Facing a paradox: Earnshaw's theorem in one dimensionDoes this example contradict Earnshaw's theorem in one dimension?Classify equilibrium points and find bifurcation points of a non-linear dynamic systemEarnshaw's theorem and springsEarnshaw's theorem for extended conducting bodiesPotential due to charge over infinite grounded plane conductor using the method of imagesRelation between electric field and dipole momentEarnshaw's theorm and Effective potentialDielectric liquid sucked up between two cylinders with a voltage differenceElectrostatics: Induced Boundary Dipole LayerWhy do we assume simply connected domains and continuously differentiable fields in electromagnetism theory?Does this example contradict Earnshaw's theorem in one dimension?

If a Gelatinous Cube takes up the entire space of a Pit Trap, what happens when a creature falls into the trap but succeeds on the saving throw?

How to draw the figure with four pentagons?

I Accidentally Deleted a Stock Terminal Theme

Watching something be written to a file live with tail

If human space travel is limited by the G force vulnerability, is there a way to counter G forces?

Can I use a neutral wire from another outlet to repair a broken neutral?

Stopping power of mountain vs road bike

How much of data wrangling is a data scientist's job?

Do I have a twin with permutated remainders?

Doing something right before you need it - expression for this?

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

What is the most common color to indicate the input-field is disabled?

Would Slavery Reparations be considered Bills of Attainder and hence Illegal?

What do you call someone who asks many questions?

Why can't we play rap on piano?

Western buddy movie with a supernatural twist where a woman turns into an eagle at the end

Can a virus destroy the BIOS of a modern computer?

Did Shadowfax go to Valinor?

How is it possible to have an ability score that is less than 3?

Fully-Firstable Anagram Sets

How to take photos in burst mode, without vibration?

How to show the equivalence between the regularized regression and their constraint formulas using KKT

How to model explosives?

Is the Joker left-handed?



Facing a paradox: Earnshaw's theorem in one dimension


Does this example contradict Earnshaw's theorem in one dimension?Classify equilibrium points and find bifurcation points of a non-linear dynamic systemEarnshaw's theorem and springsEarnshaw's theorem for extended conducting bodiesPotential due to charge over infinite grounded plane conductor using the method of imagesRelation between electric field and dipole momentEarnshaw's theorm and Effective potentialDielectric liquid sucked up between two cylinders with a voltage differenceElectrostatics: Induced Boundary Dipole LayerWhy do we assume simply connected domains and continuously differentiable fields in electromagnetism theory?Does this example contradict Earnshaw's theorem in one dimension?













6












$begingroup$


Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?






electrostatics mathematical-physics potential classical-electrodynamics equilibrium







share|cite|improve this question











$endgroup$
















    6












    $begingroup$


    Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?






    electrostatics mathematical-physics potential classical-electrodynamics equilibrium







    share|cite|improve this question











    $endgroup$














      6












      6








      6


      1



      $begingroup$


      Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?






      electrostatics mathematical-physics potential classical-electrodynamics equilibrium







      share|cite|improve this question











      $endgroup$




      Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?






      electrostatics mathematical-physics potential classical-electrodynamics equilibrium




      electrostatics mathematical-physics potential classical-electrodynamics equilibrium






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 9 hours ago









      Aaron Stevens

      14.2k42252




      14.2k42252










      asked 12 hours ago









      SRSSRS

      6,513433123




      6,513433123




















          2 Answers
          2






          active

          oldest

          votes


















          10












          $begingroup$

          Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
            $endgroup$
            – SRS
            11 hours ago










          • $begingroup$
            @SRS Yes, that's true.
            $endgroup$
            – knzhou
            11 hours ago










          • $begingroup$
            I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
            $endgroup$
            – SRS
            11 hours ago










          • $begingroup$
            @SRS The potential at a point charge is not defined (or you could say infinite)
            $endgroup$
            – Aaron Stevens
            10 hours ago










          • $begingroup$
            I have to think more about it and I'll get back.
            $endgroup$
            – SRS
            10 hours ago


















          3












          $begingroup$

          So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$



          This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "151"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470522%2ffacing-a-paradox-earnshaws-theorem-in-one-dimension%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              11 hours ago










            • $begingroup$
              I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              10 hours ago










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              10 hours ago















            10












            $begingroup$

            Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              11 hours ago










            • $begingroup$
              I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              10 hours ago










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              10 hours ago













            10












            10








            10





            $begingroup$

            Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.






            share|cite|improve this answer











            $endgroup$



            Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 9 hours ago









            Aaron Stevens

            14.2k42252




            14.2k42252










            answered 12 hours ago









            knzhouknzhou

            46.2k11124223




            46.2k11124223











            • $begingroup$
              Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              11 hours ago










            • $begingroup$
              I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              10 hours ago










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              10 hours ago
















            • $begingroup$
              Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              11 hours ago










            • $begingroup$
              I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
              $endgroup$
              – SRS
              11 hours ago










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              10 hours ago










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              10 hours ago















            $begingroup$
            Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
            $endgroup$
            – SRS
            11 hours ago




            $begingroup$
            Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
            $endgroup$
            – SRS
            11 hours ago












            $begingroup$
            @SRS Yes, that's true.
            $endgroup$
            – knzhou
            11 hours ago




            $begingroup$
            @SRS Yes, that's true.
            $endgroup$
            – knzhou
            11 hours ago












            $begingroup$
            I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
            $endgroup$
            – SRS
            11 hours ago




            $begingroup$
            I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
            $endgroup$
            – SRS
            11 hours ago












            $begingroup$
            @SRS The potential at a point charge is not defined (or you could say infinite)
            $endgroup$
            – Aaron Stevens
            10 hours ago




