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Does a point charge inside a conducting shell cause redistribution of charge in the shell?
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A point charge Q is placed inside a conducting spherical shell at a random place (non-centre).
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
Question: How can there be a force on the electrons in the shell from Q making them redistribute, but no force on Q from the electrons in the shell? Is it true that a charge inside a conducting shell induces a redistribution of charge on the shell? It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
electromagnetism electric-fields gauss-law coulombs-law
asked Aug 17 at 16:11
Elly
738
add a comment |Â
up vote
3
down vote
favorite
A point charge Q is placed inside a conducting spherical shell at a random place (non-centre).
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
Question: How can there be a force on the electrons in the shell from Q making them redistribute, but no force on Q from the electrons in the shell? Is it true that a charge inside a conducting shell induces a redistribution of charge on the shell? It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
electromagnetism electric-fields gauss-law coulombs-law
asked Aug 17 at 16:11
Elly
738
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A point charge Q is placed inside a conducting spherical shell at a random place (non-centre).
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
Question: How can there be a force on the electrons in the shell from Q making them redistribute, but no force on Q from the electrons in the shell? Is it true that a charge inside a conducting shell induces a redistribution of charge on the shell? It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
electromagnetism electric-fields gauss-law coulombs-law
asked Aug 17 at 16:11
Elly
738
A point charge Q is placed inside a conducting spherical shell at a random place (non-centre).
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
Question: How can there be a force on the electrons in the shell from Q making them redistribute, but no force on Q from the electrons in the shell? Is it true that a charge inside a conducting shell induces a redistribution of charge on the shell? It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
electromagnetism electric-fields gauss-law coulombs-law
asked Aug 17 at 16:11
Elly
738
edited Aug 17 at 16:26
asked Aug 17 at 16:11
Elly
738
asked Aug 17 at 16:11
Elly
738
asked Aug 17 at 16:11
Elly
738
738
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
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up vote
10
down vote
accepted
You're confusing two phenomena here.
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
This is (a paraphrase of) the usual argument for the fact that $vecE$ vanishes inside a spherical shell with a uniform charge distribution. If we were to place a charge $Q$ inside the shell, and held all the charges on the shell fixed (for example, if it was insulating), then $Q$ would feel no force.
Critically, the usual argument relies on the fact that the charge distribution is uniform over the surface of the shell. If the charge on the shell isn't evenly distributed, the electric field inside the shell will not be zero.
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
If, on the other hand, the sphere is conductive, the charges can move around on it. Placing the charge $Q$ inside the sphere then causes them to experience a force, and they will redistribute themselves so that they are in equilibrium with $Q$ and with the other charges on the sphere.
It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
The key difference here is that you're talking about two different distributions of charge on the shell. Since the shell charges are in different places here, we shouldn't expect the forces on $Q$ to be the same in each case.
As an analogy: suppose there is a charged bead $q$ that can slide along a rod placed along the $x$-axis. I then fix a second charge $Q$ in place somewhere along the $y$-axis. At the moment I put $Q$ in place, it feels some particular force. But $q$ can slide along the rod, and so it gets pushed away from $Q$, which causes the force on $Q$ to change.
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
1
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
add a comment |Â
up vote
4
down vote
The force between every electron and the charge is the same according to the Newton's third law. When you add together the forces from all electrons on the charge, the resulting force is zero for the reason in your first bullet. However, the force from the charge on every electron is still there. Thus the answer is yes.
answered Aug 17 at 16:50
safesphere
6,37111237
add a comment |Â
up vote
2
down vote
There can't be non-zero electric field in a conductor. No matter how the charge distributed inside or outside of a conductor, electric field inside (in the conducting material) is always zero. So if you place a point charge inside a conducting shell the inner side of the conducting shell will have same amount but opposite charge distributed so that a Gaussian surface in side the conducting material have zero enclosed charge, which is required to make the electric field zero in the conducting material. Now that the total amount of charge has to be conserved the outside of the surface of the conducting shell will have same amount of charge as the point charge inside to compensate the extra charge in the interior side of the shell; so that the net charge from outside of the conducting shell is just the point charge placed inside the shell.
answered Aug 17 at 18:08
Arafat Hossen
212
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
add a comment |Â
up vote
0
down vote
The charges induced on the inner surface will apply some force on the charge placed inside the cavity but the charge on the outer surface of conducter won't interact with the charge inside the cavity. It's explanation is given in I.E Irodov Basic laws of Electromagnetism ,page 52-53
answered Aug 18 at 3:06
user204283
91
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
You're confusing two phenomena here.
