What is the electric potential inside a point charge?
$begingroup$
We know that electric potential of spherical charge is inverse proportion with $r$ from $V = frac{kq}{r}$ , So if $r$ is getting less then electric potential will be higher.
But , What about point charge ? Is it infinity inside it ? My teacher told me that it's zero but I am not believing that.
electrostatics charge potential
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
We know that electric potential of spherical charge is inverse proportion with $r$ from $V = frac{kq}{r}$ , So if $r$ is getting less then electric potential will be higher.
But , What about point charge ? Is it infinity inside it ? My teacher told me that it's zero but I am not believing that.
electrostatics charge potential
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What do you mean by spherical charge and point charge? Could you draw a diagram of what you mean?
$endgroup$
– KV18
Apr 14 at 13:57
1
$begingroup$
@KV18 , I drew .
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:08
$begingroup$
Despite the answers, the question on the technical side does not make sense. There is no inside of point. That's the whole point of point. It's a pretty clear indication that the theory doesn't work there, and some other description is needed. At close distances the point approximation will start to deviate from reality.
$endgroup$
– luk32
Apr 15 at 8:12
add a comment |
$begingroup$
We know that electric potential of spherical charge is inverse proportion with $r$ from $V = frac{kq}{r}$ , So if $r$ is getting less then electric potential will be higher.
But , What about point charge ? Is it infinity inside it ? My teacher told me that it's zero but I am not believing that.
electrostatics charge potential
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We know that electric potential of spherical charge is inverse proportion with $r$ from $V = frac{kq}{r}$ , So if $r$ is getting less then electric potential will be higher.
But , What about point charge ? Is it infinity inside it ? My teacher told me that it's zero but I am not believing that.
electrostatics charge potential
electrostatics charge potential
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
edited Apr 14 at 16:51
Qmechanic♦
108k122001249
108k122001249
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
asked Apr 14 at 13:55
Mohammad AlshareefMohammad Alshareef
134
134
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mohammad Alshareef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What do you mean by spherical charge and point charge? Could you draw a diagram of what you mean?
$endgroup$
– KV18
Apr 14 at 13:57
1
$begingroup$
@KV18 , I drew .
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:08
$begingroup$
Despite the answers, the question on the technical side does not make sense. There is no inside of point. That's the whole point of point. It's a pretty clear indication that the theory doesn't work there, and some other description is needed. At close distances the point approximation will start to deviate from reality.
$endgroup$
– luk32
Apr 15 at 8:12
add a comment |
$begingroup$
What do you mean by spherical charge and point charge? Could you draw a diagram of what you mean?
$endgroup$
– KV18
Apr 14 at 13:57
1
$begingroup$
@KV18 , I drew .
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:08
$begingroup$
Despite the answers, the question on the technical side does not make sense. There is no inside of point. That's the whole point of point. It's a pretty clear indication that the theory doesn't work there, and some other description is needed. At close distances the point approximation will start to deviate from reality.
$endgroup$
– luk32
Apr 15 at 8:12
$begingroup$
What do you mean by spherical charge and point charge? Could you draw a diagram of what you mean?
$endgroup$
– KV18
Apr 14 at 13:57
$begingroup$
What do you mean by spherical charge and point charge? Could you draw a diagram of what you mean?
$endgroup$
– KV18
Apr 14 at 13:57
1
1
$begingroup$
@KV18 , I drew .
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:08
$begingroup$
@KV18 , I drew .
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:08
$begingroup$
Despite the answers, the question on the technical side does not make sense. There is no inside of point. That's the whole point of point. It's a pretty clear indication that the theory doesn't work there, and some other description is needed. At close distances the point approximation will start to deviate from reality.
$endgroup$
– luk32
Apr 15 at 8:12
$begingroup$
Despite the answers, the question on the technical side does not make sense. There is no inside of point. That's the whole point of point. It's a pretty clear indication that the theory doesn't work there, and some other description is needed. At close distances the point approximation will start to deviate from reality.
$endgroup$
– luk32
Apr 15 at 8:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Believe it or not, electric potential is zero (or rather it's a constant). You can prove it via shell theorem https://en.wikipedia.org/wiki/Shell_theorem same math applies to electric potential, see inside shell section (Force is gradient of potential, zero force means constant potential) or http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html for more detailed math.
