1) Electric potential inside the sphere: 
2) Ratio Vcenter/Vsurface: 3/2
3) Find graph in attachment
Explanation:
1)
The electric field inside the sphere is given by

where
is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
r is the distance from the centre at which we compute the field
For a radial field,

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

The potential at the surface, V(R), is that of a point charge, so

Therefore we can find the potential inside the sphere, V(r):
2)
At the center,
r = 0
Therefore the potential at the center of the sphere is:

On the other hand, the potential at the surface is

Therefore, the ratio V(center)/V(surface) is:

3)
The graph of V versus r can be found in attachment.
We observe the following:
- At r = 0, the value of the potential is
, as found in part b) (where
)
- Between r and R, the potential decreases as 
- Then at r = R, the potential is 
- Between r = R and r = 3R, the potential decreases as
, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (
)
Learn more about electric fields and potential:
brainly.com/question/8960054
brainly.com/question/4273177
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