Earnshaw’s Theorem and Ring of Charge The Next CEO of Stack OverflowWhy can't charge be in a stable equilibrium in electrostatic field?Why can't I contain a charged particle electrostatically in stable equilibrium? Referring to Earnshaw's TheoremEarnshaw's theorem and springsIs electric charge truly conserved for bosonic matter?Why can't charge be in a stable equilibrium in electrostatic field?Work done by the battery in series with capacitor with changing dielectricEquilibrium in ElectrostaticsProving Earnshaw's theorem is subtle in three-dimensions!Dilemma of classical physics: stationary particles that can't be in stable equilibriumElectric field produced by ring of charge (opposite charge on each semi circle)Earnshaw's theorem and stability of charge inside a hollowed cavity of conductorIs (DC/battery) voltage a result of charge? or energy? or both?
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Earnshaw’s Theorem and Ring of Charge
The Next CEO of Stack OverflowWhy can't charge be in a stable equilibrium in electrostatic field?Why can't I contain a charged particle electrostatically in stable equilibrium? Referring to Earnshaw's TheoremEarnshaw's theorem and springsIs electric charge truly conserved for bosonic matter?Why can't charge be in a stable equilibrium in electrostatic field?Work done by the battery in series with capacitor with changing dielectricEquilibrium in ElectrostaticsProving Earnshaw's theorem is subtle in three-dimensions!Dilemma of classical physics: stationary particles that can't be in stable equilibriumElectric field produced by ring of charge (opposite charge on each semi circle)Earnshaw's theorem and stability of charge inside a hollowed cavity of conductorIs (DC/battery) voltage a result of charge? or energy? or both?
$begingroup$
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 Mar 24 at 5:59
Qmechanic♦
107k121981230
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
$endgroup$
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 Mar 24 at 5:59
Qmechanic♦
107k121981230
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
$endgroup$
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
Mar 24 at 2:59
$begingroup$
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
Mar 24 at 3:43
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
Mar 24 at 3:44
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 Mar 24 at 5:59
Qmechanic♦
107k121981230
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
$endgroup$
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 Mar 24 at 5:59
Qmechanic♦
107k121981230
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
edited Mar 24 at 5:59
Qmechanic♦
107k121981230
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
edited Mar 24 at 5:59
Qmechanic♦
107k121981230
edited Mar 24 at 5:59
Qmechanic♦
107k121981230
edited Mar 24 at 5:59
Qmechanic♦
107k121981230
107k121981230
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
asked Mar 24 at 2:06
W.P.McBlainW.P.McBlain
182
182
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
Mar 24 at 2:59
$begingroup$
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
Mar 24 at 3:43
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
Mar 24 at 3:44
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
Mar 24 at 2:59
$begingroup$
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
Mar 24 at 3:43
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
Mar 24 at 3:44
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
Mar 24 at 2:59
$begingroup$
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
Mar 24 at 2:59
$begingroup$
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
Mar 24 at 3:43
$begingroup$
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
Mar 24 at 3:43
1
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
Mar 24 at 3:44
$begingroup$
Also read this and this.
$endgroup$
– harshit54
Mar 24 at 3:44
add a comment |
3 Answers
3
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 Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
$endgroup$
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 Mar 24 at 3:11
QuantumnessQuantumness
461117
$endgroup$
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 Mar 24 at 3:18
my2ctsmy2cts
5,7522719
$endgroup$
add a comment |
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3 Answers
3
active
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 Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
$endgroup$
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 Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
$endgroup$
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 Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
$endgroup$
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 Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
answered Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
answered Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
answered Mar 24 at 4:01
dmckee♦dmckee
75.1k6135273
75.1k6135273
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 Mar 24 at 3:11
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 Mar 24 at 3:11
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 Mar 24 at 3:11
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 Mar 24 at 3:11
QuantumnessQuantumness
461117
answered Mar 24 at 3:11
QuantumnessQuantumness
461117
answered Mar 24 at 3:11
QuantumnessQuantumness
461117
answered Mar 24 at 3:11
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 Mar 24 at 3:18
my2ctsmy2cts
5,7522719
$endgroup$
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 Mar 24 at 3:18
my2ctsmy2cts
5,7522719
$endgroup$
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 Mar 24 at 3:18
my2ctsmy2cts
5,7522719
$endgroup$
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 Mar 24 at 3:18
my2ctsmy2cts
5,7522719
answered Mar 24 at 3:18
my2ctsmy2cts
5,7522719
answered Mar 24 at 3:18
my2ctsmy2cts
5,7522719
answered Mar 24 at 3:18
my2ctsmy2cts
5,7522719
5,7522719
add a comment |
add a comment |
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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
Mar 24 at 2:59
$begingroup$
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
Mar 24 at 3:43
1
$begingroup$
Also read this and this.
$endgroup$
– harshit54
Mar 24 at 3:44