Why do we use a cylinder as a Gaussian surface for infinitely long charged wire?Why we cannot use Gauss's Law to find the Electric Field of a finite-length charged wire?Gaussian surface in a charged infinite plance sheetElectric field of an infinitely long (thin) metal cylinderPotential of an infinitely long cylinderWhy is Gauss's law used to compute electric field only for infinitely large objects in my physics textbook?Gauss's Law for inside a long solid cylinder of uniform charge density?Gauss law ambiguityGauss' law: Infinitely long cylinder approximationWhile deriving the electric field around a infinitely long wire why do we consider a cylinder as an Gaussian plane?Why can't Gauss surface be a cube?
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Why do we use a cylinder as a Gaussian surface for infinitely long charged wire?
Why we cannot use Gauss's Law to find the Electric Field of a finite-length charged wire?Gaussian surface in a charged infinite plance sheetElectric field of an infinitely long (thin) metal cylinderPotential of an infinitely long cylinderWhy is Gauss's law used to compute electric field only for infinitely large objects in my physics textbook?Gauss's Law for inside a long solid cylinder of uniform charge density?Gauss law ambiguityGauss' law: Infinitely long cylinder approximationWhile deriving the electric field around a infinitely long wire why do we consider a cylinder as an Gaussian plane?Why can't Gauss surface be a cube?
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$begingroup$
Why do we use a cylinder as a Gaussian surface for infinitely long charged wire and not some other shape like cube?
electrostatics electric-fields symmetry gauss-law
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
$endgroup$
|
show 1 more comment
$begingroup$
Why do we use a cylinder as a Gaussian surface for infinitely long charged wire and not some other shape like cube?
electrostatics electric-fields symmetry gauss-law
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
$endgroup$
8
$begingroup$
Have you tried doing it with a cube?
$endgroup$
– knzhou
Jun 25 at 11:59
$begingroup$
As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!
$endgroup$
– Josh Hoffmann
Jun 25 at 12:03
$begingroup$
@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:57
$begingroup$
@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.
$endgroup$
– Joshua
Jun 25 at 19:40
$begingroup$
@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?
$endgroup$
– K.T.
Jun 26 at 14:07
|
show 1 more comment
$begingroup$
Why do we use a cylinder as a Gaussian surface for infinitely long charged wire and not some other shape like cube?
electrostatics electric-fields symmetry gauss-law
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
$endgroup$
Why do we use a cylinder as a Gaussian surface for infinitely long charged wire and not some other shape like cube?
electrostatics electric-fields symmetry gauss-law
electrostatics electric-fields symmetry gauss-law
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
edited Jun 25 at 11:56
Qmechanic♦
111k12 gold badges211 silver badges1298 bronze badges
111k12 gold badges211 silver badges1298 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
asked Jun 25 at 10:00
Pranav KPranav K
245 bronze badges
245 bronze badges
8
$begingroup$
Have you tried doing it with a cube?
$endgroup$
– knzhou
Jun 25 at 11:59
$begingroup$
As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!
$endgroup$
– Josh Hoffmann
Jun 25 at 12:03
$begingroup$
@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:57
$begingroup$
@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.
$endgroup$
– Joshua
Jun 25 at 19:40
$begingroup$
@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?
$endgroup$
– K.T.
Jun 26 at 14:07
|
show 1 more comment
8
$begingroup$
Have you tried doing it with a cube?
$endgroup$
– knzhou
Jun 25 at 11:59
$begingroup$
As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!
$endgroup$
– Josh Hoffmann
Jun 25 at 12:03
$begingroup$
@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:57
$begingroup$
@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.
$endgroup$
– Joshua
Jun 25 at 19:40
$begingroup$
@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?
$endgroup$
– K.T.
Jun 26 at 14:07
8
8
$begingroup$
Have you tried doing it with a cube?
$endgroup$
– knzhou
Jun 25 at 11:59
$begingroup$
Have you tried doing it with a cube?
$endgroup$
– knzhou
Jun 25 at 11:59
$begingroup$
As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!
