Capacity of an isolated spherical conductor For an spherical conductor shell of radius R, it is known that
$V (r)=  \dfrac {q}{4 \pi \epsilon_o r}$ for r>R
$V(r) =  \dfrac {q}{4 \pi \epsilon_o R}$ for r<=R
My texbook gives the capacitance as 
$C =  \dfrac {q}{V}= 4 \pi \epsilon_o R$ 
Why is the V in the surface been used for it? If I used, the expresion for r>R , lets say r=2R, And calling V the potencial at the surface(r=R) and V' the potential at r=2R then
$V'= V (r=2R)=  \dfrac {q}{4 \pi \epsilon_o 2R}= \dfrac {1}{2}(\dfrac {q}{4 \pi \epsilon_o R})=\dfrac {V}{2}$
then since the charge is constant but the potential is different, the capacitance would be: $ C'= \dfrac{q}{ V'}=2 \dfrac{q}{ V}=2C$, which  makes the capacitance dependent on the potential... Why is this wrong? How do I make sense of the capacitance of an isolated object, I thought one always needes a pair of objects placed one close to the other to have a capacitance. The book does not specify if its a solid sphere or an spherical surface(as I am assuming). In case it is a solid sphere the inner potential would be quadratic, would the result  be different ?
 A: The capacitance measures the charged stored ($Q$) per unit voltage between two oppositely-charged conductors (carrying $Q$ and $-Q$).  The charge of an isolated conductor is a special case of this, where the second conductor is taken to be located an infinite distance away.  (You can imagine the second conductor being a sphere of radius $r\rightarrow\infty$, although the shape of the distant conductor does not actually matter.)
So for an isolated conductor, the inverse capacitance ($C^{-1}$) is the voltage difference between the conductor and infinity, divided by $Q$.  The voltage at infinity is zero by convention*, and the voltage at the other conductor is the voltage of the sphere in your case, $V(R)=Q/(4\pi\epsilon_{0}R)$. So the relevant voltage difference is just $V(R)$ and $C=4\pi\epsilon_{0}R$.
*Even when there is taken to a total charge $-Q$ at infinity, the potential at $r\rightarrow\infty$ is still vanishing, because the charge is spread out over a sphere with infinite area.  You can see this by taking the outer sphere to be at a finite radius $b$; then the voltage difference between the conductors at $R$ and $b$ is $Q/(4\pi\epsilon_{0}R)-Q/(4\pi\epsilon_{0}b)$ which goes to just $V(R)$ as $b\rightarrow\infty$.
