Electrostatic potential on the surface of a conductor

[rbnphlp]

Homework Statement

At the surface of a conductor the potential is constant .. I can't get my head around this ..

Homework Equations

E=-grad V

The Attempt at a Solution

The only reason I can think about this is Electric field is perpendicular at the surface , hence no work is done when moving a point charge around the surface , hence potential is constant ..

But if V is constant , the surely E=-grad (V)-> E=0 , which is not the case since there are charges on the surface of the conductor, so surely must provide a E field normal to the surface ..

Thanks for any help

 

[rbnphlp]

Noone ? Does no one understand what I am saying or does no one know the anwer?

 

[vela]

Say the surface lies in the xy-plane, for example. You'd have V(x,y,0)=V₀ so that ∂V/∂x=∂V/∂y=0, but that doesn't imply that ∂V/∂z=0.

 

[rbnphlp]

Say the surface lies in the xy-plane, for example. You'd have V(x,y,0)=V₀ so that ∂V/∂x=∂V/∂y=0, but that doesn't imply that ∂V/∂z=0.

Cheers :D

 

[paranormal]

Your reasoning is correct. The electrostatic potential on the surface of a conductor is constant because the electric field is perpendicular to the surface. This means that there is no change in potential as you move along the surface, and therefore the potential remains constant. However, this does not mean that the electric field is zero. As you mentioned, there are charges on the surface of the conductor which create an electric field that is normal to the surface. This electric field is responsible for keeping the charges on the surface in equilibrium, and it is also responsible for the phenomenon of electrostatic shielding, where the electric field inside a conductor is zero. So, even though the potential is constant, there is still an electric field present on the surface of the conductor.