$endgroup$
– Josh Hoffmann
Jun 25 at 12:03
$begingroup$
As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!
$endgroup$
– Josh Hoffmann
Jun 25 at 12:03
$begingroup$
@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:57
$begingroup$
@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:57
$begingroup$
@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.
$endgroup$
– Joshua
Jun 25 at 19:40
$begingroup$
@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.
$endgroup$
– Joshua
Jun 25 at 19:40
$begingroup$
@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?
$endgroup$
– K.T.
Jun 26 at 14:07
$begingroup$
@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?
$endgroup$
– K.T.
Jun 26 at 14:07
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Gauss' theorem would apply to a cube or other shape. You can write
$$
int bf E cdot dbf S = fracQepsilon_0
$$
where the surface has any shape you like. However, the next step is to do the integral. How can we evaluate $bf E cdot dbf S$ if we don't even know the angle between $bf E$ and $dbf S$? Nor do we know how the size of $bf E$ relates to its distance from the middle of the cube or whatever. This is where the cylinder comes into play. If we choose a cylinder centred on the line charge then we can argue from the rotational and translational symmmetry that $bf E$ must be radially outwards at the curved surface, and also that the size of $bf E$ will be the same everywhere around the curved surface. It is only because we can make such a claim that we can proceed to do the integral. This neat way to perform the integral wouldn't work for a some other shape of surface.
Of course, if you were treating a different problem then you would choose your surface to suit that problem.
By the way, it is also interesting to note that Gauss's law on its own does not tell you the whole solution here, because it cannot rule out that the field may also have a further contribution having zero divergence so not contributing to the flux integral. To be specific, it does not rule out there may be a non-zero component of $bf E$ in the direction in loops around the line charge. To rule that out you must invoke some further information, which could be either Coulomb's law, or the third Maxwell equation which describes Faraday's law of induction. In the present case there is no changing magnetic flux and therefore the line integral of $bf E$ around any loop is zero, which rules out the possibility of a further component to $bf E$. If your professor has omitted to mention this further piece of reasoning then you may like to inquire about it!
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
$endgroup$
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
add a comment |
$begingroup$
As mentioned by others, any Gaussian surface can be used. There are some rules of thumb that are helpful. If you have some element of your system that extends uniformly and infinitely in some direction, then you want your Gaussian surface to be uniform in that direction as well. This suggests you want a cylinder of some cross section. The math is easier if the tangents of the surface are at a constant angle relative to the vectors pointing radially outward from the wire. This suggests you want the tangent to be perpendicular to the radii, so a circular cross section.
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
$endgroup$
add a comment |
$begingroup$
As a matter of fact you can choose any arbitrary shape to be your gaussian surface, as long as the charges are inside it. But just because we can, should we? Gauss Law can be thought of geometric approach to coulombs law. We use the geometric symmetries of the charge distribution to make our problem simpler.
Remember, Gauss' Law has a dot product of E and dS. If you want, you can take a cube centered around the line charge, but then you have to integrate over all the angles of E and dS making it more cumbersome. You can reduce this effort, by cunningly choosing such a surface where the electric field is always parallel to dS or perpendicular to the surface. That way, you don't have to include any angle consideration as before.
So in conclusion, you can choose any gaussian surface. But carefully choosing one makes our life easy. That's all.
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
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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
$begingroup$
Gauss' theorem would apply to a cube or other shape. You can write
$$
int bf E cdot dbf S = fracQepsilon_0
$$
where the surface has any shape you like. However, the next step is to do the integral. How can we evaluate $bf E cdot dbf S$ if we don't even know the angle between $bf E$ and $dbf S$? Nor do we know how the size of $bf E$ relates to its distance from the middle of the cube or whatever. This is where the cylinder comes into play. If we choose a cylinder centred on the line charge then we can argue from the rotational and translational symmmetry that $bf E$ must be radially outwards at the curved surface, and also that the size of $bf E$ will be the same everywhere around the curved surface. It is only because we can make such a claim that we can proceed to do the integral. This neat way to perform the integral wouldn't work for a some other shape of surface.
