Earnshaw’s Theorem and Ring of Charge
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A classic problem in determining the motion of a negative charge when displaced from a positively charged ring shows that the charge oscillates.
However, Earnshaw’s theorem states that (quoting Griffiths) ‘A charged particle cannot be held in stable equillibrium by electrostatic forces alone’. However, the system above seems to be stable. What causes this seemingly contradictory results?
homework-and-exercises electrostatics charge stability
edited yesterday
Qmechanic♦
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asked 2 days ago
W.P.McBlainW.P.McBlain
182
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A classic problem in determining the motion of a negative charge when displaced from a positively charged ring shows that the charge oscillates.
However, Earnshaw’s theorem states that (quoting Griffiths) ‘A charged particle cannot be held in stable equillibrium by electrostatic forces alone’. However, the system above seems to be stable. What causes this seemingly contradictory results?
homework-and-exercises electrostatics charge stability
edited yesterday
Qmechanic♦
106k121971227
asked 2 days ago
W.P.McBlainW.P.McBlain
182
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1
$begingroup$
If you displace the particle along the axis that passes through the center of the ring and perpendicular to the plane of the ring, you get a restoring force. However, it seems likely that if you displace the particle instead in the plane of the ring towards one part of the ring, there is no restoring force; instead, there's a force pointed toward the nearest part of the ring. That's my guess.
$endgroup$
– march
2 days ago
$begingroup$
Earnshaw's Theorem according to wikipedia is applicable on point charges. However even in this case, why do you think that the ring will remain stationary? You will require a force to hold it in place.
$endgroup$
– harshit54
2 days ago
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
2 days ago
add a comment |
$begingroup$
A classic problem in determining the motion of a negative charge when displaced from a positively charged ring shows that the charge oscillates.
However, Earnshaw’s theorem states that (quoting Griffiths) ‘A charged particle cannot be held in stable equillibrium by electrostatic forces alone’. However, the system above seems to be stable. What causes this seemingly contradictory results?
homework-and-exercises electrostatics charge stability
edited yesterday
Qmechanic♦
106k121971227
asked 2 days ago
W.P.McBlainW.P.McBlain
182
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A classic problem in determining the motion of a negative charge when displaced from a positively charged ring shows that the charge oscillates.
However, Earnshaw’s theorem states that (quoting Griffiths) ‘A charged particle cannot be held in stable equillibrium by electrostatic forces alone’. However, the system above seems to be stable. What causes this seemingly contradictory results?
homework-and-exercises electrostatics charge stability
homework-and-exercises electrostatics charge stability
edited yesterday
Qmechanic♦
106k121971227
asked 2 days ago
W.P.McBlainW.P.McBlain
182
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Qmechanic♦
106k121971227
asked 2 days ago
W.P.McBlainW.P.McBlain
182
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Qmechanic♦
106k121971227
edited yesterday
Qmechanic♦
106k121971227
edited yesterday
Qmechanic♦
106k121971227
106k121971227
asked 2 days ago
W.P.McBlainW.P.McBlain
182
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
W.P.McBlainW.P.McBlain
182
asked 2 days ago
W.P.McBlainW.P.McBlain
182
182
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
W.P.McBlain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
If you displace the particle along the axis that passes through the center of the ring and perpendicular to the plane of the ring, you get a restoring force. However, it seems likely that if you displace the particle instead in the plane of the ring towards one part of the ring, there is no restoring force; instead, there's a force pointed toward the nearest part of the ring. That's my guess.
$endgroup$
– march
2 days ago
$begingroup$
Earnshaw's Theorem according to wikipedia is applicable on point charges. However even in this case, why do you think that the ring will remain stationary? You will require a force to hold it in place.
$endgroup$
– harshit54
2 days ago
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
2 days ago
add a comment |
1
$begingroup$
If you displace the particle along the axis that passes through the center of the ring and perpendicular to the plane of the ring, you get a restoring force. However, it seems likely that if you displace the particle instead in the plane of the ring towards one part of the ring, there is no restoring force; instead, there's a force pointed toward the nearest part of the ring. That's my guess.
