Why does free charge reside on the surface of a conductor?

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Homework Help Overview

The discussion revolves around the concept of charge distribution on conductors, specifically addressing why free charge resides on the surface of a conductor. The original poster attempts to prove this using Gauss's law and laws of capacitance, focusing on a spherical conductor model.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss's law and capacitance principles to demonstrate charge distribution. There are questions about the validity of the original poster's reasoning, particularly regarding the assumption that charge can be treated as a spherical shell and the implications for non-spherical conductors.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning the completeness of the original argument. Some suggest that the reasoning may not apply to insulators, prompting further exploration of the differences between conductors and insulators. There is no explicit consensus yet, but the discussion is probing deeper into the underlying principles.

Contextual Notes

Participants note that the problem does not specify the shape of the conductor, raising concerns about the generalizability of the arguments presented. Additionally, there is a focus on the nature of free charge versus bound charge in insulators, which adds complexity to the discussion.

kaustavdgreat
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1.Prove that free charge resides on the surface of a conductor.(Using Gauss's law or laws of capacitance)

Homework Equations


The Attempt at a Solution


I could not figure out how to prove this using Gauss's law,though i tried to solve it using the laws of capacitance.This is what i did-

Let us consider a spherical conductor of radius R having charge Q.
The potential of the conductor (V)=Q/C
If the charges are spread in such a way that it forms a spherical shell of radius r then, C=r
where [tex]r\leq{R}[/tex]
Now the maximum value of r should be R
We know that every system tends to attain a state of minimum potential.
V will be min, when C is max or, C=R
So, r=R
Hence, the charge will always reside on the surface.

Thnx in advance :wink:
 
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Try using a Gaussian surface that is just inside the conductor. What is the electric field on this Gaussian surface?
 
kaustavdgreat said:
1.Prove that free charge resides on the surface of a conductor.(Using Gauss's law or laws of capacitance)


Homework Equations





The Attempt at a Solution


I could not figure out how to prove this using Gauss's law,though i tried to solve it using the laws of capacitance.This is what i did-

Let us consider a spherical conductor of radius R having charge Q.
The potential of the conductor (V)=Q/C
If the charges are spread in such a way that it forms a spherical shell of radius r then, C=r
where [tex]r\leq{R}[/tex]
Now the maximum value of r should be R
We know that every system tends to attain a state of minimum potential.
V will be min, when C is max or, C=R
So, r=R
Hence, the charge will always reside on the surface.

Thnx in advance :wink:


This is not enough (or complete).
With the same argument you'll have the charge distributed on the surface of an insulator (or not?). Where is the conductivity in this picture?
 
nasu said:
This is not enough (or complete).
With the same argument you'll have the charge distributed on the surface of an insulator (or not?). Where is the conductivity in this picture?


The same argument can't be used in case of an insulator as you know, insulators don't have any (or any significant amount of) free charge.
 
kaustavdgreat said:
The same argument can't be used in case of an insulator as you know, insulators don't have any (or any significant amount of) free charge.

So you mean that in insulators the charge cannot take the configuration of minimum potential energy, right? (Because it's not free to do it.) Then this will make the difference, I think.
However assuming that the free charge distributes itself on a spherical shell (even of arbitrary radius) is not really proven or obvious. And what if the conductor is not spherical? The problem does not say so.
 
Last edited:
nasu said:
However assuming that the free charge distributes itself on a spherical shell (even of arbitrary radius) is not really proven or obvious.
I think [tex]C\propto R[/tex] is applicable for any other curved geometrical shape(R=radius of curvature);so, you can assume any shape.

nasu said:
And what if the conductor is not spherical? The problem does not say so.
You are right, one can prove it for any curved conductor but the problem arises when we are dealing with cubical(or such other shapes) conductors, as radius of curvature of a plane surface is infinity.I think, it can be proved more convincingly using Gauss's law,I am still working with it.

nasu said:
So you mean that in insulators the charge cannot take the configuration of minimum potential energy, right? (Because it's not free to do it.) Then this will make the difference, I think.
If you have placed a ball on the roof of a building, it will always tend to fall on the ground to minimize its gravitational potential;but this will happen only when you allow it to fall.The same thing happens in case of bound electrons in insulators,they cannot take the configuration of minimum potential just because they are not allowed to do it due to the strong nuclear force.The configuration they attain is that of minimum potential available for them.
Moreover,there is no point in arguing with the insulator issue since first of all charge doesn't reside on the surface in case of an insulator and secondly as I told you earlier insulators don't have any free charge.

PLZ HELP ME SOLVE IT ANYONE.THNX
 

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