Flux received by a negative charge The Next CEO of Stack OverflowGauss' law and an external chargeWhat is meant by “net charge”?Effect of introducing an external charge beside a Gaussian surfaceGauss Law - charges outside the gaussian surfaceGauss's Law for Non-Uniform Electric Fields enclosed within Gaussian SurfacesWhy do outside charges do not contribute to net flux of a Gaussian Surface?Gauss' law with no charge distribution inside a volumeElectric field on Gaussian surface due to external chargeWhy do outside charges not contribute to net flux through a surface, pt 2Are charges outside a conducting shell relevant to electric field inside the shell?
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Flux received by a negative charge
The Next CEO of Stack OverflowGauss' law and an external chargeWhat is meant by “net charge”?Effect of introducing an external charge beside a Gaussian surfaceGauss Law - charges outside the gaussian surfaceGauss's Law for Non-Uniform Electric Fields enclosed within Gaussian SurfacesWhy do outside charges do not contribute to net flux of a Gaussian Surface?Gauss' law with no charge distribution inside a volumeElectric field on Gaussian surface due to external chargeWhy do outside charges not contribute to net flux through a surface, pt 2Are charges outside a conducting shell relevant to electric field inside the shell?
$begingroup$
Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?
As according to gauss law
Source of image: Britannica
The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.
electrostatics electric-fields charge gauss-law
asked Mar 24 at 11:21
user72730user72730
838
$endgroup$
add a comment |
$begingroup$
Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?
As according to gauss law
Source of image: Britannica
The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.
electrostatics electric-fields charge gauss-law
asked Mar 24 at 11:21
user72730user72730
838
$endgroup$
4
$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
Mar 24 at 14:34
add a comment |
$begingroup$
Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?
As according to gauss law
Source of image: Britannica
The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.
electrostatics electric-fields charge gauss-law
asked Mar 24 at 11:21
user72730user72730
838
$endgroup$
Consider two charges $+q$ and $-Q$ placed at a distance, note- charge q and Q are different In terms of magnitude.
My question: is number of flux lines received by $-Q$ proportional to its own charge, or does $+q$ charge have anything to say at all?
As according to gauss law
Source of image: Britannica
The LHS is dependant of field external to Gaussian surface and and RHS of equation depends on charge enclosed within the Gaussian surface.
electrostatics electric-fields charge gauss-law
electrostatics electric-fields charge gauss-law
asked Mar 24 at 11:21
user72730user72730
838
asked Mar 24 at 11:21
user72730user72730
838
edited Mar 25 at 9:39
user72730
asked Mar 24 at 11:21
user72730user72730
838
asked Mar 24 at 11:21
user72730user72730
838
asked Mar 24 at 11:21
user72730user72730
838
838
4
$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
Mar 24 at 14:34
add a comment |
4
$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
Mar 24 at 14:34
4
4
$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
Mar 24 at 14:34
$begingroup$
The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
$endgroup$
– Xerxes
Mar 24 at 14:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.
See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
(image from Chegg Study: physics questions and answers)
This is in accordance with Gauss's law for the electric field:
- Draw a closed surface around the left charge ($+Q$) only.
There are 6 field lines coming out of this surface - Draw a closed surface around the right charge ($-3Q$) only.
There are 18 field lines going into this surface. - Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
There are 12 field lines going into this big surface. - Draw a closed surface which does not enclose any of the charges.
There are $n$ field lines going into and the same $n$ field lines
coming out of this surface, thus giving a sum of zero.
In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
$endgroup$
add a comment |
$begingroup$
If you draw a Gaussian surface that only includes the $-Q$ charge,
the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
You could move that external charge to infinity and not change that total flux.
So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.
update to the address the OP's follow-up question in the comment
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
... So what does the LHS of the gauss signify then?
Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.
Now, as I reposition the charge [which is easier to do with this visualization],
note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]
When I move the charge outside the Gaussian surface,
the sign of the flux changes for the patches near the charge.
The total flux drops to zero.
So, while external charges contribute to the local flux through a patch,
their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.
Note the total electric flux is not just "$EA$",
it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.
