Earnshaw’s Theorem and Ring of Charge












3












$begingroup$


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







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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


















3












$begingroup$


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









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.







$endgroup$








  • 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
















3












3








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









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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited yesterday









Qmechanic♦

106k121971227




106k121971227






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asked 2 days ago









W.P.McBlainW.P.McBlain

182




182




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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
















  • 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












3 Answers
3






active

oldest

votes


















3












$begingroup$

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









$endgroup$





















    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









    $endgroup$





















      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









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        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









        $endgroup$


















          3












          $begingroup$

          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









          $endgroup$
















            3












            3








            3





            $begingroup$

            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









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            dmckee♦dmckee

            75k6135272




            75k6135272























                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









                $endgroup$


















                  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









                  $endgroup$
















                    1












                    1








                    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









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    QuantumnessQuantumness

                    461117




                    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









                        $endgroup$


















                          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









                          $endgroup$
















                            0












                            0








                            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









                            $endgroup$



                            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.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            my2ctsmy2cts

                            5,7222719




                            5,7222719






















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