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












2












$begingroup$


Now consider a charged non conducting 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: so 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 https://www.youtube.com/watch?v=H18bYW_Do0k , textbook etc.



What am I missing here ?






electrostatics electric-fields charge gauss-law







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  • $begingroup$
    If the sphere has uniform charge density, why would you assume no charge is enclosed?
    $endgroup$
    – noah
    8 hours ago






  • 1




    $begingroup$
    Because it a cavity!!, nothing is present
    $endgroup$
    – user72730
    8 hours ago










  • $begingroup$
    So you mean the surface has a uniform surface charge denisty?
    $endgroup$
    – noah
    8 hours ago










  • $begingroup$
    On rest of sphere
    $endgroup$
    – user72730
    8 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
    8 hours ago
















2












$begingroup$


Now consider a charged non conducting 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: so 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 https://www.youtube.com/watch?v=H18bYW_Do0k , 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.







$endgroup$












  • $begingroup$
    If the sphere has uniform charge density, why would you assume no charge is enclosed?
    $endgroup$
    – noah
    8 hours ago






  • 1




    $begingroup$
    Because it a cavity!!, nothing is present
    $endgroup$
    – user72730
    8 hours ago










  • $begingroup$
    So you mean the surface has a uniform surface charge denisty?
    $endgroup$
    – noah
    8 hours ago










  • $begingroup$
    On rest of sphere
    $endgroup$
    – user72730
    8 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
    8 hours ago














2












2








2





$begingroup$


Now consider a charged non conducting 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: so 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 https://www.youtube.com/watch?v=H18bYW_Do0k , 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.







$endgroup$




Now consider a charged non conducting 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: so 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 https://www.youtube.com/watch?v=H18bYW_Do0k , textbook etc.



What am I missing here ?






electrostatics electric-fields charge gauss-law




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 51 mins ago









Qmechanic♦

106k121961225




106k121961225






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 9 hours ago









user72730user72730

173




173




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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.












  • $begingroup$
    If the sphere has uniform charge density, why would you assume no charge is enclosed?
    $endgroup$
    – noah
    8 hours ago






  • 1




    $begingroup$
    Because it a cavity!!, nothing is present
    $endgroup$
    – user72730
    8 hours ago










  • $begingroup$
    So you mean the surface has a uniform surface charge denisty?
    $endgroup$
    – noah
    8 hours ago










  • $begingroup$
    On rest of sphere
    $endgroup$
    – user72730
    8 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
    8 hours ago


















  • $begingroup$
    If the sphere has uniform charge density, why would you assume no charge is enclosed?
    $endgroup$
    – noah
    8 hours ago






  • 1




    $begingroup$
    Because it a cavity!!, nothing is present
    $endgroup$
    – user72730
    8 hours ago










  • $begingroup$
    So you mean the surface has a uniform surface charge denisty?
    $endgroup$
    – noah
    8 hours ago










  • $begingroup$
    On rest of sphere
    $endgroup$
    – user72730
    8 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
    8 hours ago
















$begingroup$
If the sphere has uniform charge density, why would you assume no charge is enclosed?
$endgroup$
– noah
8 hours ago




$begingroup$
If the sphere has uniform charge density, why would you assume no charge is enclosed?
$endgroup$
– noah
8 hours ago




1




1




$begingroup$
Because it a cavity!!, nothing is present
$endgroup$
– user72730
8 hours ago




$begingroup$
Because it a cavity!!, nothing is present
$endgroup$
– user72730
8 hours ago












$begingroup$
So you mean the surface has a uniform surface charge denisty?
$endgroup$
– noah
8 hours ago




$begingroup$
So you mean the surface has a uniform surface charge denisty?
$endgroup$
– noah
8 hours ago












$begingroup$
On rest of sphere
$endgroup$
– user72730
8 hours ago




$begingroup$
On rest of sphere
$endgroup$
– user72730
8 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
8 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
8 hours ago










4 Answers
4






active

oldest

votes


















4












$begingroup$

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









$endgroup$





















    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 $vec{E} = E_0hat{x}$ 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



    $sum_{n=1}^6 vec{E}cdot vec{A}_{n} = E_0hat{x}cdothat{x} + E_0hat{x}cdot-hat{x} = 0$



    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









    $endgroup$





















      1












      $begingroup$

      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









      $endgroup$





















        0












        $begingroup$

        The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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









          $endgroup$


















            4












            $begingroup$

            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









            $endgroup$
















              4












              4








              4





              $begingroup$

              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 8 hours ago









              noahnoah

              4,01311226




              4,01311226























                  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 $vec{E} = E_0hat{x}$ 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



                  $sum_{n=1}^6 vec{E}cdot vec{A}_{n} = E_0hat{x}cdothat{x} + E_0hat{x}cdot-hat{x} = 0$



                  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 $vec{E} = E_0hat{x}$ 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



                    $sum_{n=1}^6 vec{E}cdot vec{A}_{n} = E_0hat{x}cdothat{x} + E_0hat{x}cdot-hat{x} = 0$



                    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 $vec{E} = E_0hat{x}$ 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



                      $sum_{n=1}^6 vec{E}cdot vec{A}_{n} = E_0hat{x}cdothat{x} + E_0hat{x}cdot-hat{x} = 0$



                      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 $vec{E} = E_0hat{x}$ 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



                      $sum_{n=1}^6 vec{E}cdot vec{A}_{n} = E_0hat{x}cdothat{x} + E_0hat{x}cdot-hat{x} = 0$



                      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 8 hours ago









                      user3518839user3518839

                      764




                      764























                          1












                          $begingroup$

                          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









                          $endgroup$


















                            1












                            $begingroup$

                            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









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              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 4 hours ago









                              FarcherFarcher

                              51.3k339107




                              51.3k339107























                                  0












                                  $begingroup$

                                  The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.






                                      share|cite|improve this answer









                                      $endgroup$



                                      The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 8 hours ago









                                      my2ctsmy2cts

                                      5,7072718




                                      5,7072718






















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