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

In summary, the charge on a conductor will always reside on its surface due to the tendency of the system to attain a state of minimum potential. This can be proven using Gauss's law by considering a Gaussian surface just inside the conductor and observing the electric field on this surface. This concept does not apply to insulators, as they do not have any significant amount of free charge and cannot attain a state of minimum potential due to the strong nuclear force.
  • #1
kaustavdgreat
4
0
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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  • #2
Try using a Gaussian surface that is just inside the conductor. What is the electric field on this Gaussian surface?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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:
  • #6
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
 

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

Free charge resides on the surface of a conductor because of the principle of electrostatic equilibrium. In this state, the electric field inside a conductor is zero, and any excess charge will distribute itself on the surface in order to minimize repulsion between charges.

2. How does the distribution of free charge on a conductor's surface affect its electric field?

The distribution of free charge on a conductor's surface creates an electric field that is perpendicular to the surface. This electric field is strongest at sharp points or edges on the surface and weakest at flat areas.

3. What happens to the distribution of free charge on a conductor's surface when an external electric field is applied?

When an external electric field is applied, the distribution of free charge on a conductor's surface will adjust in order to cancel out the external field inside the conductor. This is known as electrostatic shielding.

4. Does the type of conductor affect the distribution of free charge on its surface?

Yes, the type of conductor does affect the distribution of free charge on its surface. Conductors with higher conductivity will have a more even distribution of charge on their surface, while conductors with lower conductivity will have a more uneven distribution.

5. Can the distribution of free charge on a conductor's surface be manipulated?

Yes, the distribution of free charge on a conductor's surface can be manipulated by using methods such as electroplating, where an external electric field is used to deposit a layer of metal on the surface of the conductor. This can change the overall charge distribution on the surface of the conductor.

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