            $begingroup$
            @SRS The potential at a point charge is not defined (or you could say infinite)
            $endgroup$
            – Aaron Stevens
            10 hours ago












            $begingroup$
            I have to think more about it and I'll get back.
            $endgroup$
            – SRS
            10 hours ago




            $begingroup$
            I have to think more about it and I'll get back.
            $endgroup$
            – SRS
            10 hours ago











            3












            $begingroup$

            So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$



            This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$



              This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$



                This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.






                share|cite|improve this answer









                $endgroup$



                So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$



                This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 12 hours ago









                Aaron StevensAaron Stevens

                14.2k42252




                14.2k42252



























                    draft saved

                    draft discarded
















































                    Thanks for contributing an answer to Physics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470522%2ffacing-a-paradox-earnshaws-theorem-in-one-dimension%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Reverse int within the 32-bit signed integer range: [−2^31, 2^31 − 1]Combining two 32-bit integers into one 64-bit integerDetermine if an int is within rangeLossy packing 32 bit integer to 16 bitComputing the square root of a 64-bit integerKeeping integer addition within boundsSafe multiplication of two 64-bit signed integersLeetcode 10: Regular Expression MatchingSigned integer-to-ascii x86_64 assembler macroReverse the digits of an Integer“Add two numbers given in reverse order from a linked list”

                    Image
                    C++ debug of nlohmann json using GDB Is this toilet slogan correct usage of the English language? Symbol used to indicate indivisibility What if a revenant (monster) gains fire resistance? Is aluminum electrical wire used on aircraft? Closed-form expression for certain product Strong empirical falsification of quantum mechanics based on vacuum energy density How should I respond when I lied about my education and the company finds out through background check? Pre-mixing cryogenic fuels and using only one fuel tank How do you respond to a colleague from another team when they're wrongly expecting that you'll help them? How can I block email signup overlays or javascript popups in Safari? How to implement a feedback to keep the DC gain at zero for this conceptual passive filter? Count the occurrence of each unique word in the file Why electric field inside a cavity of a non-conducting sphere not zero? If infinitesimal transformations commute why dont ...
                    Read more

                    Category:Fedor von Bock Media in category "Fedor von Bock"Navigation menuUpload mediaISNI: 0000 0000 5511 3417VIAF ID: 24712551GND ID: 119294796Library of Congress authority ID: n96068363BnF ID: 12534305fSUDOC authorities ID: 034604189Open Library ID: OL338253ANKCR AUT ID: jn19990000869National Library of Israel ID: 000514068National Thesaurus for Author Names ID: 341574317ReasonatorScholiaStatistics

                    Image
                    1880 births1945 deathsFedor (given name)Recipients of the Pour le Mérite (military class)Recipients of the Military Merit Cross (Austria-Hungary)Knight's Cross of the Iron Cross recipientsRecipients of the Order of Michael the BraveRecipients of the Order of the Crown (Prussia)Recipients of the Order of the Crown (Württemberg)Recipients of the House Order of HohenzollernRecipients of the Order of the Zähringer LionRecipients of the Hanseatic CrossRecipients of the House Order of the Wendish CrownRecipients of the Order of the Iron CrownGrand Master of the Order of Military Merit (Bulgaria)Recipients of the Order of the Yugoslav CrownRecipients of the Order of Merit of the Republic of HungaryRecipients of the Cross of HonorKnights Grand Cross of the Order of the Crown of ItalyBock (surname)19th-century men of Germany20th-century men of GermanyField Marshals of the Heer (Wehrmacht)Generals of the ReichswehrHeer Knight's Cross of the Iron Cross recipientsMen of Germany by na...
                    Read more

                    Kiel Indholdsfortegnelse Historie | Transport og færgeforbindelser | Sejlsport og anden sport | Kultur | Kendte personer fra Kiel | Noter | Litteratur | Eksterne henvisninger | Navigationsmenuwww.kiel.de54°19′31″N 10°8′26″Ø / 54.32528°N 10.14056°Ø / 54.32528; 10.14056Oberbürgermeister Dr. Ulf Kämpferwww.statistik-nord.deDen danske Stats StatistikKiels hjemmesiderrrWorldCat312794080n790547494030481-4

                    Image
                    DitmarskenFlensborgLauenburgKielLübeckNeumünsterNordfrislandØstholstenPinnebergPlönRendsborg-EgernførdeSlesvig-FlensborgSegebergSteinburgStormarnStuttgartMannheimKarlsruheFreiburg im BreisgauHeidelbergHeilbronnUlmPforzheimReutlingenBerlinBremenBremerhavenFrankfurt am MainWiesbadenKasselDarmstadtOffenbach am MainHannoverBraunschweigOsnabrückOldenburgWolfsburgGöttingenHildesheimSalzgitterMainzLudwigshafen am RheinKoblenzTrierDresdenLeipzigChemnitzKielLübeckKaunasKlaipėdaPärnuRigaTallinnTartuViljandiPolotskElblagGdanskKrakówStargard SzczecinskiSzczecinTorunWroclawBerwick-upon-TweedBostonGreat YarmouthKing's LynnKingston upon HullIpswichLeithLondonNewcastle upon TyneYork Byer i Slesvig-HolstenKielHanse NordtysklandhovedstadendelstatenSlesvig-HolstenTysklandsKielerfjordenØstersøenLangelandÆrøHamborgEjderenVesterhavetdansktyskslavisk1865Kielerkanalen1918novemberrevolutionenoprør blandt matrosernekejserlige marineW...
                    Read more