Why is electric field inside a cavity of a non-conducting sphere not zero?Net flux through insulating cylinderElectric Field from charged sphere within another charged sphere does not reinforce?Gauss Law Not Working Inside Cavityelectric field inside a conducting bodyHow does Gauss's Law imply that the electric field is zero inside a hollow sphere?What is the electric field intensity inside a conducting sphere when charges are asymetrically distributed over the surface?Electric field in non-uniformly charged hollow sphereIs the electric field inside a conductor zero a result of Gauss's Law?Why is the electric field inside the hole non-zero?Electric field outside and inside of a sphereAre charges outside a conducting shell relevant to electric field inside the shell?

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Why is electric field inside a cavity of a non-conducting sphere not zero?


Net flux through insulating cylinderElectric Field from charged sphere within another charged sphere does not reinforce?Gauss Law Not Working Inside Cavityelectric field inside a conducting bodyHow does Gauss's Law imply that the electric field is zero inside a hollow sphere?What is the electric field intensity inside a conducting sphere when charges are asymetrically distributed over the surface?Electric field in non-uniformly charged hollow sphereIs the electric field inside a conductor zero a result of Gauss's Law?Why is the electric field inside the hole non-zero?Electric field outside and inside of a sphereAre charges outside a conducting shell relevant to electric field inside the shell?













4












$begingroup$


Consider a charged non conducting solid sphere of uniform charge density, and it as hole of some radius at the centre.



Now suppose I apply Gauss law.
enter image description here



As there is no charge inside the cavity, no charge is enclosed by Gaussian sphere, so electric flux is zero, hence electric field is zero.



But this is not true according to sources like this, textbook etc..



What am I missing here?










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.







$endgroup$











  • $begingroup$
    So let me clarify. There is a sphere with a uniform volumetric charge density, and you cut out a cavity that has no charge in it?
    $endgroup$
    – noah
    yesterday










  • $begingroup$
    Yes that is correct interpretation
    $endgroup$
    – user72730
    yesterday










  • $begingroup$
    closely related to physics.stackexchange.com/questions/461620/… Simply because the flux is $0$ doesn't mean the field is $0$.
    $endgroup$
    – ZeroTheHero
    yesterday











  • $begingroup$
    Please make this post self contained by giving an explicit formulation of the question rather than a YouTube link or reference to an unnamed textbook.
    $endgroup$
    – my2cts
    17 hours ago










  • $begingroup$
    @ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
    $endgroup$
    – noah
    16 hours ago















4












$begingroup$


Consider a charged non conducting solid sphere of uniform charge density, and it as hole of some radius at the centre.



Now suppose I apply Gauss law.
enter image description here



As there is no charge inside the cavity, no charge is enclosed by Gaussian sphere, so electric flux is zero, hence electric field is zero.



But this is not true according to sources like this, textbook etc..



What am I missing here?










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.







$endgroup$











  • $begingroup$
    So let me clarify. There is a sphere with a uniform volumetric charge density, and you cut out a cavity that has no charge in it?
    $endgroup$
    – noah
    yesterday










  • $begingroup$
    Yes that is correct interpretation
    $endgroup$
    – user72730
    yesterday










  • $begingroup$
    closely related to physics.stackexchange.com/questions/461620/… Simply because the flux is $0$ doesn't mean the field is $0$.
    $endgroup$
    – ZeroTheHero
    yesterday











  • $begingroup$
    Please make this post self contained by giving an explicit formulation of the question rather than a YouTube link or reference to an unnamed textbook.
    $endgroup$
    – my2cts
    17 hours ago










  • $begingroup$
    @ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
    $endgroup$
    – noah
    16 hours ago













4












4








4


1



$begingroup$


Consider a charged non conducting solid sphere of uniform charge density, and it as hole of some radius at the centre.



Now suppose I apply Gauss law.
enter image description here



As there is no charge inside the cavity, no charge is enclosed by Gaussian sphere, so electric flux is zero, hence electric field is zero.



But this is not true according to sources like this, textbook etc..



What am I missing here?










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 a charged non conducting solid sphere of uniform charge density, and it as hole of some radius at the centre.



Now suppose I apply Gauss law.
enter image description here



As there is no charge inside the cavity, no charge is enclosed by Gaussian sphere, so electric flux is zero, hence electric field is zero.



But this is not true according to sources like this, textbook etc..



What am I missing here?







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 16 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

566




566




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.





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.











