Capacitance of a spherical capacitor? Derivation

As we know, to determine the electric field at any point on the surface of a charged spherical conductor of radius r applying Gauss's theorem, 
{tex}E = \frac { q } { 4 \pi \varepsilon _ { 0 } r ^ { 2 } }{/tex}
Hence electrostatic potential, {tex}V = -\int \mathbf { E } \cdot d \mathbf { r } = \frac { q } { 4 \pi \varepsilon _ { 0 } \cdot r }{/tex}
where, {tex}\varepsilon _ { 0 } = 8.854 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } \mathrm { N } ^ { - 1 } \mathrm { m } ^ { - 2}{/tex} = permittivity of free space or vacuum
The capacitance of the spherical conductor situated in vacuum is given by
{tex}C = \frac { q } { V } = \frac { q } { \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \cdot \frac { q } { r } } \Rightarrow C = 4 \pi \varepsilon _ { 0 } r{/tex}
Hence, the capacitance of an isolated spherical conductor situated in a vacuum is always {tex}4 \pi \varepsilon _ { 0 }{/tex} times of its radius.