Flux received by a negative charge












7












$begingroup$


Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
like this



My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?



As according to gauss law



guass law
Source of image: Britannica



The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.










share|cite|improve this question









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  • 4




    $begingroup$
    The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
    $endgroup$
    – Xerxes
    yesterday
















7












$begingroup$


Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
like this



My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?



As according to gauss law



guass law
Source of image: Britannica



The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.










share|cite|improve this question









New contributor




user72730 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
    $endgroup$
    – Xerxes
    yesterday














7












7








7


3



$begingroup$


Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
like this



My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?



As according to gauss law



guass law
Source of image: Britannica



The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.










share|cite|improve this question









New contributor




user72730 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
like this



My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?



As according to gauss law



guass law
Source of image: Britannica



The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.







electrostatics electric-fields charge gauss-law






share|cite|improve this question









New contributor




user72730 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




user72730 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 19 hours ago







user72730













New contributor




user72730 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









user72730user72730

717




717




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user72730 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





user72730 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user72730 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    $begingroup$
    The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
    $endgroup$
    – Xerxes
    yesterday














  • 4




    $begingroup$
    The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
    $endgroup$
    – Xerxes
    yesterday








4




4




$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
yesterday




$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
yesterday










3 Answers
3






active

oldest

votes


















11












$begingroup$

The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.



See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
image

(image from Chegg Study: physics questions and answers)



This is in accordance with Gauss's law for the electric field:




  • Draw a closed surface around the left charge ($+Q$) only.

    There are 6 field lines coming out of this surface

  • Draw a closed surface around the right charge ($-3Q$) only.

    There are 18 field lines going into this surface.

  • Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).

    There are 12 field lines going into this big surface.

  • Draw a closed surface which does not enclose any of the charges.

    There are $n$ field lines going into and the same $n$ field lines
    coming out of this surface, thus giving a sum of zero.


In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    If you draw a Gaussian surface that only includes the $-Q$ charge,
    the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
    You could move that external charge to infinity and not change that total flux.
    So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.



    update to the address the OP's follow-up question in the comment




    But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
    ... So what does the LHS of the gauss signify then?




    Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.



    Glowscript MatterAndInteractions 13-fields emptyGlowscript MatterAndInteractions 13-fields centered



    Now, as I reposition the charge [which is easier to do with this visualization],
    note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]



    Glowscript MatterAndInteractions 13-fields off-centered



    When I move the charge outside the Gaussian surface,
    the sign of the flux changes for the patches near the charge.
    The total flux drops to zero.

    So, while external charges contribute to the local flux through a patch,
    their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.



    Glowscript MatterAndInteractions 13-fields external



    Note the total electric flux is not just "$EA$",


    it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.


    In integral form, it's $oint vec Ecdot dvec A$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
      $endgroup$
      – user72730
      yesterday










    • $begingroup$
      Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
      $endgroup$
      – robphy
      yesterday










    • $begingroup$
      So what does the LHS of the gauss signify then?
      $endgroup$
      – user72730
      yesterday










    • $begingroup$
      I added some visualizations in my answer to address your follow-up questions.
      $endgroup$
      – robphy
      yesterday



















    0












    $begingroup$

    The answer would be "yes, but...".



    In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.



    Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.



    However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
      $endgroup$
      – user72730
      yesterday










    • $begingroup$
      @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
      $endgroup$
      – Pere
      20 hours ago












    • $begingroup$
      I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
      $endgroup$
      – user72730
      19 hours ago











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    11












    $begingroup$

    The number of flux lines of each charge is proportional to its own charge.
    The other charge has nothing to do with that.



    See this image with two unequal charges. The right negative charge ($-3Q$)
    has three times the size of the left positive charge ($+Q$):
    image

    (image from Chegg Study: physics questions and answers)



    This is in accordance with Gauss's law for the electric field:




    • Draw a closed surface around the left charge ($+Q$) only.

      There are 6 field lines coming out of this surface

    • Draw a closed surface around the right charge ($-3Q$) only.

      There are 18 field lines going into this surface.

    • Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).

      There are 12 field lines going into this big surface.

    • Draw a closed surface which does not enclose any of the charges.

      There are $n$ field lines going into and the same $n$ field lines
      coming out of this surface, thus giving a sum of zero.


    In all cases the number of field lines (i.e. the electric flux)
    through the closed surface is proportional
    to the charge inside the surface.






    share|cite|improve this answer











    $endgroup$


















      11












      $begingroup$

      The number of flux lines of each charge is proportional to its own charge.
      The other charge has nothing to do with that.



      See this image with two unequal charges. The right negative charge ($-3Q$)
      has three times the size of the left positive charge ($+Q$):
      image

      (image from Chegg Study: physics questions and answers)



      This is in accordance with Gauss's law for the electric field:




      • Draw a closed surface around the left charge ($+Q$) only.

        There are 6 field lines coming out of this surface

      • Draw a closed surface around the right charge ($-3Q$) only.

        There are 18 field lines going into this surface.

      • Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).

        There are 12 field lines going into this big surface.

      • Draw a closed surface which does not enclose any of the charges.

        There are $n$ field lines going into and the same $n$ field lines
        coming out of this surface, thus giving a sum of zero.


      In all cases the number of field lines (i.e. the electric flux)
      through the closed surface is proportional
      to the charge inside the surface.






      share|cite|improve this answer











      $endgroup$
















        11












        11








        11





        $begingroup$

        The number of flux lines of each charge is proportional to its own charge.
        The other charge has nothing to do with that.



        See this image with two unequal charges. The right negative charge ($-3Q$)
        has three times the size of the left positive charge ($+Q$):
        image

        (image from Chegg Study: physics questions and answers)



        This is in accordance with Gauss's law for the electric field:




        • Draw a closed surface around the left charge ($+Q$) only.

          There are 6 field lines coming out of this surface

        • Draw a closed surface around the right charge ($-3Q$) only.

          There are 18 field lines going into this surface.

        • Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).

          There are 12 field lines going into this big surface.

        • Draw a closed surface which does not enclose any of the charges.

          There are $n$ field lines going into and the same $n$ field lines
          coming out of this surface, thus giving a sum of zero.


        In all cases the number of field lines (i.e. the electric flux)
        through the closed surface is proportional
        to the charge inside the surface.






        share|cite|improve this answer











        $endgroup$



        The number of flux lines of each charge is proportional to its own charge.
        The other charge has nothing to do with that.



        See this image with two unequal charges. The right negative charge ($-3Q$)
        has three times the size of the left positive charge ($+Q$):
        image

        (image from Chegg Study: physics questions and answers)



        This is in accordance with Gauss's law for the electric field:




        • Draw a closed surface around the left charge ($+Q$) only.

          There are 6 field lines coming out of this surface

        • Draw a closed surface around the right charge ($-3Q$) only.

          There are 18 field lines going into this surface.

        • Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).

          There are 12 field lines going into this big surface.

        • Draw a closed surface which does not enclose any of the charges.

          There are $n$ field lines going into and the same $n$ field lines
          coming out of this surface, thus giving a sum of zero.


        In all cases the number of field lines (i.e. the electric flux)
        through the closed surface is proportional
        to the charge inside the surface.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 20 hours ago

























        answered yesterday









        Thomas FritschThomas Fritsch

        1,334415




        1,334415























            4












            $begingroup$

            If you draw a Gaussian surface that only includes the $-Q$ charge,
            the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
            You could move that external charge to infinity and not change that total flux.
            So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.



            update to the address the OP's follow-up question in the comment




            But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
            ... So what does the LHS of the gauss signify then?




            Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.



            Glowscript MatterAndInteractions 13-fields emptyGlowscript MatterAndInteractions 13-fields centered



            Now, as I reposition the charge [which is easier to do with this visualization],
            note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]



            Glowscript MatterAndInteractions 13-fields off-centered



            When I move the charge outside the Gaussian surface,
            the sign of the flux changes for the patches near the charge.
            The total flux drops to zero.

            So, while external charges contribute to the local flux through a patch,
            their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.



            Glowscript MatterAndInteractions 13-fields external



            Note the total electric flux is not just "$EA$",


            it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.


            In integral form, it's $oint vec Ecdot dvec A$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
              $endgroup$
              – robphy
              yesterday










            • $begingroup$
              So what does the LHS of the gauss signify then?
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              I added some visualizations in my answer to address your follow-up questions.
              $endgroup$
              – robphy
              yesterday
















            4












            $begingroup$

            If you draw a Gaussian surface that only includes the $-Q$ charge,
            the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
            You could move that external charge to infinity and not change that total flux.
            So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.



            update to the address the OP's follow-up question in the comment




            But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
            ... So what does the LHS of the gauss signify then?




            Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.



            Glowscript MatterAndInteractions 13-fields emptyGlowscript MatterAndInteractions 13-fields centered



            Now, as I reposition the charge [which is easier to do with this visualization],
            note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]



            Glowscript MatterAndInteractions 13-fields off-centered



            When I move the charge outside the Gaussian surface,
            the sign of the flux changes for the patches near the charge.
            The total flux drops to zero.

            So, while external charges contribute to the local flux through a patch,
            their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.



            Glowscript MatterAndInteractions 13-fields external



            Note the total electric flux is not just "$EA$",


            it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.


            In integral form, it's $oint vec Ecdot dvec A$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
              $endgroup$
              – robphy
              yesterday










            • $begingroup$
              So what does the LHS of the gauss signify then?
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              I added some visualizations in my answer to address your follow-up questions.
              $endgroup$
              – robphy
              yesterday














            4












            4








            4





            $begingroup$

            If you draw a Gaussian surface that only includes the $-Q$ charge,
            the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
            You could move that external charge to infinity and not change that total flux.
            So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.



            update to the address the OP's follow-up question in the comment




            But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
            ... So what does the LHS of the gauss signify then?




            Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.