In integral form, it's $oint vec Ecdot dvec A$.
answered Mar 24 at 11:39
robphyrobphy
2,182248
$endgroup$
$begingroup$
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
$endgroup$
– user72730
Mar 24 at 11:40
$begingroup$
Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
$endgroup$
– robphy
Mar 24 at 11:49
$begingroup$
So what does the LHS of the gauss signify then?
$endgroup$
– user72730
Mar 24 at 11:55
$begingroup$
I added some visualizations in my answer to address your follow-up questions.
$endgroup$
– robphy
Mar 24 at 12:20
add a comment |
$begingroup$
The answer would be "yes, but...".
In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.
Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.
However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.
answered Mar 24 at 23:02
PerePere
1,309148
$endgroup$
$begingroup$
@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
$endgroup$
– Pere
Mar 25 at 9:19
$begingroup$
I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.
See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
(image from Chegg Study: physics questions and answers)
This is in accordance with Gauss's law for the electric field:
- Draw a closed surface around the left charge ($+Q$) only.
There are 6 field lines coming out of this surface - Draw a closed surface around the right charge ($-3Q$) only.
There are 18 field lines going into this surface. - Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
There are 12 field lines going into this big surface. - Draw a closed surface which does not enclose any of the charges.
There are $n$ field lines going into and the same $n$ field lines
coming out of this surface, thus giving a sum of zero.
In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
$endgroup$
add a comment |
$begingroup$
The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.
See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
(image from Chegg Study: physics questions and answers)
This is in accordance with Gauss's law for the electric field:
- Draw a closed surface around the left charge ($+Q$) only.
There are 6 field lines coming out of this surface - Draw a closed surface around the right charge ($-3Q$) only.
There are 18 field lines going into this surface. - Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
There are 12 field lines going into this big surface. - Draw a closed surface which does not enclose any of the charges.
There are $n$ field lines going into and the same $n$ field lines
coming out of this surface, thus giving a sum of zero.
In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
$endgroup$
add a comment |
$begingroup$
The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.
See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
(image from Chegg Study: physics questions and answers)
This is in accordance with Gauss's law for the electric field:
- Draw a closed surface around the left charge ($+Q$) only.
There are 6 field lines coming out of this surface - Draw a closed surface around the right charge ($-3Q$) only.
There are 18 field lines going into this surface. - Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
There are 12 field lines going into this big surface. - Draw a closed surface which does not enclose any of the charges.
There are $n$ field lines going into and the same $n$ field lines
coming out of this surface, thus giving a sum of zero.
In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
$endgroup$
The number of flux lines of each charge is proportional to its own charge.
The other charge has nothing to do with that.
See this image with two unequal charges. The right negative charge ($-3Q$)
has three times the size of the left positive charge ($+Q$):
(image from Chegg Study: physics questions and answers)
This is in accordance with Gauss's law for the electric field:
- Draw a closed surface around the left charge ($+Q$) only.
There are 6 field lines coming out of this surface - Draw a closed surface around the right charge ($-3Q$) only.
There are 18 field lines going into this surface. - Draw a big closed surface around both charges together ($+Q-3Q = -2Q$).
There are 12 field lines going into this big surface. - Draw a closed surface which does not enclose any of the charges.
There are $n$ field lines going into and the same $n$ field lines
coming out of this surface, thus giving a sum of zero.
In all cases the number of field lines (i.e. the electric flux)
through the closed surface is proportional
to the charge inside the surface.
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
edited Mar 25 at 8:51
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
answered Mar 24 at 11:44
Thomas FritschThomas Fritsch
1,403415
1,403415
add a comment |
add a comment |
$begingroup$
If you draw a Gaussian surface that only includes the $-Q$ charge,
the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
You could move that external charge to infinity and not change that total flux.
So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.
update to the address the OP's follow-up question in the comment
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
... So what does the LHS of the gauss signify then?
Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.
Now, as I reposition the charge [which is easier to do with this visualization],
note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]
When I move the charge outside the Gaussian surface,
the sign of the flux changes for the patches near the charge.
The total flux drops to zero.
So, while external charges contribute to the local flux through a patch,
their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.
Note the total electric flux is not just "$EA$",
it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.
In integral form, it's $oint vec Ecdot dvec A$.
answered Mar 24 at 11:39
robphyrobphy
2,182248
$endgroup$
$begingroup$
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
$endgroup$
– user72730
Mar 24 at 11:40
$begingroup$
Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
$endgroup$
– robphy
Mar 24 at 11:49
$begingroup$
So what does the LHS of the gauss signify then?