Note that, $V=kQ/r$ is only applicable outside of an sphere, so there is no infinity for potential (You may ask why only outside?, well you can see it yourself, just consider a sphere and calculate electric potential outside of it, by Laplace equation or direct calculation of potential's integral. Then try to calculate the same potential, but this time inside of shell).
Also point charges (i.e electron, etc) in classical mechanics are not really point like, they have in fact radius https://en.wikipedia.org/wiki/Classical_electron_radius .
answered Apr 14 at 17:33
paradoxyparadoxy
1095
$endgroup$
1
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
add a comment |
$begingroup$
This post does not really answer your question, but still it is worth reading what I would like to highlight.
Point charge refers to electrons or protons.
But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge.
A point is something that has no dimensions at all. So there is no inside.
ON YOUR COMMENT:
There is no inside for a point charge, so there is no point of using $V=kQ/r$ because mathematics break down over there.
Meanwhile, for a spherical charge distribution on a conducting shell like you have drawn there, the potential anywhere inside the shell is the same as the potential on the surface - $V=kQ/R$, where $r=R$ (radius of the sphere).
Check mathematically why this is the case!
answered Apr 14 at 14:15
KV18KV18
1,205516
$endgroup$
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
5
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
add a comment |
$begingroup$
When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different.
The electron is a charged point particle in the standard model of particle physics, but when interacting with a proton, the potential near the point location is not accessible. What happens the electron is bound to the proton forming the hydrogen atom, and it cannot fall on the proton as the classical 1/r potential indicates. There is a lowest energy state, and it stays there in orbitals, which are instead of a track, probable locations where it may be found.
So it has no meaning for quantum dimensions to be asking the value of the 1/r potential. Its functional form has been used to calculate the orbitals, and even when there is a probability of overlap, as seen in the S orbitals, there are quantum number conservation rules and energy conservation rules that keep the orbitals stable.
Actually, the stability of the atom is one of the reasons that quantum mechanics had to be invented. In classical physics because of the 1/r attraction the electrons would fall on the nucleus, neutralizing them, and no chemistry would exist.
The proton cannot become neutral when the electron overlaps with it in space, because quantum mechanically the neutral particle is the neutron, which is heavier , and also electron lepton number has to be conserved. ( there is electron capture in nuclei where there is extra energy to allow for the reaction , but that is another story)
One needs to study quantum mechanics and nuclear/particle physics at the realm of point particles.
answered Apr 14 at 14:40
anna vanna v
162k8154457
$endgroup$
1
$begingroup$
I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
$endgroup$
– luk32
Apr 15 at 8:30
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
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
$begingroup$
Believe it or not, electric potential is zero (or rather it's a constant). You can prove it via shell theorem https://en.wikipedia.org/wiki/Shell_theorem same math applies to electric potential, see inside shell section (Force is gradient of potential, zero force means constant potential) or http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html for more detailed math.
Note that, $V=kQ/r$ is only applicable outside of an sphere, so there is no infinity for potential (You may ask why only outside?, well you can see it yourself, just consider a sphere and calculate electric potential outside of it, by Laplace equation or direct calculation of potential's integral. Then try to calculate the same potential, but this time inside of shell).
Also point charges (i.e electron, etc) in classical mechanics are not really point like, they have in fact radius https://en.wikipedia.org/wiki/Classical_electron_radius .
answered Apr 14 at 17:33
paradoxyparadoxy
1095
$endgroup$
1
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
add a comment |
$begingroup$
Believe it or not, electric potential is zero (or rather it's a constant). You can prove it via shell theorem https://en.wikipedia.org/wiki/Shell_theorem same math applies to electric potential, see inside shell section (Force is gradient of potential, zero force means constant potential) or http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html for more detailed math.
Note that, $V=kQ/r$ is only applicable outside of an sphere, so there is no infinity for potential (You may ask why only outside?, well you can see it yourself, just consider a sphere and calculate electric potential outside of it, by Laplace equation or direct calculation of potential's integral. Then try to calculate the same potential, but this time inside of shell).
Also point charges (i.e electron, etc) in classical mechanics are not really point like, they have in fact radius https://en.wikipedia.org/wiki/Classical_electron_radius .
answered Apr 14 at 17:33
paradoxyparadoxy
1095
$endgroup$
1
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
add a comment |
$begingroup$
Believe it or not, electric potential is zero (or rather it's a constant). You can prove it via shell theorem https://en.wikipedia.org/wiki/Shell_theorem same math applies to electric potential, see inside shell section (Force is gradient of potential, zero force means constant potential) or http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html for more detailed math.