Of course, if you were treating a different problem then you would choose your surface to suit that problem.
By the way, it is also interesting to note that Gauss's law on its own does not tell you the whole solution here, because it cannot rule out that the field may also have a further contribution having zero divergence so not contributing to the flux integral. To be specific, it does not rule out there may be a non-zero component of $bf E$ in the direction in loops around the line charge. To rule that out you must invoke some further information, which could be either Coulomb's law, or the third Maxwell equation which describes Faraday's law of induction. In the present case there is no changing magnetic flux and therefore the line integral of $bf E$ around any loop is zero, which rules out the possibility of a further component to $bf E$. If your professor has omitted to mention this further piece of reasoning then you may like to inquire about it!
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
$endgroup$
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
add a comment |
$begingroup$
Gauss' theorem would apply to a cube or other shape. You can write
$$
int bf E cdot dbf S = fracQepsilon_0
$$
where the surface has any shape you like. However, the next step is to do the integral. How can we evaluate $bf E cdot dbf S$ if we don't even know the angle between $bf E$ and $dbf S$? Nor do we know how the size of $bf E$ relates to its distance from the middle of the cube or whatever. This is where the cylinder comes into play. If we choose a cylinder centred on the line charge then we can argue from the rotational and translational symmmetry that $bf E$ must be radially outwards at the curved surface, and also that the size of $bf E$ will be the same everywhere around the curved surface. It is only because we can make such a claim that we can proceed to do the integral. This neat way to perform the integral wouldn't work for a some other shape of surface.
Of course, if you were treating a different problem then you would choose your surface to suit that problem.
By the way, it is also interesting to note that Gauss's law on its own does not tell you the whole solution here, because it cannot rule out that the field may also have a further contribution having zero divergence so not contributing to the flux integral. To be specific, it does not rule out there may be a non-zero component of $bf E$ in the direction in loops around the line charge. To rule that out you must invoke some further information, which could be either Coulomb's law, or the third Maxwell equation which describes Faraday's law of induction. In the present case there is no changing magnetic flux and therefore the line integral of $bf E$ around any loop is zero, which rules out the possibility of a further component to $bf E$. If your professor has omitted to mention this further piece of reasoning then you may like to inquire about it!
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
$endgroup$
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
add a comment |
$begingroup$
Gauss' theorem would apply to a cube or other shape. You can write
$$
int bf E cdot dbf S = fracQepsilon_0
$$
where the surface has any shape you like. However, the next step is to do the integral. How can we evaluate $bf E cdot dbf S$ if we don't even know the angle between $bf E$ and $dbf S$? Nor do we know how the size of $bf E$ relates to its distance from the middle of the cube or whatever. This is where the cylinder comes into play. If we choose a cylinder centred on the line charge then we can argue from the rotational and translational symmmetry that $bf E$ must be radially outwards at the curved surface, and also that the size of $bf E$ will be the same everywhere around the curved surface. It is only because we can make such a claim that we can proceed to do the integral. This neat way to perform the integral wouldn't work for a some other shape of surface.
Of course, if you were treating a different problem then you would choose your surface to suit that problem.
By the way, it is also interesting to note that Gauss's law on its own does not tell you the whole solution here, because it cannot rule out that the field may also have a further contribution having zero divergence so not contributing to the flux integral. To be specific, it does not rule out there may be a non-zero component of $bf E$ in the direction in loops around the line charge. To rule that out you must invoke some further information, which could be either Coulomb's law, or the third Maxwell equation which describes Faraday's law of induction. In the present case there is no changing magnetic flux and therefore the line integral of $bf E$ around any loop is zero, which rules out the possibility of a further component to $bf E$. If your professor has omitted to mention this further piece of reasoning then you may like to inquire about it!