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
This is (a paraphrase of) the usual argument for the fact that $vecE$ vanishes inside a spherical shell with a uniform charge distribution. If we were to place a charge $Q$ inside the shell, and held all the charges on the shell fixed (for example, if it was insulating), then $Q$ would feel no force.
Critically, the usual argument relies on the fact that the charge distribution is uniform over the surface of the shell. If the charge on the shell isn't evenly distributed, the electric field inside the shell will not be zero.
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
If, on the other hand, the sphere is conductive, the charges can move around on it. Placing the charge $Q$ inside the sphere then causes them to experience a force, and they will redistribute themselves so that they are in equilibrium with $Q$ and with the other charges on the sphere.
It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
The key difference here is that you're talking about two different distributions of charge on the shell. Since the shell charges are in different places here, we shouldn't expect the forces on $Q$ to be the same in each case.
As an analogy: suppose there is a charged bead $q$ that can slide along a rod placed along the $x$-axis. I then fix a second charge $Q$ in place somewhere along the $y$-axis. At the moment I put $Q$ in place, it feels some particular force. But $q$ can slide along the rod, and so it gets pushed away from $Q$, which causes the force on $Q$ to change.
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
1
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
add a comment |Â
up vote
10
down vote
accepted
You're confusing two phenomena here.
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
This is (a paraphrase of) the usual argument for the fact that $vecE$ vanishes inside a spherical shell with a uniform charge distribution. If we were to place a charge $Q$ inside the shell, and held all the charges on the shell fixed (for example, if it was insulating), then $Q$ would feel no force.
Critically, the usual argument relies on the fact that the charge distribution is uniform over the surface of the shell. If the charge on the shell isn't evenly distributed, the electric field inside the shell will not be zero.
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
If, on the other hand, the sphere is conductive, the charges can move around on it. Placing the charge $Q$ inside the sphere then causes them to experience a force, and they will redistribute themselves so that they are in equilibrium with $Q$ and with the other charges on the sphere.
It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
The key difference here is that you're talking about two different distributions of charge on the shell. Since the shell charges are in different places here, we shouldn't expect the forces on $Q$ to be the same in each case.
As an analogy: suppose there is a charged bead $q$ that can slide along a rod placed along the $x$-axis. I then fix a second charge $Q$ in place somewhere along the $y$-axis. At the moment I put $Q$ in place, it feels some particular force. But $q$ can slide along the rod, and so it gets pushed away from $Q$, which causes the force on $Q$ to change.
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
1
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
You're confusing two phenomena here.
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
This is (a paraphrase of) the usual argument for the fact that $vecE$ vanishes inside a spherical shell with a uniform charge distribution. If we were to place a charge $Q$ inside the shell, and held all the charges on the shell fixed (for example, if it was insulating), then $Q$ would feel no force.
Critically, the usual argument relies on the fact that the charge distribution is uniform over the surface of the shell. If the charge on the shell isn't evenly distributed, the electric field inside the shell will not be zero.
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
If, on the other hand, the sphere is conductive, the charges can move around on it. Placing the charge $Q$ inside the sphere then causes them to experience a force, and they will redistribute themselves so that they are in equilibrium with $Q$ and with the other charges on the sphere.
It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
The key difference here is that you're talking about two different distributions of charge on the shell. Since the shell charges are in different places here, we shouldn't expect the forces on $Q$ to be the same in each case.
As an analogy: suppose there is a charged bead $q$ that can slide along a rod placed along the $x$-axis. I then fix a second charge $Q$ in place somewhere along the $y$-axis. At the moment I put $Q$ in place, it feels some particular force. But $q$ can slide along the rod, and so it gets pushed away from $Q$, which causes the force on $Q$ to change.
