Field and Potential from Conducting Spheres
We know what the electric field and potential from a point charge look like:
E 
= 
kQ
 
r^{2}



and 
V 
= 
kQ
 
r


Consider a charged sphere with a symmetrical distribution of charge. Gauss' Law tells us that the electric field outside the sphere is the same as that from a point charge. This implies that outside the sphere the potential also looks like the potential from a point charge.
What about inside the sphere? If the sphere is a conductor we know the field inside the sphere is zero. What about the potential?
Moving from a point on the surface of the sphere to a point inside, the potential changes by an amount:
ΔV = ∫ E • ds
Because E = 0, we can only conclude that ΔV is also zero, so V is constant and equal to the value of the potential at the outer surface of the sphere.
Potential near an Insulating Sphere
Now consider a solid insulating sphere of radius R with charge uniformly distributed throughout its volume. Once again, outside the sphere both the electric field and the electric potential are identical to the field and potential from a point charge.
What happens inside the sphere? Now the potential is not constant because there is a field inside the sphere. Using Gauss' Law we showed that the field inside a uniformly charged insulator is:
E 
= 
k Q r
 
R^{3}



Use this to calculate the potential inside the sphere. Starting from some point a distance r from the center and moving out to the edge of the sphere, the potential changes by an amount:
ΔV = V(R)  V(r) 
= 
– 
∫ 

= 
– 
∫ 

= 
–kQ
 
R^{3}


∫ 

Integrating gives:
V(R) – V(r) 
= 
– kQ
 
2R^{3}


(R^{2} – r^{2}) 
V(R) – V(r) 
= 
– kQ
 
2R


( 1 – 
r^{2}
 
R^{2}

 ) 
V(R) is simply kQ/R, which can be written as 2kQ/2R, so:
V(r) 
= 
2kQ
 
2R


+ 
kQ
 
2R


( 1 – 
r^{2}
 
R^{2}

 ) 
Therefore, for r < R,
V(r) 
= 
kQ
 
2R


( 3 – 
r^{2}
 
R^{2}

 ) 