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
$endgroup$
Gauss' theorem would apply to a cube or other shape. You can write
$$
int bf E cdot dbf S = fracQepsilon_0
$$
where the surface has any shape you like. However, the next step is to do the integral. How can we evaluate $bf E cdot dbf S$ if we don't even know the angle between $bf E$ and $dbf S$? Nor do we know how the size of $bf E$ relates to its distance from the middle of the cube or whatever. This is where the cylinder comes into play. If we choose a cylinder centred on the line charge then we can argue from the rotational and translational symmmetry that $bf E$ must be radially outwards at the curved surface, and also that the size of $bf E$ will be the same everywhere around the curved surface. It is only because we can make such a claim that we can proceed to do the integral. This neat way to perform the integral wouldn't work for a some other shape of surface.
Of course, if you were treating a different problem then you would choose your surface to suit that problem.
By the way, it is also interesting to note that Gauss's law on its own does not tell you the whole solution here, because it cannot rule out that the field may also have a further contribution having zero divergence so not contributing to the flux integral. To be specific, it does not rule out there may be a non-zero component of $bf E$ in the direction in loops around the line charge. To rule that out you must invoke some further information, which could be either Coulomb's law, or the third Maxwell equation which describes Faraday's law of induction. In the present case there is no changing magnetic flux and therefore the line integral of $bf E$ around any loop is zero, which rules out the possibility of a further component to $bf E$. If your professor has omitted to mention this further piece of reasoning then you may like to inquire about it!
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
edited Jun 25 at 15:26
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
answered Jun 25 at 10:11
Andrew SteaneAndrew Steane
8,0811 gold badge9 silver badges43 bronze badges
8,0811 gold badge9 silver badges43 bronze badges
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
add a comment |
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
$endgroup$
– K.T.
Jun 25 at 14:56
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
Note that you also need reflection symmetry (not just rotational and translational) to argue that $mathbfE$ is radial; otherwise, it could have a component parallel to the wire.
$endgroup$
– Michael Seifert
Jun 25 at 17:39
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
$begingroup$
@MichaelSeifert yes: quite right, thanks; I forgot to mention it
$endgroup$
– Andrew Steane
Jun 25 at 20:51
add a comment |
$begingroup$
As mentioned by others, any Gaussian surface can be used. There are some rules of thumb that are helpful. If you have some element of your system that extends uniformly and infinitely in some direction, then you want your Gaussian surface to be uniform in that direction as well. This suggests you want a cylinder of some cross section. The math is easier if the tangents of the surface are at a constant angle relative to the vectors pointing radially outward from the wire. This suggests you want the tangent to be perpendicular to the radii, so a circular cross section.
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
$endgroup$
add a comment |
$begingroup$
As mentioned by others, any Gaussian surface can be used. There are some rules of thumb that are helpful. If you have some element of your system that extends uniformly and infinitely in some direction, then you want your Gaussian surface to be uniform in that direction as well. This suggests you want a cylinder of some cross section. The math is easier if the tangents of the surface are at a constant angle relative to the vectors pointing radially outward from the wire. This suggests you want the tangent to be perpendicular to the radii, so a circular cross section.
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
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add a comment |
$begingroup$
As mentioned by others, any Gaussian surface can be used. There are some rules of thumb that are helpful. If you have some element of your system that extends uniformly and infinitely in some direction, then you want your Gaussian surface to be uniform in that direction as well. This suggests you want a cylinder of some cross section. The math is easier if the tangents of the surface are at a constant angle relative to the vectors pointing radially outward from the wire. This suggests you want the tangent to be perpendicular to the radii, so a circular cross section.
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
$endgroup$
As mentioned by others, any Gaussian surface can be used. There are some rules of thumb that are helpful. If you have some element of your system that extends uniformly and infinitely in some direction, then you want your Gaussian surface to be uniform in that direction as well. This suggests you want a cylinder of some cross section. The math is easier if the tangents of the surface are at a constant angle relative to the vectors pointing radially outward from the wire. This suggests you want the tangent to be perpendicular to the radii, so a circular cross section.