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
You're confusing two phenomena here.
I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').
This is (a paraphrase of) the usual argument for the fact that $vecE$ vanishes inside a spherical shell with a uniform charge distribution. If we were to place a charge $Q$ inside the shell, and held all the charges on the shell fixed (for example, if it was insulating), then $Q$ would feel no force.
Critically, the usual argument relies on the fact that the charge distribution is uniform over the surface of the shell. If the charge on the shell isn't evenly distributed, the electric field inside the shell will not be zero.
But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.
If, on the other hand, the sphere is conductive, the charges can move around on it. Placing the charge $Q$ inside the sphere then causes them to experience a force, and they will redistribute themselves so that they are in equilibrium with $Q$ and with the other charges on the sphere.
It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
The key difference here is that you're talking about two different distributions of charge on the shell. Since the shell charges are in different places here, we shouldn't expect the forces on $Q$ to be the same in each case.
As an analogy: suppose there is a charged bead $q$ that can slide along a rod placed along the $x$-axis. I then fix a second charge $Q$ in place somewhere along the $y$-axis. At the moment I put $Q$ in place, it feels some particular force. But $q$ can slide along the rod, and so it gets pushed away from $Q$, which causes the force on $Q$ to change.
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
answered Aug 17 at 17:33
Michael Seifert
13.6k12650
13.6k12650
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
1
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
add a comment |Â
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
1
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
Ok (and thanks), so if the electrons are stuck in a uniform distribution there is no force felt by charge Q. But if it is a conducting shell the electrons feel a force from the charge and move. How can the charge Q feel no force from the electrons in the shell (at the initial point before any redistribution, but the electrons in the shell feel a force from Q at this point to redistribute?
– Elly
Aug 17 at 20:14
1
1
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
The fact that the net force on $Q$ from all the surface charges is zero doesn't imply that the force on $Q$ from any individual patch of the surface is zero. And if the force on $Q$ from some patch of the surface is non-zero, then the force on the charges in that patch from $Q$ will also be non-zero, and the charges on that path will move around in response to it.
– Michael Seifert
Aug 17 at 20:18
add a comment |Â
up vote
4
down vote
The force between every electron and the charge is the same according to the Newton's third law. When you add together the forces from all electrons on the charge, the resulting force is zero for the reason in your first bullet. However, the force from the charge on every electron is still there. Thus the answer is yes.
answered Aug 17 at 16:50
safesphere
6,37111237
add a comment |Â
up vote
4
down vote
The force between every electron and the charge is the same according to the Newton's third law. When you add together the forces from all electrons on the charge, the resulting force is zero for the reason in your first bullet. However, the force from the charge on every electron is still there. Thus the answer is yes.
answered Aug 17 at 16:50
safesphere
6,37111237
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The force between every electron and the charge is the same according to the Newton's third law. When you add together the forces from all electrons on the charge, the resulting force is zero for the reason in your first bullet. However, the force from the charge on every electron is still there. Thus the answer is yes.
answered Aug 17 at 16:50
safesphere
6,37111237
The force between every electron and the charge is the same according to the Newton's third law. When you add together the forces from all electrons on the charge, the resulting force is zero for the reason in your first bullet. However, the force from the charge on every electron is still there. Thus the answer is yes.
answered Aug 17 at 16:50
safesphere
6,37111237
answered Aug 17 at 16:50
safesphere
6,37111237
answered Aug 17 at 16:50
safesphere
6,37111237
answered Aug 17 at 16:50
safesphere
6,37111237
6,37111237
add a comment |Â
add a comment |Â
up vote
2
down vote
There can't be non-zero electric field in a conductor. No matter how the charge distributed inside or outside of a conductor, electric field inside (in the conducting material) is always zero. So if you place a point charge inside a conducting shell the inner side of the conducting shell will have same amount but opposite charge distributed so that a Gaussian surface in side the conducting material have zero enclosed charge, which is required to make the electric field zero in the conducting material. Now that the total amount of charge has to be conserved the outside of the surface of the conducting shell will have same amount of charge as the point charge inside to compensate the extra charge in the interior side of the shell; so that the net charge from outside of the conducting shell is just the point charge placed inside the shell.