Note that, $V=kQ/r$ is only applicable outside of an sphere, so there is no infinity for potential (You may ask why only outside?, well you can see it yourself, just consider a sphere and calculate electric potential outside of it, by Laplace equation or direct calculation of potential's integral. Then try to calculate the same potential, but this time inside of shell).
Also point charges (i.e electron, etc) in classical mechanics are not really point like, they have in fact radius https://en.wikipedia.org/wiki/Classical_electron_radius .
answered Apr 14 at 17:33
paradoxyparadoxy
1095
$endgroup$
Believe it or not, electric potential is zero (or rather it's a constant). You can prove it via shell theorem https://en.wikipedia.org/wiki/Shell_theorem same math applies to electric potential, see inside shell section (Force is gradient of potential, zero force means constant potential) or http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html for more detailed math.
Note that, $V=kQ/r$ is only applicable outside of an sphere, so there is no infinity for potential (You may ask why only outside?, well you can see it yourself, just consider a sphere and calculate electric potential outside of it, by Laplace equation or direct calculation of potential's integral. Then try to calculate the same potential, but this time inside of shell).
Also point charges (i.e electron, etc) in classical mechanics are not really point like, they have in fact radius https://en.wikipedia.org/wiki/Classical_electron_radius .
answered Apr 14 at 17:33
paradoxyparadoxy
1095
answered Apr 14 at 17:33
paradoxyparadoxy
1095
answered Apr 14 at 17:33
paradoxyparadoxy
1095
answered Apr 14 at 17:33
paradoxyparadoxy
1095
1095
1
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
add a comment |
1
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
1
1
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
$begingroup$
Is the classical electron radius actually used in classical mechanics? You're implying that it is, but I've never heard of that.
$endgroup$
– Tanner Swett
Apr 15 at 3:48
add a comment |
$begingroup$
This post does not really answer your question, but still it is worth reading what I would like to highlight.
Point charge refers to electrons or protons.
But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge.
A point is something that has no dimensions at all. So there is no inside.
ON YOUR COMMENT:
There is no inside for a point charge, so there is no point of using $V=kQ/r$ because mathematics break down over there.
Meanwhile, for a spherical charge distribution on a conducting shell like you have drawn there, the potential anywhere inside the shell is the same as the potential on the surface - $V=kQ/R$, where $r=R$ (radius of the sphere).
Check mathematically why this is the case!
answered Apr 14 at 14:15
KV18KV18
1,205516
$endgroup$
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
5
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
add a comment |
$begingroup$
This post does not really answer your question, but still it is worth reading what I would like to highlight.
Point charge refers to electrons or protons.
But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge.
A point is something that has no dimensions at all. So there is no inside.
ON YOUR COMMENT:
There is no inside for a point charge, so there is no point of using $V=kQ/r$ because mathematics break down over there.
Meanwhile, for a spherical charge distribution on a conducting shell like you have drawn there, the potential anywhere inside the shell is the same as the potential on the surface - $V=kQ/R$, where $r=R$ (radius of the sphere).
Check mathematically why this is the case!
answered Apr 14 at 14:15
KV18KV18
1,205516
$endgroup$
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
5
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
add a comment |
$begingroup$
This post does not really answer your question, but still it is worth reading what I would like to highlight.
Point charge refers to electrons or protons.
But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge.
A point is something that has no dimensions at all. So there is no inside.
ON YOUR COMMENT:
There is no inside for a point charge, so there is no point of using $V=kQ/r$ because mathematics break down over there.
Meanwhile, for a spherical charge distribution on a conducting shell like you have drawn there, the potential anywhere inside the shell is the same as the potential on the surface - $V=kQ/R$, where $r=R$ (radius of the sphere).
Check mathematically why this is the case!
answered Apr 14 at 14:15
KV18KV18
1,205516
$endgroup$
This post does not really answer your question, but still it is worth reading what I would like to highlight.
Point charge refers to electrons or protons.
But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge.
A point is something that has no dimensions at all. So there is no inside.