$endgroup$
– user72730
Mar 24 at 11:55
$begingroup$
I added some visualizations in my answer to address your follow-up questions.
$endgroup$
– robphy
Mar 24 at 12:20
add a comment |
$begingroup$
If you draw a Gaussian surface that only includes the $-Q$ charge,
the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
You could move that external charge to infinity and not change that total flux.
So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.
update to the address the OP's follow-up question in the comment
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
... So what does the LHS of the gauss signify then?
Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.
Now, as I reposition the charge [which is easier to do with this visualization],
note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]
When I move the charge outside the Gaussian surface,
the sign of the flux changes for the patches near the charge.
The total flux drops to zero.
So, while external charges contribute to the local flux through a patch,
their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.
Note the total electric flux is not just "$EA$",
it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.
In integral form, it's $oint vec Ecdot dvec A$.
answered Mar 24 at 11:39
robphyrobphy
2,182248
$endgroup$
$begingroup$
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
$endgroup$
– user72730
Mar 24 at 11:40
$begingroup$
Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
$endgroup$
– robphy
Mar 24 at 11:49
$begingroup$
So what does the LHS of the gauss signify then?
$endgroup$
– user72730
Mar 24 at 11:55
$begingroup$
I added some visualizations in my answer to address your follow-up questions.
$endgroup$
– robphy
Mar 24 at 12:20
add a comment |
$begingroup$
If you draw a Gaussian surface that only includes the $-Q$ charge,
the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
You could move that external charge to infinity and not change that total flux.
So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.
update to the address the OP's follow-up question in the comment
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
... So what does the LHS of the gauss signify then?
Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.
Now, as I reposition the charge [which is easier to do with this visualization],
note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]
When I move the charge outside the Gaussian surface,
the sign of the flux changes for the patches near the charge.
The total flux drops to zero.
So, while external charges contribute to the local flux through a patch,
their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.
Note the total electric flux is not just "$EA$",
it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.
In integral form, it's $oint vec Ecdot dvec A$.
answered Mar 24 at 11:39
robphyrobphy
2,182248
$endgroup$
If you draw a Gaussian surface that only includes the $-Q$ charge,
the total electric flux through that surface is proportional to the enclosed charged $-Q$ by Gauss' Law. This doesn't depend on the charge $+q$ external to that Gaussian surface.
You could move that external charge to infinity and not change that total flux.
So, the number of flux lines into $-Q$ is proportional to $-Q$ alone.
update to the address the OP's follow-up question in the comment
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
... So what does the LHS of the gauss signify then?
Using https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/13-fields (and selecting Measurement type: "Gauss's law"), draw a closed surface then introduce a positive charge inside. Observe the outward flux through each patch.
Now, as I reposition the charge [which is easier to do with this visualization],
note that the flux through each patch changes... but the total remains constant [suggested by Gauss's Law, a physical law that says the total electric flux through a Gaussian surface is equal to the enclosed charge divided by $epsilon_0$.]
When I move the charge outside the Gaussian surface,
the sign of the flux changes for the patches near the charge.
The total flux drops to zero.
So, while external charges contribute to the local flux through a patch,
their net [total] contribution to the flux is zero through a Gaussian surface that doesn't enclose those external charges.
Note the total electric flux is not just "$EA$",
it's $sum vec E_i cdot Delta vec A_i$ summing over all patches.
In integral form, it's $oint vec Ecdot dvec A$.
answered Mar 24 at 11:39
robphyrobphy
2,182248
edited Mar 24 at 12:28
answered Mar 24 at 11:39
robphyrobphy
2,182248
answered Mar 24 at 11:39
robphyrobphy
2,182248
answered Mar 24 at 11:39
robphyrobphy
2,182248
2,182248
$begingroup$
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
$endgroup$
– user72730
Mar 24 at 11:40
$begingroup$
Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
$endgroup$
– robphy
Mar 24 at 11:49
$begingroup$
So what does the LHS of the gauss signify then?
$endgroup$
– user72730
Mar 24 at 11:55
$begingroup$
I added some visualizations in my answer to address your follow-up questions.
