Doubts in understanding some concepts of potential energy
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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.
electrostatics charge potential-energy coulombs-law
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add a comment |
$begingroup$
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.
electrostatics charge potential-energy coulombs-law
$endgroup$
add a comment |
$begingroup$
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.
electrostatics charge potential-energy coulombs-law
$endgroup$
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.
electrostatics charge potential-energy coulombs-law
electrostatics charge potential-energy coulombs-law
edited 4 hours ago
Aethenosity
1033
1033
asked 14 hours ago
Rifat SafinRifat Safin
816
816
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5 Answers
5
active
oldest
votes
$begingroup$
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 )
$endgroup$
add a comment |
$begingroup$
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.
$$U=-frac{1}{4piepsilon_0}sum_ifrac{Qq_i}{r_i}$$
where we are summing over all charges except for the one in question. $r_i$ is the distance from charge $Q$ to charge $q_i$
$endgroup$
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
The potential energy of a system of point-like charges $q_i$ at positions ${bf r}_i$ is
$$
U=-frac{1}{4 pi epsilon_0}sum_isum_{j>i} frac{q_i q_j}{|{bf r}_i-{bf r}_j|} $$
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 $q_1$.
$endgroup$
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
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.
$endgroup$
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.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@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.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
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.
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add a comment |
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5 Answers
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active
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5 Answers
5
active
oldest
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$begingroup$
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 )
$endgroup$
add a comment |
$begingroup$
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 )
$endgroup$
add a comment |
$begingroup$
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 )
$endgroup$
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 )
answered 13 hours ago
StarboyStarboy
716
716
add a comment |
add a comment |
$begingroup$
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.
$$U=-frac{1}{4piepsilon_0}sum_ifrac{Qq_i}{r_i}$$
where we are summing over all charges except for the one in question. $r_i$ is the distance from charge $Q$ to charge $q_i$
$endgroup$
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
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.
$$U=-frac{1}{4piepsilon_0}sum_ifrac{Qq_i}{r_i}$$
where we are summing over all charges except for the one in question. $r_i$ is the distance from charge $Q$ to charge $q_i$
$endgroup$
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
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.
$$U=-frac{1}{4piepsilon_0}sum_ifrac{Qq_i}{r_i}$$
where we are summing over all charges except for the one in question. $r_i$ is the distance from charge $Q$ to charge $q_i$
$endgroup$
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.
$$U=-frac{1}{4piepsilon_0}sum_ifrac{Qq_i}{r_i}$$
where we are summing over all charges except for the one in question. $r_i$ is the distance from charge $Q$ to charge $q_i$
answered 13 hours ago
Aaron StevensAaron Stevens
12.8k42248
12.8k42248
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
Am I right in thinking that the potential energy for the system isn't the sum of the potential energy of the individual charges?
$endgroup$
– Harry Johnston
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
@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 $2$.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
The potential energy of a system of point-like charges $q_i$ at positions ${bf r}_i$ is
$$
U=-frac{1}{4 pi epsilon_0}sum_isum_{j>i} frac{q_i q_j}{|{bf r}_i-{bf r}_j|} $$
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 $q_1$.
$endgroup$
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
The potential energy of a system of point-like charges $q_i$ at positions ${bf r}_i$ is
$$
U=-frac{1}{4 pi epsilon_0}sum_isum_{j>i} frac{q_i q_j}{|{bf r}_i-{bf r}_j|} $$
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 $q_1$.
$endgroup$
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
The potential energy of a system of point-like charges $q_i$ at positions ${bf r}_i$ is
$$
U=-frac{1}{4 pi epsilon_0}sum_isum_{j>i} frac{q_i q_j}{|{bf r}_i-{bf r}_j|} $$
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 $q_1$.
$endgroup$
The potential energy of a system of point-like charges $q_i$ at positions ${bf r}_i$ is
$$
U=-frac{1}{4 pi epsilon_0}sum_isum_{j>i} frac{q_i q_j}{|{bf r}_i-{bf r}_j|} $$
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 $q_1$.
edited 5 hours ago
answered 12 hours ago
GiorgioPGiorgioP
3,7051526
3,7051526
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
$begingroup$
Beautiful answer.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
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.
$endgroup$
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.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@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.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
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.
$endgroup$
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.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@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.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 14 hours ago
garypgaryp
16.9k13064
16.9k13064
2
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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.
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– Aaron Stevens
13 hours ago
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@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.
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– garyp
19 mins ago
add a comment |
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.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@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.
$endgroup$
– garyp
19 mins ago
2
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.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
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.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@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.
$endgroup$
– garyp
19 mins ago
$begingroup$
@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.
$endgroup$
– garyp
19 mins ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 12 hours ago
answered 13 hours ago
TechDroidTechDroid
57212
57212
add a comment |
add a comment |
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