Why is it necessary that the field lines from a point placed in a visinity of a conductor must be normal to the surface of the conductor at every point?
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I am answering because Suggested Edits to a previous answer, which lays the argument in similar lines, were not considered.
The point charge in question is creating an electric field which is radially outwards from the point charge. Now you have placed a conducting metal near it. The electric field vectors now get bent in such a way that now they are normal to the surface of the conductor at every point on the surface. The question is why.
A conductor means it has ample free charges: electrons. This is what sets them apart from other materials like dielectrics where all the electrons are bound to the parent atoms. In conductors, a good fraction of the electrons in the material are free to move around. What that means is that, once you have an electric field anywhere inside the conductor, the free charges move around and nullify it. Now imagine having two components of an external electric field (in this case produced by the point charge outside) on the surface: one tangential to the surface and the other perpendicular to it. Due to the same reason as above, the mobile charges will nullify the tangential component. However, they cannot move outof the surface (they are bound to the conductor as a whole, by a threshold energy, called the ‘work function’), i.e. motions in the directions to the surfaces are not allowed. Hence this component of the field remains, and the ‘field lines’ are perpendicular.
There are two caveats in the above answer:
I am talking about ‘motions’ of electrons, which is not strictly true in the accurate description of electrons, which is quantum mechanics. However, the general picture I have painted is valid even in that picture, the description being via wave-functions of the ‘electron cloud’ and things like that.The ‘motion’ of the electrons (accurately, adjustments of the wave-functions of the electron-cloud) happen in finite amount of time, but they are so small that for all practical purposes, such time-scales are negligible.
The point charge in question is creating an electric field which is radially outwards from the point charge. Now you have placed a conducting metal near it. The electric field vectors now get bent in such a way that now they are normal to the surface of the conductor at every point on the surface. The question is why.
A conductor means it has ample free charges: electrons. This is what sets them apart from other materials like dielectrics where all the electrons are bound to the parent atoms. In conductors, a good fraction of the electrons in the material are free to move around. What that means is that, once you have an electric field anywhere inside the conductor, the free charges move around and nullify it. Now imagine having two components of an external electric field (in this case produced by the point charge outside) on the surface: one tangential to the surface and the other perpendicular to it. Due to the same reason as above, the mobile charges will nullify the tangential component. However, they cannot move outof the surface (they are bound to the conductor as a whole, by a threshold energy, called the ‘work function’), i.e. motions in the directions to the surfaces are not allowed. Hence this component of the field remains, and the ‘field lines’ are perpendicular.
There are two caveats in the above answer:
I am talking about ‘motions’ of electrons, which is not strictly true in the accurate description of electrons, which is quantum mechanics. However, the general picture I have painted is valid even in that picture, the description being via wave-functions of the ‘electron cloud’ and things like that.The ‘motion’ of the electrons (accurately, adjustments of the wave-functions of the electron-cloud) happen in finite amount of time, but they are so small that for all practical purposes, such time-scales are negligible.
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