# 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 ?

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 $$b\rightarrow\infty$$.