ON YOUR COMMENT:
There is no inside for a point charge, so there is no point of using $V=kQ/r$ because mathematics break down over there.
Meanwhile, for a spherical charge distribution on a conducting shell like you have drawn there, the potential anywhere inside the shell is the same as the potential on the surface - $V=kQ/R$, where $r=R$ (radius of the sphere).
Check mathematically why this is the case!
answered Apr 14 at 14:15
KV18KV18
1,205516
edited Apr 14 at 14:40
answered Apr 14 at 14:15
KV18KV18
1,205516
answered Apr 14 at 14:15
KV18KV18
1,205516
answered Apr 14 at 14:15
KV18KV18
1,205516
1,205516
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
5
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
add a comment |
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
5
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
$begingroup$
If we use the relation $V = frac {kq}{r}$ then we will get $V = Infinity$ , But this is physically forbidden . There is no any answer except zero , But how ?
$endgroup$
– Mohammad Alshareef
Apr 14 at 14:27
5
5
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
$begingroup$
This seems to be the only answer that properly challenges the frame of the question. A point has no inside,so "Inside a point" is completely meaningless.
$endgroup$
– James K
Apr 14 at 20:06
add a comment |
$begingroup$
When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different.
The electron is a charged point particle in the standard model of particle physics, but when interacting with a proton, the potential near the point location is not accessible. What happens the electron is bound to the proton forming the hydrogen atom, and it cannot fall on the proton as the classical 1/r potential indicates. There is a lowest energy state, and it stays there in orbitals, which are instead of a track, probable locations where it may be found.
So it has no meaning for quantum dimensions to be asking the value of the 1/r potential. Its functional form has been used to calculate the orbitals, and even when there is a probability of overlap, as seen in the S orbitals, there are quantum number conservation rules and energy conservation rules that keep the orbitals stable.
Actually, the stability of the atom is one of the reasons that quantum mechanics had to be invented. In classical physics because of the 1/r attraction the electrons would fall on the nucleus, neutralizing them, and no chemistry would exist.
The proton cannot become neutral when the electron overlaps with it in space, because quantum mechanically the neutral particle is the neutron, which is heavier , and also electron lepton number has to be conserved. ( there is electron capture in nuclei where there is extra energy to allow for the reaction , but that is another story)
One needs to study quantum mechanics and nuclear/particle physics at the realm of point particles.
answered Apr 14 at 14:40
anna vanna v
162k8154457
$endgroup$
1
$begingroup$
I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
$endgroup$
– luk32
Apr 15 at 8:30
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
add a comment |
$begingroup$
When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different.
The electron is a charged point particle in the standard model of particle physics, but when interacting with a proton, the potential near the point location is not accessible. What happens the electron is bound to the proton forming the hydrogen atom, and it cannot fall on the proton as the classical 1/r potential indicates. There is a lowest energy state, and it stays there in orbitals, which are instead of a track, probable locations where it may be found.
So it has no meaning for quantum dimensions to be asking the value of the 1/r potential. Its functional form has been used to calculate the orbitals, and even when there is a probability of overlap, as seen in the S orbitals, there are quantum number conservation rules and energy conservation rules that keep the orbitals stable.
Actually, the stability of the atom is one of the reasons that quantum mechanics had to be invented. In classical physics because of the 1/r attraction the electrons would fall on the nucleus, neutralizing them, and no chemistry would exist.
The proton cannot become neutral when the electron overlaps with it in space, because quantum mechanically the neutral particle is the neutron, which is heavier , and also electron lepton number has to be conserved. ( there is electron capture in nuclei where there is extra energy to allow for the reaction , but that is another story)
One needs to study quantum mechanics and nuclear/particle physics at the realm of point particles.
answered Apr 14 at 14:40
anna vanna v
162k8154457
$endgroup$
1
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I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
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– luk32
Apr 15 at 8:30
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
add a comment |
$begingroup$
When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different.
The electron is a charged point particle in the standard model of particle physics, but when interacting with a proton, the potential near the point location is not accessible. What happens the electron is bound to the proton forming the hydrogen atom, and it cannot fall on the proton as the classical 1/r potential indicates. There is a lowest energy state, and it stays there in orbitals, which are instead of a track, probable locations where it may be found.
