Let us consider a system of charges in space. The potential energy of the system of charges is determined by the amount of work done by the external force to assimilate the charges in that manner. But what is the potential energy of a particular charge in that system of charges? Does that question makes any sense? Because potential can't be defined for a single charge, it is always defined for a system, as far I know. If there is some definition about the potential energy of a single charge then please mention it.
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5 Answers
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The potential energy of a system of point-like charges $q_i$ at positions ${\bf r}_i$ is $$ U=-\frac{1}{4 \pi \epsilon_0}\sum_i\sum_{j>i} \frac{q_i q_j}{|{\bf r}_i-{\bf r}_j|} $$
Such formula can be intepreted (and derived) as the work done by the Coulomb forces (better to avoid to introduce additional forces in the definition) to bring together the charges from an infinite relative distance, to their positions.
It turns out, looking at the formula, that this work can be interpreted as well as the sum of the work to assemble the first pair, summed to the work to add a third charge to the first pair, plus the work required to add a fourth particle to the first triple, plus ..., plus the work required to add the $N$-th charge to the previous $N-1$.
Does this observation allow to say that the energy of the system of $N$ charges coincides with the energy of one of the charges interacting with the other $N-1$ ?
Yes, because the previous formula says that. But one has to be careful to understand what is implicit in the formulae, if we would like to exploit them.
What has to be very clear with formula for potential energy is that in any case the potential energy remains a property of the whole system. This should be evident, if we think what would happen in a system of just two charges. We could fix one of them (say $q_1$) at its final position and then we evaluate the work done on the second charge (say $q_2$), when it is moved from infinity to its position. Even though we could speak of the work done by the force due to particle $1$ on particle $2$, and then speak about the potential energy of charge $q_2$, it is clear that it is a potential energy $U_{12}$ of the two-charge system. Indeed, if, after assembling the system, we fix charge $q_2$ and we free charge $q_1$, it starts to move according the to work-energy theorem, keeping fixed $K_1+U_{12}$, where $K_1$ is the kinetic energy of charge $q_1$.
answered Mar 10, 2019 at 13:26
GiorgioP-DoomsdayClockIsAt-90GiorgioP-DoomsdayClockIsAt-90
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Strictly speaking,
Potential energy of a charged particle at a point ( $\vec r $ ) is the amount of work done by the external force in bringing that charge from infinity to that particular point
Obviously, if there are no charges around (including static and in motion), the work done would be zero as the other charge would not experience any force.
Potential energy of a particular charge of the system ( q ) means you already had the other charges of your system already in place and then you bring the concerned charge q whose P.E. you want to find from infinity to that point.It can also be calculated by subtracting the potential energy of the system of other charges (excluding the charge q ) and subtracting it from the potential energy of the whole system ( including q )
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Because potential energy can't be defined for a single charge, it is always defined for a system.
This is the issue. You can look at the energy contained in the system, or you can just look at a single charge $Q$. All you have to do is calculate.
$$U=-\frac{1}{4\pi\epsilon_0}\sum_i\frac{Qq_i}{r_i}$$ where we are summing over all charges except for the one in question. $r_i$ is the distance from charge $Q$ to charge $q_i$
I have been having some discussions in the comments of another answer about the validity of this. I want to address that I am not saying that potential energy is purely a property of a single body. I am also not saying that the idea of defining the potential energy contained in the entire system is incorrect. All I am saying is that it is not unreasonable to talk about the potential energy of a single body due to all of its interactions.
We typically do this in introductory physics when we use $mgy$ for the potential energy of an object in the nearly uniform gravitational field near the Earth's surface. Of course by using this we are not saying that the potential energy is only a property of the object of mass $m$. But I don't see an issue with saying the object has potential energy $mgy$ and therefore the force it experiences is $F=-\frac{\text d U}{\text d y}=-mg$.
Of course there is the issue mentioned in the comments of this answer of relating the potential energy of each charged object to the total potential energy of the system. There is a simple fix of just adding up each individual energy and then dividing by $2$ to address the double counting. I have seen this done in more than one text book.
I am not saying that this is the only "definition", or even that it is better than only considering things in terms of interactions. I am just trying to say that it is possible to use potential energy in this way. I think it works fine, and I don't think we should say we can't do this.
answered Mar 10, 2019 at 12:16
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$\begingroup$ Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges? $\endgroup$ Commented Mar 10, 2019 at 20:57
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$\begingroup$ @HarryJohnston Kind of. It is the sum of each interaction. If you added up the energy of each charge then you would have actually double counted, so the easy fix would to add everything up and then just divide by $2$. $\endgroup$ Commented Mar 10, 2019 at 21:11
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Potential energy of a single object is not defined.
Introductory expositions often begin with the potential energy of an object in Earth's gravitational field near the Earth's surface $U=mgh$. Students often keep that initial picture in mind even when a more proper definition is presented.
A slightly better definition is $\Delta U = -W_\mathrm{internal}$, the work done by internal conservative forces. This definition makes it clear that two objects are required, allows for the grouping of potential energy by source, and removes any ambiguities that might be caused by external forces, which otherwise might be thought to contribute to PE. Furthermore, this definition forces a clear distinction between the system and its environment. Finally, it makes explicit that only changes in energy are physically significant.
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2$\begingroup$ I don't understand. Just because potential energy depends on the interaction between two bodies doesn't mean it isn't defined for a single body. $\endgroup$ Commented Mar 10, 2019 at 12:11
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$\begingroup$ @AaronStevens Potential energy is the energy of configuration (or position) of a pair of interacting objects. It is defined by internal work, which requires at least two objects. For examples, look at the explicit expressions that appear in the other answers to this question. GirogioP has the best answer here; study it carefully. The "definition" of PE as the work to assemble the system can fail if there are non-conservative forces either internally or externally. $\endgroup$– garypCommented Mar 11, 2019 at 1:11
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$\begingroup$ Yeah I completely understand those definitions. Those are fine and I'm not saying those are wrong at all. I also agree that potential energy is due to interactions. I'm just saying I don't see why you can't define the potential energy that a single charge has due to all of the interactions it has with the rest of the configuration. $\endgroup$ Commented Mar 11, 2019 at 1:55
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$\begingroup$ @AaronStevens (I'm surprised!) If a 1 kg masschanges position by 1 m above Earth's surface, which mass would change by 9.8 J, the 1 kg or the $6\times 10^{24}$ kg mass? Neither. You can't assign the energy change to either mass. It belongs to the system because the configuration of the system changed. The Earth's change of position in the field of the 1 kg mass is just as valid (though less convenient) as the 1 kg in Earth's field. To respond to your last sentence, we don't define potential energy that way. $\endgroup$– Bill NCommented Mar 11, 2019 at 2:07
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$\begingroup$ @BillN I think it's still valid. For example, we can say that an object experiences a force $F$, even though the force is an interaction between two things. And we can take the conservative force that acts on that object and then say the potential energy of the object is related to this force by $F=-\nabla U$. We even talk about the potential energy of a charge based on the potential at the location by $U=qV$ $\endgroup$ Commented Mar 11, 2019 at 2:19
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To your question, it does make sense for a single charge in an electric field to have its own potential energy. As you have effectively defined but didn't understand, the potential energy of anything is the amount of work done against the attraction or repulsion force to move that thing $l$ distance closer or away from the source of the force. In the context of a charge in an electric field, the potential energy of a single charge is equal to the amount to work done against the coulomb force, either repulsion or attraction, to move the charge $l$ distance closer or away from the source of electric field.
