at 18:05
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When a charge is moved from infinity to $r$, the EPE is equal to the negative work done by the field. In this argument, is it also true that there is positive work done by the force to get from infinity to $r$. Or is it just that one exists.

Its both.

An external agent does positive work and the electric field simultaneously does an equal amount of negative work. But you need to be careful of your wording when you say "the EPE is equal to the negative work done by the field". The work done by the field is itself negative since the force of the field is in the opposite direction as the displacement of the charge. So the change in EPE then equals the negative of the negative work done by the field, meaning the change in EPE is positive.

Or equivalently you can say the change in EPE equals the positive work done by the external agent, since its force is in the same direction as the displacement. Either way there is a positive change in EPE.

Bottom line: You can say change in EPE is the negative of the negative work done by the electric field or, equivalently, it is the positive work done by the external agent. Either way, the result is a positive change in EPE. The electric field takes the work done by the external agent and stores it as electrical potential energy.

Hope this helps.

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Suppose you have a charge $q$ and an electric field $E(r)$ in space. By integration of the electric field, also an electrical potential $V(r)= \int E(r) dr$ can be found.

The electrical potential energy of the charge at position $r$ is given by $\mathcal{E}(r_1) = qV(r_1)$. When the charge moves to another place by the electrical force $F(r)=qE(r)$, it will have another potential energy $\mathcal{E}(r_2)= qV(r_2)$. The potential energy difference of the charge is equal to the work done by the electric field on the charge because of the conservation of energy, this is $$\mathcal{E}(r_1) - \mathcal{E}(r_2) = \int_{r_1}^{r_2} F(r) dr \,.$$

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  • $\begingroup$ Ok that makes sense, however in situations where a force is used to move the charge and negative work done by the field, how does conservation of energy work then? $\endgroup$ – Mudit Tulsianey Mar 29 at 13:33
  • $\begingroup$ If the work done by the field is negative, then the potential difference will be positive. If the work done by the field is positive, then the potential difference will be negative. $\endgroup$ – Frederic Mar 29 at 13:41
  • $\begingroup$ My confusion comes from this: two interpretation of the EPE- it s equal to the work done by the electric field or work done by an external force. I am assuming these can not be combined, I just have issues with the electric field or the external force doing negative work as in reality why would that happen $\endgroup$ – Mudit Tulsianey Mar 29 at 14:03
  • $\begingroup$ These two concepts are the same. The electric field will exert the "external" force on the charge $F=qE$. In the system considered here, the force acting on the charge is the force generated by the electric field and thus the electric field delivers the work. $\endgroup$ – Frederic Mar 29 at 14:13
  • $\begingroup$ If the field exerts the force, in the situation of the two positive charges how can the distance r between them be reduced, increasing the EPE $\endgroup$ – Mudit Tulsianey Mar 29 at 14:51
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I think there's a big conceptual issue here.

1- A field does not gain or lose energy, a material system gains or loses it.

If a charge moves, it might gain energy, or lose energy. But we're talking about a particle. Energy belogns to systems, not to forces or fields.

2- A usual confusion:

$W_{cons}=-\Delta E_p$

The work done by the electrostatic force is "minus the increment of EPE". Yes, work done... by the electric force!

But the work done by ME is the opposite. The work done by people is always the oone pposing to a field. If you want to move a charge, you have to "overcome" the work done by the nature.

So $W_{me}=+\Delta E_p$

If I have to do a negative work, it's because the field is doing it for me, that's good. If I have to do a possitive work, then I am fighting against nature.

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  • $\begingroup$ A field does not gain or lose EPE. It is the combined charge-field system that gains or loses EPE. It's the same thing for gravitational potential energy (GPE). An object alone does not possess GPE. Its the object-earth system that possesses GPE. $\endgroup$ – Bob D Mar 29 at 18:06
  • $\begingroup$ See the comment by Bill N above. $\endgroup$ – Bob D Mar 29 at 18:08
  • $\begingroup$ You're right, I agree, but I wanted to simplify it. If you lift a rock, the rock gains potential energy. You're right that the energy would dissappear if gravity is gone, but I'm presuming it will not do so. I was assuming that the field is somehow "fixed" $\endgroup$ – FGSUZ Mar 29 at 19:59
  • $\begingroup$ No problem. It is common and convenient to say the rock gains potential energy but technically the it is the rock-earth system that gains the energy. The distinction becomes important when we apply the work energy theorem because people get confused when we say the net work done on the rock is zero. The only way to clarify why is to point out the GPE is not possessed by the rock alone. $\endgroup$ – Bob D Mar 29 at 20:05

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