Help understanding voltage within a conductor

[channel1]

Homework Statement

Help me understand why electric potential can ever be a nonzero value at a point within a charged conductor? I need conceptual reasons to help me make sense of the integration "flaw". (See below.)

Homework Equations

The electric field inside a charged spherical conductor is zero.
The electric potential at any point within a charged conductor is the same as the electric potential on the sphere's surface.

The Attempt at a Solution

But if the equation for electric potential is the integral of the electric field dotted with the displacement vector and E is zero within the conductor---then V=Eds=0 mathematically...So help me understand why electric potential can ever be a nonzero value at a point within a charged conductor? Please use integration to demonstrate your point!

 

[issacnewton]

hi the potential difference is given by

$$V_B-V_A=-\int_A^B \vec{E}\bullet\vec{dl}$$

inside the conductor E=0 , so

$$V_B-V_A = 0$$

which means that potential is same everywhere in the conductor...does it answer your question

 

[gneill]

To find the potential you want to integrate from "infinity" (the ultimate reference point) to the point where you wish to know the potential. In order to do this you have to first integrate E.ds from infinite distance up to the surface of the conductor (thus accounting for the surface potential of the conductor). After that the field is zero and no further contribution is made to the potential. So inside it's all at the surface potential.

 

[channel1]

i guess its still not really clicking...

V_A = kq/infinity
V_B = kq/r (where r less than R and R is the radius to the surface of the sphere)

kq/infinity - kq/r = 0
kq/r - 0 = 0

i understand that when you let V_B = kq/R then E is a nonzero value...but from the algebra of the equation why isn't V zero inside the conductor?

 

[SammyS]

... After that the field is zero and no further contribution is made to the potential. So inside it's all at the surface potential.

What gneill is referring to here is that within the conductor the E field is zero (as you mentioned), so the electric potential is the same throughout the conducting material. The electric potential difference for any two points in or on the conductor is zero.

As IssacNewton posted, the following integral gives the (electric) potential difference from one point to another.
$$V_B-V_A=-\int_A^B \vec{E}\cdot\vec{dl}$$
So, in a conductor the difference in electric potential from point A to point B is zero, because the E field is zero. That doesn't mean that the electric potential is zero, it means that the electric potential is the same value throughout the conducting material.

 

[channel1]

i feel like what you said didnt address my second post, or maybe I am just not seeing the connection...

 

[gneill]

If the potential inside were not everywhere the same as that of the surface of the sphere, there would be a potential difference and thus an E-field that would drive a current of charge carriers to "level" the potential.

To make a crude analogy, gravitational potential rises as we climb a hill. If we reach the rim a plateau at the top the potential has reached its maximum value. Passing on to the plateau we don't say that the potential on the plateau is zero even though it's flat and nothing their wants to roll spontaneously. It has the same potential as the rim on the side of the hill.