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
answered Jun 25 at 16:29
R. RomeroR. Romero
87511 bronze badges
87511 bronze badges
add a comment |
add a comment |
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As a matter of fact you can choose any arbitrary shape to be your gaussian surface, as long as the charges are inside it. But just because we can, should we? Gauss Law can be thought of geometric approach to coulombs law. We use the geometric symmetries of the charge distribution to make our problem simpler.
Remember, Gauss' Law has a dot product of E and dS. If you want, you can take a cube centered around the line charge, but then you have to integrate over all the angles of E and dS making it more cumbersome. You can reduce this effort, by cunningly choosing such a surface where the electric field is always parallel to dS or perpendicular to the surface. That way, you don't have to include any angle consideration as before.
So in conclusion, you can choose any gaussian surface. But carefully choosing one makes our life easy. That's all.
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
$endgroup$
add a comment |
$begingroup$
As a matter of fact you can choose any arbitrary shape to be your gaussian surface, as long as the charges are inside it. But just because we can, should we? Gauss Law can be thought of geometric approach to coulombs law. We use the geometric symmetries of the charge distribution to make our problem simpler.
Remember, Gauss' Law has a dot product of E and dS. If you want, you can take a cube centered around the line charge, but then you have to integrate over all the angles of E and dS making it more cumbersome. You can reduce this effort, by cunningly choosing such a surface where the electric field is always parallel to dS or perpendicular to the surface. That way, you don't have to include any angle consideration as before.
So in conclusion, you can choose any gaussian surface. But carefully choosing one makes our life easy. That's all.
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
$endgroup$
add a comment |
$begingroup$
As a matter of fact you can choose any arbitrary shape to be your gaussian surface, as long as the charges are inside it. But just because we can, should we? Gauss Law can be thought of geometric approach to coulombs law. We use the geometric symmetries of the charge distribution to make our problem simpler.
Remember, Gauss' Law has a dot product of E and dS. If you want, you can take a cube centered around the line charge, but then you have to integrate over all the angles of E and dS making it more cumbersome. You can reduce this effort, by cunningly choosing such a surface where the electric field is always parallel to dS or perpendicular to the surface. That way, you don't have to include any angle consideration as before.
So in conclusion, you can choose any gaussian surface. But carefully choosing one makes our life easy. That's all.
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
$endgroup$
As a matter of fact you can choose any arbitrary shape to be your gaussian surface, as long as the charges are inside it. But just because we can, should we? Gauss Law can be thought of geometric approach to coulombs law. We use the geometric symmetries of the charge distribution to make our problem simpler.
Remember, Gauss' Law has a dot product of E and dS. If you want, you can take a cube centered around the line charge, but then you have to integrate over all the angles of E and dS making it more cumbersome. You can reduce this effort, by cunningly choosing such a surface where the electric field is always parallel to dS or perpendicular to the surface. That way, you don't have to include any angle consideration as before.
So in conclusion, you can choose any gaussian surface. But carefully choosing one makes our life easy. That's all.
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
edited Jun 27 at 7:24
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
answered Jun 25 at 10:35
Siddhartha DamSiddhartha Dam
17012 bronze badges
17012 bronze badges
add a comment |
add a comment |
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8
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Have you tried doing it with a cube?
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– knzhou
Jun 25 at 11:59
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As has just been eluded to by @knzhou, try doing it with some other shape and you will very quickly see why we choose a surface with cylindrical symmetry!
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– Josh Hoffmann
Jun 25 at 12:03
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@knzhou, do you have a link to read someone's solution for finding electric field due to infinitely long wire? Or can you do it so we can see? It might be tedious, but it'll be fun to see how someone approaches that kind of mathematics.
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– K.T.
Jun 25 at 14:57
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@K.T.: We did this in physics class another way. We proceeded through several steps of the kinematics, eventually proving the shape of the surface doesn't actually matter.
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– Joshua
Jun 25 at 19:40
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@Joshua I tried proving it(and failed), but don't see how it's connected to kinematics. Care to elaborate, please?
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– K.T.
Jun 26 at 14:07