Why is the electric potential constant inside a spherical conducting shell?

[Corneo]

Consider a spherical conducting shell where all the charges reside on the surface. Using Gauss's Law, it can be found that the electric field inside the shell is zero. But why is the electric potential a constant? More over, why is the potential same as the potential on the surface of the shell? Please refer to the voltage plots in the following link.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Thank you.

 

[Timbuqtu]

The electric field is just the derivative/gradient of the potential:

E = \nabla V

So where the electric field vanishes, V must be constant. And V must be continuous everywhere (unless E is infinite).

 

[Dr.Brain]

Consider a spherical conducting shell where all the charges reside on the surface. Using Gauss's Law, it can be found that the electric field inside the shell is zero. But why is the electric potential a constant? More over, why is the potential same as the potential on the surface of the shell? Please refer to the voltage plots in the following link.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Thank you.

The relation between Electric Field and Potential is given by:

<br /> <br /> E=- \frac{dV}{dR}<br />

When E =0 , then from the above expression the potential has to constant.
Because everywhere inside the shell the electric field is zero, therefore everywhere inside it , potential is constant and same . If there are two different potentials between two different points, then due to potential difference the charges on the sphere might start moving, which is not the case when E=0.