$endgroup$
– march
2 days ago
$begingroup$
Earnshaw's Theorem according to wikipedia is applicable on point charges. However even in this case, why do you think that the ring will remain stationary? You will require a force to hold it in place.
$endgroup$
– harshit54
2 days ago
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
2 days ago
1
1
$begingroup$
If you displace the particle along the axis that passes through the center of the ring and perpendicular to the plane of the ring, you get a restoring force. However, it seems likely that if you displace the particle instead in the plane of the ring towards one part of the ring, there is no restoring force; instead, there's a force pointed toward the nearest part of the ring. That's my guess.
$endgroup$
– march
2 days ago
$begingroup$
If you displace the particle along the axis that passes through the center of the ring and perpendicular to the plane of the ring, you get a restoring force. However, it seems likely that if you displace the particle instead in the plane of the ring towards one part of the ring, there is no restoring force; instead, there's a force pointed toward the nearest part of the ring. That's my guess.
$endgroup$
– march
2 days ago
$begingroup$
Earnshaw's Theorem according to wikipedia is applicable on point charges. However even in this case, why do you think that the ring will remain stationary? You will require a force to hold it in place.
$endgroup$
– harshit54
2 days ago
$begingroup$
Earnshaw's Theorem according to wikipedia is applicable on point charges. However even in this case, why do you think that the ring will remain stationary? You will require a force to hold it in place.
$endgroup$
– harshit54
2 days ago
1
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
2 days ago
$begingroup$
Also read this and this.
$endgroup$
– harshit54
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
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Displacement along the symmetry axis results in a restoring force along the symmetry axis as you have calculated.
Now ask what happens if it is displaced radially?
Answer: the charge is unstable to radial displacements.
The theorem is proved in 3D space, so a problem confined to a single dimension is not subject to it. If you take the same geometry and consider it in 3D the theorem holds.
answered 2 days ago
dmckee♦dmckee
75k6135272
$endgroup$
add a comment |
$begingroup$
Classically, introducing extended objects such as a ring also introduces additional forces: The ring is held together by elastic forces working against the positive charge distribution which pushes against itself. Without it, the ring will expand indefinitely.
On the other hand, I am not sure if the statement holds quantum-mechanically, as one may argue that all "contact" forces are electromagnetic in nature.
answered 2 days ago
QuantumnessQuantumness
461117
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add a comment |
$begingroup$
A negative charge in the centre of a positively charged ring is in a metastable state, so not stable Any deviation form this perfect geometry will lead to collapse. The theorem applies.
answered 2 days ago
my2ctsmy2cts
5,7222719
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add a comment |
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3 Answers
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oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Displacement along the symmetry axis results in a restoring force along the symmetry axis as you have calculated.
Now ask what happens if it is displaced radially?
Answer: the charge is unstable to radial displacements.
The theorem is proved in 3D space, so a problem confined to a single dimension is not subject to it. If you take the same geometry and consider it in 3D the theorem holds.
answered 2 days ago
dmckee♦dmckee
75k6135272
$endgroup$
add a comment |
$begingroup$
Displacement along the symmetry axis results in a restoring force along the symmetry axis as you have calculated.
Now ask what happens if it is displaced radially?
Answer: the charge is unstable to radial displacements.
The theorem is proved in 3D space, so a problem confined to a single dimension is not subject to it. If you take the same geometry and consider it in 3D the theorem holds.
answered 2 days ago
dmckee♦dmckee
75k6135272
$endgroup$
add a comment |
$begingroup$
Displacement along the symmetry axis results in a restoring force along the symmetry axis as you have calculated.
Now ask what happens if it is displaced radially?
Answer: the charge is unstable to radial displacements.
The theorem is proved in 3D space, so a problem confined to a single dimension is not subject to it. If you take the same geometry and consider it in 3D the theorem holds.
answered 2 days ago
dmckee♦dmckee
75k6135272
$endgroup$
Displacement along the symmetry axis results in a restoring force along the symmetry axis as you have calculated.
Now ask what happens if it is displaced radially?
Answer: the charge is unstable to radial displacements.