answered Aug 17 at 18:08
Arafat Hossen
212
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
add a comment |Â
up vote
2
down vote
There can't be non-zero electric field in a conductor. No matter how the charge distributed inside or outside of a conductor, electric field inside (in the conducting material) is always zero. So if you place a point charge inside a conducting shell the inner side of the conducting shell will have same amount but opposite charge distributed so that a Gaussian surface in side the conducting material have zero enclosed charge, which is required to make the electric field zero in the conducting material. Now that the total amount of charge has to be conserved the outside of the surface of the conducting shell will have same amount of charge as the point charge inside to compensate the extra charge in the interior side of the shell; so that the net charge from outside of the conducting shell is just the point charge placed inside the shell.
answered Aug 17 at 18:08
Arafat Hossen
212
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There can't be non-zero electric field in a conductor. No matter how the charge distributed inside or outside of a conductor, electric field inside (in the conducting material) is always zero. So if you place a point charge inside a conducting shell the inner side of the conducting shell will have same amount but opposite charge distributed so that a Gaussian surface in side the conducting material have zero enclosed charge, which is required to make the electric field zero in the conducting material. Now that the total amount of charge has to be conserved the outside of the surface of the conducting shell will have same amount of charge as the point charge inside to compensate the extra charge in the interior side of the shell; so that the net charge from outside of the conducting shell is just the point charge placed inside the shell.
answered Aug 17 at 18:08
Arafat Hossen
212
There can't be non-zero electric field in a conductor. No matter how the charge distributed inside or outside of a conductor, electric field inside (in the conducting material) is always zero. So if you place a point charge inside a conducting shell the inner side of the conducting shell will have same amount but opposite charge distributed so that a Gaussian surface in side the conducting material have zero enclosed charge, which is required to make the electric field zero in the conducting material. Now that the total amount of charge has to be conserved the outside of the surface of the conducting shell will have same amount of charge as the point charge inside to compensate the extra charge in the interior side of the shell; so that the net charge from outside of the conducting shell is just the point charge placed inside the shell.
answered Aug 17 at 18:08
Arafat Hossen
212
answered Aug 17 at 18:08
Arafat Hossen
212
answered Aug 17 at 18:08
Arafat Hossen
212
answered Aug 17 at 18:08
Arafat Hossen
212
212
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
add a comment |Â
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
I don't think this actually answers the question that was asked; it doesn't address the existence/non-existence of the force on the charge inside the shell.
– Michael Seifert
Aug 17 at 20:21
add a comment |Â
up vote
0
down vote
The charges induced on the inner surface will apply some force on the charge placed inside the cavity but the charge on the outer surface of conducter won't interact with the charge inside the cavity. It's explanation is given in I.E Irodov Basic laws of Electromagnetism ,page 52-53
answered Aug 18 at 3:06
user204283
91
add a comment |Â
up vote
0
down vote
The charges induced on the inner surface will apply some force on the charge placed inside the cavity but the charge on the outer surface of conducter won't interact with the charge inside the cavity. It's explanation is given in I.E Irodov Basic laws of Electromagnetism ,page 52-53
answered Aug 18 at 3:06
user204283
91
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The charges induced on the inner surface will apply some force on the charge placed inside the cavity but the charge on the outer surface of conducter won't interact with the charge inside the cavity. It's explanation is given in I.E Irodov Basic laws of Electromagnetism ,page 52-53
answered Aug 18 at 3:06
user204283
91
The charges induced on the inner surface will apply some force on the charge placed inside the cavity but the charge on the outer surface of conducter won't interact with the charge inside the cavity. It's explanation is given in I.E Irodov Basic laws of Electromagnetism ,page 52-53
answered Aug 18 at 3:06
user204283
91
answered Aug 18 at 3:06
user204283
91
answered Aug 18 at 3:06
user204283
91
answered Aug 18 at 3:06
user204283
91
91
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