  • $begingroup$
    So let me clarify. There is a sphere with a uniform volumetric charge density, and you cut out a cavity that has no charge in it?
    $endgroup$
    – noah
    yesterday










  • $begingroup$
    Yes that is correct interpretation
    $endgroup$
    – user72730
    yesterday










  • $begingroup$
    closely related to physics.stackexchange.com/questions/461620/… Simply because the flux is $0$ doesn't mean the field is $0$.
    $endgroup$
    – ZeroTheHero
    yesterday











  • $begingroup$
    Please make this post self contained by giving an explicit formulation of the question rather than a YouTube link or reference to an unnamed textbook.
    $endgroup$
    – my2cts
    17 hours ago










  • $begingroup$
    @ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
    $endgroup$
    – noah
    16 hours ago
















  • $begingroup$
    So let me clarify. There is a sphere with a uniform volumetric charge density, and you cut out a cavity that has no charge in it?
    $endgroup$
    – noah
    yesterday










  • $begingroup$
    Yes that is correct interpretation
    $endgroup$
    – user72730
    yesterday










  • $begingroup$
    closely related to physics.stackexchange.com/questions/461620/… Simply because the flux is $0$ doesn't mean the field is $0$.
    $endgroup$
    – ZeroTheHero
    yesterday











  • $begingroup$
    Please make this post self contained by giving an explicit formulation of the question rather than a YouTube link or reference to an unnamed textbook.
    $endgroup$
    – my2cts
    17 hours ago










  • $begingroup$
    @ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
    $endgroup$
    – noah
    16 hours ago















$begingroup$
So let me clarify. There is a sphere with a uniform volumetric charge density, and you cut out a cavity that has no charge in it?
$endgroup$
– noah
yesterday




$begingroup$
So let me clarify. There is a sphere with a uniform volumetric charge density, and you cut out a cavity that has no charge in it?
$endgroup$
– noah
yesterday












$begingroup$
Yes that is correct interpretation
$endgroup$
– user72730
yesterday




$begingroup$
Yes that is correct interpretation
$endgroup$
– user72730
yesterday












$begingroup$
closely related to physics.stackexchange.com/questions/461620/… Simply because the flux is $0$ doesn't mean the field is $0$.
$endgroup$
– ZeroTheHero
yesterday





$begingroup$
closely related to physics.stackexchange.com/questions/461620/… Simply because the flux is $0$ doesn't mean the field is $0$.
$endgroup$
– ZeroTheHero
yesterday













$begingroup$
Please make this post self contained by giving an explicit formulation of the question rather than a YouTube link or reference to an unnamed textbook.
$endgroup$
– my2cts
17 hours ago




$begingroup$
Please make this post self contained by giving an explicit formulation of the question rather than a YouTube link or reference to an unnamed textbook.
$endgroup$
– my2cts
17 hours ago












$begingroup$
@ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
$endgroup$
– noah
16 hours ago




$begingroup$
@ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
$endgroup$
– noah
16 hours ago










4 Answers
4






active

oldest

votes


















3












$begingroup$

Just to add to what Noah said, the law says that the total flux through the surface is zero. To make this point clear, imagine a region of uniform electric field $vecE = E_0hatx$ along the x axis. Consider a cube of unit area faces with two of its opposite faces normal to the field. From the RHS of Gauss law we know that the flux has to be zero as there are no charges enclosed by the cube. The LHS says



$sum_n=1^6 vecEcdot vecA_n = E_0hatxcdothatx + E_0hatxcdot-hatx = 0$



Thus the flux is zero by explicit calculation as well. Remember that the electric field and area are vectors. So the relative directions are very important. If flux is zero all you can be sure of is that the extent of field lines entering is equal to that of exiting.






share|cite|improve this answer









$endgroup$




















    7












    $begingroup$

    Gauss's Law only gives results for the integral over a whole, closed surface. This does not mean the electric field is zero. It simply means that all field lines entering the volume also exit it at some other point.






    share|cite|improve this answer









    $endgroup$




















      5












      $begingroup$

      In the video the electric field lines are as shown in blue in the diagram below.



      enter image description here



      Let the edge of the cavity be the Gaussian surface which has no charge within it.



      Consider small areas $Delta A$ on either side of the cavity as shown in the diagram in red.
      Because of the symmetrical nature of the situation you can imagine that the electric flux entering the cavity through area $Delta A$ is the same as the electric flux leaving the area $Delta A$ on the right hand side.

      This means that the net flux through those two surfaces is zero.



