Does insulating sphere have a capacitance? A conducting sphere with radius $r$ has capacitance $C=4\pi \epsilon_0 r$. This can be derived as follows:
$$
C=\frac{Q}{V}=\frac{Q}{\frac{Q}{4\pi \epsilon_0 r}}.
$$
Now I am wondering if this applies to a plastic sphere as well. It is always tricky to work with insulators since the value of $V$ is different at different positions. 
The whole concept of capacitance becomes doubtful: we define $C=Q/V$, but $V$ is different at different positions, so where should we measure $V$? Also: in electronics, we take $V$ to be the potential difference - but here, $V$ is the potential, NOT a potential difference.
 A: No, just for the reason you say:  The potential is not the same at every point on an insulator.  A capacitance measures the voltage difference $V$ per unit charge $Q$ between two opposite assemblages of charge, each at a unique voltage.  (In some cases, the second assemblage of charge may be taken to be on a sphere at spatial infinity, where we known $V=0$.)  Without conductors to even out the voltages involved, there is no way to choose which voltage should appear in the definition of $C$.
A: Spherical Styrofoam pellets and balloons can have capacitance (C) and a voltage gradient (V) because they can retain electrostatic energy (W).
W=1/2 CV^2
A: You do not know $V$ at all - it could be anything assuming it is an ideal insulator. In reality nothing is ideal. The equation you give assumes a uniform distribution of charge, and given enough time the charge would redistribute uniformally over the sphere due to the extremely small - but non zero - conductance of the plastic. This means eventually the equation would be a very good description.
Failing that, it's probably an okay order of magnitude approximation depending on what you're using the result for.
