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Why electric field inside a cavity of a non-conducting sphere not zero?
Electric Field from charged sphere within another charged sphere does not reinforce?Gauss's law in a uniform charge distribution extending infinitely in all directionsGauss Law Not Working Inside CavityHow 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 sphereFlux through hollow non-conducting sphereWhy 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?
$begingroup$
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 
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
edited 33 mins ago
Qmechanic♦
106k121961225
asked 8 hours ago
user72730user72730
173
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.
$endgroup$
|
show 1 more comment
$begingroup$
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
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
edited 33 mins ago
Qmechanic♦
106k121961225
asked 8 hours ago
user72730user72730
173
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
|
show 1 more comment
$begingroup$
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
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
edited 33 mins ago
Qmechanic♦
106k121961225
asked 8 hours ago
user72730user72730
173
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
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
edited 33 mins ago
Qmechanic♦
106k121961225
asked 8 hours ago
user72730user72730
173
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.
edited 33 mins ago
Qmechanic♦
106k121961225
asked 8 hours ago
user72730user72730
173
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.
edited 33 mins ago
Qmechanic♦
106k121961225
edited 33 mins ago
Qmechanic♦
106k121961225
edited 33 mins ago
Qmechanic♦
106k121961225
106k121961225
asked 8 hours ago
user72730user72730
173
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 8 hours ago
user72730user72730
173
asked 8 hours ago
user72730user72730
173
173
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.
$begingroup$
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
noahnoah
4,01311226
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
user3518839user3518839
764
$endgroup$
add a comment |
$begingroup$
In the video the electric field lines are as shown in blue in the diagram below.
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.
answered 4 hours ago
FarcherFarcher
51.3k339107
$endgroup$
add a comment |
$begingroup$
The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.
answered 8 hours ago
my2ctsmy2cts
5,6972718
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
noahnoah
4,01311226
$endgroup$
add a comment |
$begingroup$
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.
answered 8 hours ago
noahnoah
4,01311226
$endgroup$
add a comment |
$begingroup$
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.
answered 8 hours ago
noahnoah
4,01311226
$endgroup$
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.
answered 8 hours ago
noahnoah
4,01311226
answered 8 hours ago
noahnoah
4,01311226
answered 8 hours ago
noahnoah
4,01311226
answered 8 hours ago
noahnoah
4,01311226
4,01311226
add a comment |
add a comment |
$begingroup$
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.
answered 7 hours ago
user3518839user3518839
764
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
user3518839user3518839
764
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
user3518839user3518839
764
$endgroup$
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.
answered 7 hours ago
user3518839user3518839
764
answered 7 hours ago
user3518839user3518839
764
answered 7 hours ago
user3518839user3518839
764
answered 7 hours ago
user3518839user3518839
764
764
add a comment |
add a comment |
$begingroup$
In the video the electric field lines are as shown in blue in the diagram below.
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.
answered 4 hours ago
FarcherFarcher
51.3k339107
$endgroup$
add a comment |
$begingroup$
In the video the electric field lines are as shown in blue in the diagram below.
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.
answered 4 hours ago
FarcherFarcher
51.3k339107
$endgroup$
add a comment |
$begingroup$
In the video the electric field lines are as shown in blue in the diagram below.
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.
answered 4 hours ago
FarcherFarcher
51.3k339107
$endgroup$
In the video the electric field lines are as shown in blue in the diagram below.
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.
answered 4 hours ago
FarcherFarcher
51.3k339107
answered 4 hours ago
FarcherFarcher
51.3k339107
answered 4 hours ago
FarcherFarcher
51.3k339107
answered 4 hours ago
FarcherFarcher
51.3k339107
51.3k339107
add a comment |
add a comment |
$begingroup$
The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.
answered 8 hours ago
my2ctsmy2cts
5,6972718
$endgroup$
add a comment |
$begingroup$
The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.
answered 8 hours ago
my2ctsmy2cts
5,6972718
$endgroup$
add a comment |
$begingroup$
The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.
answered 8 hours ago
my2ctsmy2cts
5,6972718
$endgroup$
The field of a uniformly charged shell is zero inside the shell in agreement with Gauss's law.
answered 8 hours ago
my2ctsmy2cts
5,6972718
answered 8 hours ago
my2ctsmy2cts
5,6972718
answered 8 hours ago
my2ctsmy2cts
5,6972718
answered 8 hours ago
my2ctsmy2cts
5,6972718
5,6972718
add a comment |
add a comment |
user72730 is a new contributor. Be nice, and check out our Code of Conduct.
user72730 is a new contributor. Be nice, and check out our Code of Conduct.
user72730 is a new contributor. Be nice, and check out our Code of Conduct.
user72730 is a new contributor. Be nice, and check out our Code of Conduct.
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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