The theorem is proved in 3D space, so a problem confined to a single dimension is not subject to it. If you take the same geometry and consider it in 3D the theorem holds.
answered 2 days ago
dmckee♦dmckee
75k6135272
answered 2 days ago
dmckee♦dmckee
75k6135272
answered 2 days ago
dmckee♦dmckee
75k6135272
answered 2 days ago
dmckee♦dmckee
75k6135272
75k6135272
add a comment |
add a comment |
$begingroup$
Classically, introducing extended objects such as a ring also introduces additional forces: The ring is held together by elastic forces working against the positive charge distribution which pushes against itself. Without it, the ring will expand indefinitely.
On the other hand, I am not sure if the statement holds quantum-mechanically, as one may argue that all "contact" forces are electromagnetic in nature.
answered 2 days ago
QuantumnessQuantumness
461117
$endgroup$
add a comment |
$begingroup$
Classically, introducing extended objects such as a ring also introduces additional forces: The ring is held together by elastic forces working against the positive charge distribution which pushes against itself. Without it, the ring will expand indefinitely.
On the other hand, I am not sure if the statement holds quantum-mechanically, as one may argue that all "contact" forces are electromagnetic in nature.
answered 2 days ago
QuantumnessQuantumness
461117
$endgroup$
add a comment |
$begingroup$
Classically, introducing extended objects such as a ring also introduces additional forces: The ring is held together by elastic forces working against the positive charge distribution which pushes against itself. Without it, the ring will expand indefinitely.
On the other hand, I am not sure if the statement holds quantum-mechanically, as one may argue that all "contact" forces are electromagnetic in nature.
answered 2 days ago
QuantumnessQuantumness
461117
$endgroup$
Classically, introducing extended objects such as a ring also introduces additional forces: The ring is held together by elastic forces working against the positive charge distribution which pushes against itself. Without it, the ring will expand indefinitely.
On the other hand, I am not sure if the statement holds quantum-mechanically, as one may argue that all "contact" forces are electromagnetic in nature.
answered 2 days ago
QuantumnessQuantumness
461117
answered 2 days ago
QuantumnessQuantumness
461117
answered 2 days ago
QuantumnessQuantumness
461117
answered 2 days ago
QuantumnessQuantumness
461117
461117
add a comment |
add a comment |
$begingroup$
A negative charge in the centre of a positively charged ring is in a metastable state, so not stable Any deviation form this perfect geometry will lead to collapse. The theorem applies.
answered 2 days ago
my2ctsmy2cts
5,7222719
$endgroup$
add a comment |
$begingroup$
A negative charge in the centre of a positively charged ring is in a metastable state, so not stable Any deviation form this perfect geometry will lead to collapse. The theorem applies.
answered 2 days ago
my2ctsmy2cts
5,7222719
$endgroup$
add a comment |
$begingroup$
A negative charge in the centre of a positively charged ring is in a metastable state, so not stable Any deviation form this perfect geometry will lead to collapse. The theorem applies.
answered 2 days ago
my2ctsmy2cts
5,7222719
$endgroup$
A negative charge in the centre of a positively charged ring is in a metastable state, so not stable Any deviation form this perfect geometry will lead to collapse. The theorem applies.
answered 2 days ago
my2ctsmy2cts
5,7222719
answered 2 days ago
my2ctsmy2cts
5,7222719
answered 2 days ago
my2ctsmy2cts
5,7222719
answered 2 days ago
my2ctsmy2cts
5,7222719
5,7222719
add a comment |
add a comment |
W.P.McBlain is a new contributor. Be nice, and check out our Code of Conduct.
W.P.McBlain is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
If you displace the particle along the axis that passes through the center of the ring and perpendicular to the plane of the ring, you get a restoring force. However, it seems likely that if you displace the particle instead in the plane of the ring towards one part of the ring, there is no restoring force; instead, there's a force pointed toward the nearest part of the ring. That's my guess.
$endgroup$
– march
2 days ago
$begingroup$
Earnshaw's Theorem according to wikipedia is applicable on point charges. However even in this case, why do you think that the ring will remain stationary? You will require a force to hold it in place.
$endgroup$
– harshit54
2 days ago
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
2 days ago