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Earnshaw’s Theorem and Ring of Charge

3

1

$begingroup$

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

share|cite|improve this question

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

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 | 

3

1

$begingroup$

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

share|cite|improve this question

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

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

share|cite|improve this question

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

asked Mar 24 at 2:06

W.P.McBlainW.P.McBlain

182

$endgroup$

share|cite|improve this question

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

asked Mar 24 at 2:06

W.P.McBlainW.P.McBlain

182

share|cite|improve this question

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

asked Mar 24 at 2:06

W.P.McBlainW.P.McBlain

182

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

edited Mar 24 at 5:59

Qmechanic♦

108k122001246

asked Mar 24 at 2:06

W.P.McBlainW.P.McBlain

182

asked Mar 24 at 2:06

W.P.McBlainW.P.McBlain

182

3 Answers
3

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 3:11

QuantumnessQuantumness

461117

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

$endgroup$

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.

share|cite|improve this answer

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

$endgroup$

share|cite|improve this answer

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

answered Mar 24 at 4:01

dmckee♦dmckee

75.3k6136273

1

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 3:11

QuantumnessQuantumness

461117

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 24 at 3:11

QuantumnessQuantumness

461117

$endgroup$

share|cite|improve this answer

answered Mar 24 at 3:11

QuantumnessQuantumness

461117

answered Mar 24 at 3:11

QuantumnessQuantumness

461117

answered Mar 24 at 3:11

QuantumnessQuantumness

461117

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719

$endgroup$

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.

share|cite|improve this answer

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719

$endgroup$

share|cite|improve this answer

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719

answered Mar 24 at 3:18

my2ctsmy2cts

5,8042719