So it has no meaning for quantum dimensions to be asking the value of the 1/r potential. Its functional form has been used to calculate the orbitals, and even when there is a probability of overlap, as seen in the S orbitals, there are quantum number conservation rules and energy conservation rules that keep the orbitals stable.
Actually, the stability of the atom is one of the reasons that quantum mechanics had to be invented. In classical physics because of the 1/r attraction the electrons would fall on the nucleus, neutralizing them, and no chemistry would exist.
The proton cannot become neutral when the electron overlaps with it in space, because quantum mechanically the neutral particle is the neutron, which is heavier , and also electron lepton number has to be conserved. ( there is electron capture in nuclei where there is extra energy to allow for the reaction , but that is another story)
One needs to study quantum mechanics and nuclear/particle physics at the realm of point particles.
answered Apr 14 at 14:40
anna vanna v
162k8154457
$endgroup$
When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different.
The electron is a charged point particle in the standard model of particle physics, but when interacting with a proton, the potential near the point location is not accessible. What happens the electron is bound to the proton forming the hydrogen atom, and it cannot fall on the proton as the classical 1/r potential indicates. There is a lowest energy state, and it stays there in orbitals, which are instead of a track, probable locations where it may be found.
So it has no meaning for quantum dimensions to be asking the value of the 1/r potential. Its functional form has been used to calculate the orbitals, and even when there is a probability of overlap, as seen in the S orbitals, there are quantum number conservation rules and energy conservation rules that keep the orbitals stable.
Actually, the stability of the atom is one of the reasons that quantum mechanics had to be invented. In classical physics because of the 1/r attraction the electrons would fall on the nucleus, neutralizing them, and no chemistry would exist.
The proton cannot become neutral when the electron overlaps with it in space, because quantum mechanically the neutral particle is the neutron, which is heavier , and also electron lepton number has to be conserved. ( there is electron capture in nuclei where there is extra energy to allow for the reaction , but that is another story)
One needs to study quantum mechanics and nuclear/particle physics at the realm of point particles.
answered Apr 14 at 14:40
anna vanna v
162k8154457
edited Apr 14 at 17:01
answered Apr 14 at 14:40
anna vanna v
162k8154457
answered Apr 14 at 14:40
anna vanna v
162k8154457
answered Apr 14 at 14:40
anna vanna v
162k8154457
162k8154457
1
$begingroup$
I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
$endgroup$
– luk32
Apr 15 at 8:30
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
add a comment |
1
$begingroup$
I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
$endgroup$
– luk32
Apr 15 at 8:30
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
1
1
$begingroup$
I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
$endgroup$
– luk32
Apr 15 at 8:30
$begingroup$
I believe such narrative is wrong "Point particles belong to the realm of quantum mechanics". It gives an impression that point particle exist but that is not true. In the link you give "The electron is a charged point particle in the standard model of particle physics" there's no occurrence of word "point". And in the description of particle physics it's explained that despite the name, those are not points. Questions what's inside the point make no mathematical sense, and what's super close are beyond physical application of the model.
$endgroup$
– luk32
Apr 15 at 8:30
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
$begingroup$
@luk32 in the field theory of the standard model the electrons in the table are treated as point particles,. Experimentally there is a size limit . Certainly due to radiative corrections there is a form factor and it will acquire a size. physics.stackexchange.com/questions/397022/… A point particle has no meaning in classical mechanics, because the term particle implies mass and it is experimentally known that it has to be composed of atoms with their quantum mechanical limits.
$endgroup$
– anna v
Apr 15 at 9:58
add a comment |
Mohammad Alshareef is a new contributor. Be nice, and check out our Code of Conduct.
Mohammad Alshareef is a new contributor. Be nice, and check out our Code of Conduct.
Mohammad Alshareef is a new contributor. Be nice, and check out our Code of Conduct.
Mohammad Alshareef is a new contributor. Be nice, and check out our Code of Conduct.
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What do you mean by spherical charge and point charge? Could you draw a diagram of what you mean?
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– KV18
Apr 14 at 13:57
1
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@KV18 , I drew .
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– Mohammad Alshareef
Apr 14 at 14:08
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Despite the answers, the question on the technical side does not make sense. There is no inside of point. That's the whole point of point. It's a pretty clear indication that the theory doesn't work there, and some other description is needed. At close distances the point approximation will start to deviate from reality.
$endgroup$
– luk32
Apr 15 at 8:12