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?
$begingroup$
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.
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
asked yesterday
user72730user72730
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.
$endgroup$
|
show 2 more comments
$begingroup$
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.
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
asked yesterday
user72730user72730
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.
$endgroup$
$begingroup$
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
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
15 hours ago
|
show 2 more comments
$begingroup$
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.
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
asked yesterday
user72730user72730
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.
$endgroup$
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.
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
electrostatics electric-fields charge gauss-law
asked yesterday
user72730user72730
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.
asked yesterday
user72730user72730
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.
edited 15 hours ago
user72730
asked yesterday
user72730user72730
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.
asked yesterday
user72730user72730
566
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
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
15 hours ago
|
show 2 more comments
$begingroup$
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
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
15 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
16 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
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
15 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
15 hours ago
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
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 yesterday
user3518839user3518839
914
$endgroup$
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 yesterday
noahnoah
4,05311226
$endgroup$
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 yesterday
FarcherFarcher
51.4k339108
$endgroup$
add a comment |
$begingroup$
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.
answered yesterday
my2ctsmy2cts
5,7192719
$endgroup$
$begingroup$
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
16 hours ago
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
$begingroup$
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 yesterday
user3518839user3518839
914
$endgroup$
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 yesterday
user3518839user3518839
914
$endgroup$
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 yesterday
user3518839user3518839
914
$endgroup$
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 yesterday
user3518839user3518839
914
answered yesterday
user3518839user3518839
914
answered yesterday
user3518839user3518839
914
answered yesterday
user3518839user3518839
914
914
add a comment |
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 yesterday
noahnoah
4,05311226
$endgroup$
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 yesterday
noahnoah
4,05311226
$endgroup$
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 yesterday
noahnoah
4,05311226
$endgroup$
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 yesterday
noahnoah
4,05311226
answered yesterday
noahnoah
4,05311226
answered yesterday
noahnoah
4,05311226
answered yesterday
noahnoah
4,05311226
4,05311226
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 yesterday
FarcherFarcher
51.4k339108
$endgroup$
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 yesterday
FarcherFarcher
51.4k339108
$endgroup$
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 yesterday
FarcherFarcher
51.4k339108
$endgroup$
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 yesterday
FarcherFarcher
51.4k339108
answered yesterday
FarcherFarcher
51.4k339108
answered yesterday
FarcherFarcher
51.4k339108
answered yesterday
FarcherFarcher
51.4k339108
51.4k339108
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, 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.
answered yesterday
my2ctsmy2cts
5,7192719
$endgroup$
$begingroup$
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
16 hours ago
add a comment |
$begingroup$
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.
answered yesterday
my2ctsmy2cts
5,7192719
$endgroup$
$begingroup$
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
16 hours ago
add a comment |
$begingroup$
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.
answered yesterday
my2ctsmy2cts
5,7192719
$endgroup$
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.
answered yesterday
my2ctsmy2cts
5,7192719
edited 16 hours ago
answered yesterday
my2ctsmy2cts
5,7192719
answered yesterday
my2ctsmy2cts
5,7192719
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
16 hours ago
add a comment |
$begingroup$
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
16 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
16 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
16 hours ago
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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$begingroup$
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.
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– my2cts
16 hours ago
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@ruakh But OP is clearly talking about a solid ball with some cavity in it, but calls it a sphere.
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– noah
15 hours ago