            Glowscript MatterAndInteractions 13-fields emptyGlowscript MatterAndInteractions 13-fields centered



            Now, as I reposition the charge [which is easier to do with this visualization],
            note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]



            Glowscript MatterAndInteractions 13-fields off-centered



            When I move the charge outside the Gaussian surface,
            the sign of the flux changes for the patches near the charge.
            The total flux drops to zero.

            So, while external charges contribute to the local flux through a patch,
            their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.



            Glowscript MatterAndInteractions 13-fields external



            Note the total electric flux is not just "$EA$",


            it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.


            In integral form, it's $oint vec Ecdot dvec A$.






            share|cite|improve this answer











            $endgroup$



            If you draw a Gaussian surface that only includes the $-Q$ charge,
            the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
            You could move that external charge to infinity and not change that total flux.
            So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.



            update to the address the OP's follow-up question in the comment




            But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
            ... So what does the LHS of the gauss signify then?




            Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.



            Glowscript MatterAndInteractions 13-fields emptyGlowscript MatterAndInteractions 13-fields centered



            Now, as I reposition the charge [which is easier to do with this visualization],
            note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]



            Glowscript MatterAndInteractions 13-fields off-centered



            When I move the charge outside the Gaussian surface,
            the sign of the flux changes for the patches near the charge.
            The total flux drops to zero.

            So, while external charges contribute to the local flux through a patch,
            their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.



            Glowscript MatterAndInteractions 13-fields external



            Note the total electric flux is not just "$EA$",


            it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.


            In integral form, it's $oint vec Ecdot dvec A$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            robphyrobphy

            2,162248




            2,162248












            • $begingroup$
              But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
              $endgroup$
              – robphy
              yesterday










            • $begingroup$
              So what does the LHS of the gauss signify then?
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              I added some visualizations in my answer to address your follow-up questions.
              $endgroup$
              – robphy
              yesterday


















            • $begingroup$
              But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
              $endgroup$
              – robphy
              yesterday










            • $begingroup$
              So what does the LHS of the gauss signify then?
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              I added some visualizations in my answer to address your follow-up questions.
              $endgroup$
              – robphy
              yesterday
















            $begingroup$
            But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
            $endgroup$
            – user72730
            yesterday




            $begingroup$
            But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
            $endgroup$
            – user72730
            yesterday












            $begingroup$
            Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
            $endgroup$
            – robphy
            yesterday




            $begingroup$
            Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
            $endgroup$
            – robphy
            yesterday












            $begingroup$
            So what does the LHS of the gauss signify then?
            $endgroup$
            – user72730
            yesterday




            $begingroup$
            So what does the LHS of the gauss signify then?
            $endgroup$
            – user72730
            yesterday












            $begingroup$
            I added some visualizations in my answer to address your follow-up questions.
            $endgroup$
            – robphy
            yesterday




            $begingroup$
            I added some visualizations in my answer to address your follow-up questions.
            $endgroup$
            – robphy
            yesterday











            0












            $begingroup$

            The answer would be "yes, but...".



            In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.



            Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.



            However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
              $endgroup$
              – Pere
              20 hours ago












            • $begingroup$
              I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
              $endgroup$
              – user72730
              19 hours ago
















            0












            $begingroup$

            The answer would be "yes, but...".



            In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.



            Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.



            However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
              $endgroup$
              – Pere
              20 hours ago












            • $begingroup$
              I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
              $endgroup$
              – user72730
              19 hours ago














            0












            0








            0





            $begingroup$

            The answer would be "yes, but...".



            In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.



            Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.



            However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.






            share|cite|improve this answer









            $endgroup$



            The answer would be "yes, but...".



            In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.



            Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.



            However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            PerePere

            1,309148




            1,309148












            • $begingroup$
              I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
              $endgroup$
              – Pere
              20 hours ago












            • $begingroup$
              I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
              $endgroup$
              – user72730
              19 hours ago


















            • $begingroup$
              I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
              $endgroup$
              – user72730
              yesterday










            • $begingroup$
              @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
              $endgroup$
              – Pere
              20 hours ago












            • $begingroup$
              I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
              $endgroup$
              – user72730
              19 hours ago
















            $begingroup$
            I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
            $endgroup$
            – user72730
            yesterday




            $begingroup$
            I’m very concerned about distance between them, but I want to effect of charge difference I.e is all of the flux from positive charge goes into negative , or some of flux into it depending on charge of negative charge and rest goes away to universe.
            $endgroup$
            – user72730
            yesterday












            $begingroup$
            @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
            $endgroup$
            – Pere
            20 hours ago






            $begingroup$
            @user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
            $endgroup$
            – Pere
            20 hours ago














            $begingroup$
            I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
            $endgroup$
            – user72730
            19 hours ago




            $begingroup$
            I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
            $endgroup$
            – user72730
            19 hours ago










            user72730 is a new contributor. Be nice, and check out our Code of Conduct.










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            user72730 is a new contributor. Be nice, and check out our Code of Conduct.
















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