      Doing the same for the whole Gaussian surface leads to the result that the net flux though the surface is zero commensurate with the fact that there is no charge within the surface.






      share|cite|improve this answer









      $endgroup$




















        -3












        $begingroup$

        The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law, unless there is an external field.
        A uniformly charged sphere with a concentric spherical cavity is a concentric sum of such shells, so the field in the cavity vanishes.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
          $endgroup$
          – Cosmas Zachos
          17 hours ago










        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Just to add to what Noah said, the law says that the total flux through the surface is zero. To make this point clear, imagine a region of uniform electric field $vecE = E_0hatx$ along the x axis. Consider a cube of unit area faces with two of its opposite faces normal to the field. From the RHS of Gauss law we know that the flux has to be zero as there are no charges enclosed by the cube. The LHS says



        $sum_n=1^6 vecEcdot vecA_n = E_0hatxcdothatx + E_0hatxcdot-hatx = 0$



        Thus the flux is zero by explicit calculation as well. Remember that the electric field and area are vectors. So the relative directions are very important. If flux is zero all you can be sure of is that the extent of field lines entering is equal to that of exiting.






        share|cite|improve this answer









        $endgroup$

















          3












          $begingroup$

          Just to add to what Noah said, the law says that the total flux through the surface is zero. To make this point clear, imagine a region of uniform electric field $vecE = E_0hatx$ along the x axis. Consider a cube of unit area faces with two of its opposite faces normal to the field. From the RHS of Gauss law we know that the flux has to be zero as there are no charges enclosed by the cube. The LHS says



          $sum_n=1^6 vecEcdot vecA_n = E_0hatxcdothatx + E_0hatxcdot-hatx = 0$



          Thus the flux is zero by explicit calculation as well. Remember that the electric field and area are vectors. So the relative directions are very important. If flux is zero all you can be sure of is that the extent of field lines entering is equal to that of exiting.






          share|cite|improve this answer









          $endgroup$















            3












            3








            3





            $begingroup$

            Just to add to what Noah said, the law says that the total flux through the surface is zero. To make this point clear, imagine a region of uniform electric field $vecE = E_0hatx$ along the x axis. Consider a cube of unit area faces with two of its opposite faces normal to the field. From the RHS of Gauss law we know that the flux has to be zero as there are no charges enclosed by the cube. The LHS says



            $sum_n=1^6 vecEcdot vecA_n = E_0hatxcdothatx + E_0hatxcdot-hatx = 0$



            Thus the flux is zero by explicit calculation as well. Remember that the electric field and area are vectors. So the relative directions are very important. If flux is zero all you can be sure of is that the extent of field lines entering is equal to that of exiting.






            share|cite|improve this answer









            $endgroup$



            Just to add to what Noah said, the law says that the total flux through the surface is zero. To make this point clear, imagine a region of uniform electric field $vecE = E_0hatx$ along the x axis. Consider a cube of unit area faces with two of its opposite faces normal to the field. From the RHS of Gauss law we know that the flux has to be zero as there are no charges enclosed by the cube. The LHS says



            $sum_n=1^6 vecEcdot vecA_n = E_0hatxcdothatx + E_0hatxcdot-hatx = 0$



            Thus the flux is zero by explicit calculation as well. Remember that the electric field and area are vectors. So the relative directions are very important. If flux is zero all you can be sure of is that the extent of field lines entering is equal to that of exiting.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            user3518839user3518839

            914




            914





















                7












                $begingroup$

                Gauss's Law only gives results for the integral over a whole, closed surface. This does not mean the electric field is zero. It simply means that all field lines entering the volume also exit it at some other point.






                share|cite|improve this answer









                $endgroup$

















                  7












                  $begingroup$

                  Gauss's Law only gives results for the integral over a whole, closed surface. This does not mean the electric field is zero. It simply means that all field lines entering the volume also exit it at some other point.






                  share|cite|improve this answer









                  $endgroup$















                    7












                    7








                    7





                    $begingroup$

                    Gauss's Law only gives results for the integral over a whole, closed surface. This does not mean the electric field is zero. It simply means that all field lines entering the volume also exit it at some other point.






                    share|cite|improve this answer









                    $endgroup$



                    Gauss's Law only gives results for the integral over a whole, closed surface. This does not mean the electric field is zero. It simply means that all field lines entering the volume also exit it at some other point.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    noahnoah

                    4,05311226




                    4,05311226





















                        5












                        $begingroup$

                        In the video the electric field lines are as shown in blue in the diagram below.



