Why are the electrostatic field lines normal to a charged conductor?

  • #1
kirito
54
8
Homework Statement
understanding basics of conductors
Relevant Equations
gauss law , work, potential

I understand the following .a conductor is made of atoms and atoms always strive to be at equilibrium and that's why the electric field inside a conductor is zero because the electros distribute themselves in such a way so that they are in equilibrium , yet they do produce an electric field outside the conductor . When an external electric field is applied, these free electrons redistribute themselves within the conductor in such a way that cancels out the field inside the material, establishing an equilibrium.

and since divE=density/e0 we get that total density inside the conductor is zero if they are distributed somewhere its on the outer shell ,
and since the electric field is the negative gradient of the electric potential (E=−∇V). then the potential is constant , I just can't seem to connect what does this have to do with the electric field lines being perpendicular to the conductor at any point ,

I tried to think of work force and electric field to connect the dots but I this is the farthest I reached​

 
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  • #2
If there were a component of the field parallel to the surface of the conductor, what would the charges there do?
 
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  • #3
the charge would experience a force in that direction leading it to move and change the potential of the system?
 
  • #4
I would like to ask something that I am confused with ,when we talk about a surface its direction is described by the normal to it , so here parallel to the surface does that mean in the direction of one of the vectors in the plane or in the direction of the normal , I assume I should stick to the first definition but I do admit getting confused about what is meant
 
  • #5
kirito said:
I would like to ask something that I am confused with ,when we talk about a surface its direction is described by the normal to it , so here parallel to the surface does that mean in the direction of one of the vectors in the plane or in the direction of the normal , I assume I should stick to the first definition but I do admit getting confused about what is meant
A curved surface's direction ('orientation' might be a better word) at a point P is the direction of the normal at P.

We can construct the tangent-plane at P. Any line which passes through P and is in this plane is parallel to the surface at P.

1717842175373.png


https://upload.wikimedia.org/wikipe...t-plane.svg/330px-Image_Tangent-plane.svg.png

If it helps, you can imagine looking at the region near P through an infinitely powerful magnifying glass – the region of the surface near P will look like a flat plane – this is part of P's tangent plane.

Minor edit.
 
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  • #6
kirito said:
the charge would experience a force in that direction leading it to move and change the potential of the system?
Yes. Another way of looking at it is to use the fact that a conductor is an equipotential surface. Since the field is the (negative) gradient of the potential, the field is normal to the surface.

kirito said:
I would like to ask something that I am confused with ,when we talk about a surface its direction is described by the normal to it , so here parallel to the surface does that mean in the direction of one of the vectors in the plane or in the direction of the normal , I assume I should stick to the first definition but I do admit getting confused about what is meant
Parallel to the surface means orthogonal to the surface normal.
 
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  • #7
Orodruin said:
Yes. Another way of looking at it is to use the fact that a conductor is an equipotential surface. Since the field is the (negative) gradient of the potential, the field is normal to the surface.


Parallel to the surface means orthogonal to the surface normal.
here I can't follow through why knowing that the field is the negative gradient of the potential =E=0 will lead to the conclusion that the field is normal to the surface , and thank you in advance
 
  • #8
kirito said:
here I can't follow through why knowing that the field is the negative gradient of the potential =E=0 will lead to the conclusion that the field is normal to the surface , and thank you in advance
Surface Electron.png
Consider an electron on the surface of the conductor. It is represented by the red dot e-. Assume that field E (blue arrow) is at an angle away from the surface normal. It can be resolved into a normal component En and a tangential component Et.

The tangential component implies a potential difference across the surface which means a concentration of positive charges to the right of the electron (shown in yellow) and a concentration of negative charges to the left of the electron (shown in magenta.) What will the electron do?

It will migrate to the right and, in so doing, it will diminish the potential difference and hence reduce the tangential component. Other surface electrons will do the same. So the long and short of it is that if there is a tangential component, electrons will migrate and keep on migrating until there have no reason to migrate any more. This will happen when the tangential component of the field becomes zero. Therefore, if there is an electric field at the surface, it can only have a normal component.
 
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  • #9
kuruman said:
View attachment 346626Consider an electron on the surface of the conductor. It is represented by the red dot e-. Assume that field E (blue arrow) is at an angle away from the surface normal. It can be resolved into a normal component En and a tangential component Et.

The tangential component implies a potential difference across the surface which means a concentration of positive charges to the right of the electron (shown in yellow) and a concentration of negative charges to the left of the electron (shown in magenta.) What will the electron do?

It will migrate to the right and, in so doing, it will diminish the potential difference and hence reduce the tangential component. Other surface electrons will do the same. So the long and short of it is that if there is a tangential component, electrons will migrate and keep on migrating until there have no reason to migrate any more. This will happen when the tangential component of the field becomes zero. Therefore, if there is an electric field at the surface, it can only have a normal component.
thanks a lot for both your time and the visuals , so that means that when I combine the 2 conditions of the electric field being minus the gradient of the potential , with the condition of constant potential inside the conductor it leads me to concluding If there where to be an electric field outside on the surface it must be perpendicular to the surface of the conductor, since if it is not then the potential is not constant meaning the electric field inside can differ from zero which is a contradiction,
 
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  • #10
kirito said:
here I can't follow through why knowing that the field is the negative gradient of the potential =E=0 will lead to the conclusion that the field is normal to the surface , and thank you in advance
The gradient of any function is normal to the level surfaces of that function. For any small displacement ##d\vec x##,
$$
f(\vec x_0 + d\vec x) = f(\vec x_0) + d\vec x \cdot \nabla f + \mathcal O(dx)^2
$$
For any ##d\vec x## within the level surface, it must hold that ##f(\vec x_0+d\vec x) = f(\vec x_0)## and therefore
$$
d\vec x \cdot \nabla f = 0
$$
whenever ##d\vec x## is along the level surface. In other words, the gradient ##\nabla f## is orthogonal to all displacements in the level surface.
 
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  • #11
Orodruin said:
The gradient of any function is normal to the level surfaces of that function. For any small displacement ##d\vec x##,
$$
f(\vec x_0 + d\vec x) = f(\vec x_0) + d\vec x \cdot \nabla f + \mathcal O(dx)^2
$$
For any ##d\vec x## within the level surface, it must hold that ##f(\vec x_0+d\vec x) = f(\vec x_0)## and therefore
$$
d\vec x \cdot \nabla f = 0
$$
whenever ##d\vec x## is along the level surface. In other words, the gradient ##\nabla f## is orthogonal to all displacements in the level surface.
thank you this was actually needed though I derived it before and watched a couple of explanations on it ,I did not remember ,hopefully it will stick with me from here on
 
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