$endgroup$
– robphy
Mar 24 at 12:20
add a comment |
$begingroup$
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
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– user72730
Mar 24 at 11:40
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Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
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– robphy
Mar 24 at 11:49
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So what does the LHS of the gauss signify then?
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– user72730
Mar 24 at 11:55
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I added some visualizations in my answer to address your follow-up questions.
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– robphy
Mar 24 at 12:20
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But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
$endgroup$
– user72730
Mar 24 at 11:40
$begingroup$
But on LHS Of gauss law I.e /E.A , the E is due to all the charges , this is causing the confusion!
$endgroup$
– user72730
Mar 24 at 11:40
$begingroup$
Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
$endgroup$
– robphy
Mar 24 at 11:49
$begingroup$
Consider a single [say, positive] point charge and draw its field lines. Now draw a gaussian spherical surface centered at that point. The flux through that surface is positive. Now reposition that sphere so that it doesn't include the point charge. Note that every field line that entered that repositioned sphere will leave that sphere. This suggests (and can be backed up with a more detailed calculation) the total flux through that sphere is zero [due to an external charge]. While external charges might affect the local flux through a patch of the sphere, it doesn't affect the total flux.
$endgroup$
– robphy
Mar 24 at 11:49
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So what does the LHS of the gauss signify then?
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– user72730
Mar 24 at 11:55
$begingroup$
So what does the LHS of the gauss signify then?
$endgroup$
– user72730
Mar 24 at 11:55
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I added some visualizations in my answer to address your follow-up questions.
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– robphy
Mar 24 at 12:20
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I added some visualizations in my answer to address your follow-up questions.
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– robphy
Mar 24 at 12:20
add a comment |
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The answer would be "yes, but...".
In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.
Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.
However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.
answered Mar 24 at 23:02
PerePere
1,309148
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@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
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– Pere
Mar 25 at 9:19
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I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
add a comment |
$begingroup$
The answer would be "yes, but...".
In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.
Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.
However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.
answered Mar 24 at 23:02
PerePere
1,309148
$endgroup$
$begingroup$
@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
$endgroup$
– Pere
Mar 25 at 9:19
$begingroup$
I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
add a comment |
$begingroup$
The answer would be "yes, but...".
In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.
Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.
However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.
answered Mar 24 at 23:02
PerePere
1,309148
$endgroup$
The answer would be "yes, but...".
In an universe with global neutral electrical charge, the flux through a closed surface is proportional to the charge inside the surface, but it means that it is also proportional to the charge outside the surface, because both are opposite but equal in absolute value.
Your drawing shows only two opposite charges and the rest of the universe is not supposed matter - it is not included in the model. Therefore all the flux received by the negative charge is originated in the positive one - that is, all lines that end in the negative charge start in the positive charge.
However, although you can say that the flux depends on the charge inside the closed surface or the charge outside the closed surface, it doesn't depend on how are those charges distributed. Therefore, the flux that a given negative point charge receives doesn't depend on whether an opposite negative charge is close to it or very far away.
answered Mar 24 at 23:02
PerePere
1,309148
answered Mar 24 at 23:02
PerePere
1,309148
answered Mar 24 at 23:02
PerePere
1,309148
answered Mar 24 at 23:02
PerePere
1,309148
1,309148
$begingroup$
@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
$endgroup$
– Pere
Mar 25 at 9:19
$begingroup$
I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
add a comment |
$begingroup$
@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
$endgroup$
– Pere
Mar 25 at 9:19
$begingroup$
I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
$begingroup$
@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
$endgroup$
– Pere
Mar 25 at 9:19
$begingroup$
@user72730 - If both charges are equal, the amount that goes away to universe equals the amount that comes from universe - you can prove that by applying the Gauss theorem to a closed surface enclosing both charges and nothing else. The distance between charges doesn't matter.
$endgroup$
– Pere
Mar 25 at 9:19
$begingroup$
I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
$begingroup$
I meant to say I’m not concerned about the distance because you have mentioned it the last line of your answer.
$endgroup$
– user72730
Mar 25 at 9:38
add a comment |
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The "number of flux lines" is a purely aesthetic choice made by the artist of a diagram. While there are obvious choices of number that give intuition for the physical flux density, the choice of number is not a physical phenomenon that follows any kind of laws.
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– Xerxes
Mar 24 at 14:34