                        enter image description here



                        Let the edge of the cavity be the Gaussian surface which has no charge within it.



                        Consider small areas $Delta A$ on either side of the cavity as shown in the diagram in red.
                        Because of the symmetrical nature of the situation you can imagine that the electric flux entering the cavity through area $Delta A$ is the same as the electric flux leaving the area $Delta A$ on the right hand side.

                        This means that the net flux through those two surfaces is zero.



                        Doing the same for the whole Gaussian surface leads to the result that the net flux though the surface is zero commensurate with the fact that there is no charge within the surface.






                        share|cite|improve this answer









                        $endgroup$

















                          5












                          $begingroup$

                          In the video the electric field lines are as shown in blue in the diagram below.



                          enter image description here



                          Let the edge of the cavity be the Gaussian surface which has no charge within it.



                          Consider small areas $Delta A$ on either side of the cavity as shown in the diagram in red.
                          Because of the symmetrical nature of the situation you can imagine that the electric flux entering the cavity through area $Delta A$ is the same as the electric flux leaving the area $Delta A$ on the right hand side.

                          This means that the net flux through those two surfaces is zero.



                          Doing the same for the whole Gaussian surface leads to the result that the net flux though the surface is zero commensurate with the fact that there is no charge within the surface.






                          share|cite|improve this answer









                          $endgroup$















                            5












                            5








                            5





                            $begingroup$

                            In the video the electric field lines are as shown in blue in the diagram below.



                            enter image description here



                            Let the edge of the cavity be the Gaussian surface which has no charge within it.



                            Consider small areas $Delta A$ on either side of the cavity as shown in the diagram in red.
                            Because of the symmetrical nature of the situation you can imagine that the electric flux entering the cavity through area $Delta A$ is the same as the electric flux leaving the area $Delta A$ on the right hand side.

                            This means that the net flux through those two surfaces is zero.



                            Doing the same for the whole Gaussian surface leads to the result that the net flux though the surface is zero commensurate with the fact that there is no charge within the surface.






                            share|cite|improve this answer









                            $endgroup$



                            In the video the electric field lines are as shown in blue in the diagram below.



                            enter image description here



                            Let the edge of the cavity be the Gaussian surface which has no charge within it.



                            Consider small areas $Delta A$ on either side of the cavity as shown in the diagram in red.
                            Because of the symmetrical nature of the situation you can imagine that the electric flux entering the cavity through area $Delta A$ is the same as the electric flux leaving the area $Delta A$ on the right hand side.

                            This means that the net flux through those two surfaces is zero.



                            Doing the same for the whole Gaussian surface leads to the result that the net flux though the surface is zero commensurate with the fact that there is no charge within the surface.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            FarcherFarcher

                            51.4k339108




                            51.4k339108





















                                -3












                                $begingroup$

                                The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law, unless there is an external field.
                                A uniformly charged sphere with a concentric spherical cavity is a concentric sum of such shells, so the field in the cavity vanishes.






                                share|cite|improve this answer











                                $endgroup$












                                • $begingroup$
                                  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                                  $endgroup$
                                  – Cosmas Zachos
                                  17 hours ago















                                -3












                                $begingroup$

                                The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law, unless there is an external field.
                                A uniformly charged sphere with a concentric spherical cavity is a concentric sum of such shells, so the field in the cavity vanishes.






                                share|cite|improve this answer











                                $endgroup$












                                • $begingroup$
                                  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                                  $endgroup$
                                  – Cosmas Zachos
                                  17 hours ago













                                -3












                                -3








                                -3





                                $begingroup$

                                The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law, unless there is an external field.
                                A uniformly charged sphere with a concentric spherical cavity is a concentric sum of such shells, so the field in the cavity vanishes.






                                share|cite|improve this answer











                                $endgroup$



                                The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law, unless there is an external field.
                                A uniformly charged sphere with a concentric spherical cavity is a concentric sum of such shells, so the field in the cavity vanishes.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 17 hours ago

























                                answered yesterday









                                my2ctsmy2cts

                                5,7192719




                                5,7192719











                                • $begingroup$
                                  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                                  $endgroup$
                                  – Cosmas Zachos
                                  17 hours ago
















                                • $begingroup$
                                  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                                  $endgroup$
                                  – Cosmas Zachos
                                  17 hours ago















                                $begingroup$
                                This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                                $endgroup$
                                – Cosmas Zachos
                                17 hours ago




                                $begingroup$
                                This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
                                $endgroup$
                                – Cosmas Zachos
                                17 hours ago










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









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