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I did lot of searching but couldn't find any textbook nor any webpage which could clarify my doubt. Maybe my doubt is insanely stupid and I was dumb for not realizing it at the first place.

The negative of the work done by the electrostatic field in bringing a charge from infinity to a point is called electric potential.

Let us assume that there is a positive charge at the origin. Let work done by the external force to bring a positive from infinity to a point P close to the origin be W. And hence work done by the field to bring a positive charge from infinity to the point P will be -W.

If I did work W to bring the charge from infinity or if the field brought the charge from infinity, either way the change in potential energy will be same.

In case of work done by the field, we say that the work done is stored in the form of electrostatic potential energy.

In case of work done by the external force, we say that the work done was positive and the energy was taken by some external source.

In the above case, I did work W in bringing it & the field did work -W on the charge trying to push it away so I did positive work of W & the field did negative work of W. We can say that the work done by the field was stored as potential energy.

Where did the work done by the external force (me) go?

I can summarize the whole doubt in the following line: I did work W to bring an charge towards another unlike charge, and therefore the field also did work -W, the net work done on the charge is 0. But there is still a change in potential energy. Why did the potential energy change? Isn't conservation of energy being violated here?

Please clarify my doubt (I do understand that there is a horrible conceptual error in one of my arguments but I do not know which one it is). If I am not wrong this has nothing to do with electrostatics rather has to deal with field theory/inverse square law. I believe I will be encountering the same problem again when I would be studying another force, maybe gravitational, which obeys inverse square law.

Thank You

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You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the electric force negative work, so that there is no net gain of kinetic energy in the object. It is a mistake to say that in this description the particle also has potential energy, because in doing so you be considering the work made by the electric field twice (both, as doing work and as gaining potential energy. The descriptions are equivalent, but it is either one or the other. If you chose the potential energy description then you no longer deal with the work of the electric force, as it is implicitly inside the concept of potential energy.

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You are in your reasoning overlooking something. Look at the diagram below:

Coulomb's Law.

$-q_1,+q_2$ are two point charges at distance $r$. Coulomb's Law dictates that the attractive electrostatic attraction force between them is:

$$F=k_e\frac{|q_1q_2|}{r^2}$$

And the electrostatic potential $U(r)$:

$$dU(r)=F(r)dr$$

$$U(r)=-k_e\frac{|q_1q_2|}{r}$$

Assume now that we hold $-q_1$ stationary and allow $+q_2$ to fall towards it to new distance $r'$, then $+q_2$ acquires kinetic energy as follows:

$$\frac{mv^2}{2}=k_e|q_1q_2| \bigg(\frac{1}{r'}-\frac{1}{r}\bigg),$$

where $v$ is the speed at $r'$ and $m$ the mass of $+q_2$.

Of course $\frac{mv^2}{2}$ is also the work $W$ done on $+q_2$.

But of course $+q_2$ is still moving and to stop it you will have to apply a braking force which will do work of $-W$. Then there's no net increase in kinetic energy.

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  • $\begingroup$ Found a small error in your answer. You have accidentally written U = dF/dr instead of dU = Fdr. $\endgroup$ – Yashas Nov 16 '15 at 17:16
  • $\begingroup$ @YashasSamaga: ooopsie. You are correct, thank you. Edited. $\endgroup$ – Gert Nov 16 '15 at 17:22
  • $\begingroup$ You did correct the differential equation but the solution is still wrong. You forgot to update it. $\endgroup$ – Yashas Nov 16 '15 at 17:28
  • $\begingroup$ @YashasSamaga: 'haste makes waste'. $\endgroup$ – Gert Nov 16 '15 at 17:32

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