Instead of charging the blue sphere with charges from infinity (a), we can bring the charges from a nearby sphere (b), which becomes negative.

Up till now, when we charged a conductor we brought the charge in from very far away (i.e. infinity). Suppose the charge came from a nearby source instead? This should make it easier to charge the conductor, and thereby increase the capacitance. We can do this by introducing a second conductor, initially neutral, and move charge from the second conductor onto the first conductor. We call this combination of two conductors a two-piece capacitor, and define the capacitance of the pair combined as $$C=\frac{Q}{\Delta V}$$ where \(\Delta V\) is the potential difference between the positive piece and the negative piece.

Let's look at an example, and see if the capacitance really does improve.

How can we calculate its capacitance? What do we need to figure out?

Move a charge \(Q\) from the outer shell to the inner shell, and
calculate the potential difference between the two shells. The ratio
of charge to potential difference is the capacitance.

How can I find the potential difference between the two shells?

We don't have a formula for the potential difference of spherical
shells, but we *do* know the electric field created by spherical
shells. And we can get the potential difference from the electric
field:
$$\Dl V=-\int E_r\,dr$$
where \(E_r\) is the electric field between the two shells. The
integral is from the negative shell to the positive shell.

OK, what is the electric field between the two shells?

Spherical shells only create an electric field outside themselves;
thus the red shell doesn't contribute to the electric field between
the shells at all. The blue shell creates a field outside itself
which is the same as the field of a point charge at its center. Thus
the electric field between the shells is
$$E_r=k\frac{Q}{r^2}$$
where \(r\) is the distance from the center of the shells.

What is the potential difference between the two shells?

Using the formula above
$$
\begin{eqnarray}
\Delta V&=&-\int E_r\,dr\\
&=&-\int_{R_2}^{R_1}k\frac{Q}{r^2}\,dr\\
&=&-kQ\left[-\frac1{r}\right]_{R_2}^{R_1}\\
&=& kQ\left[\frac1{R_1}-\frac1{R_2}\right]\\
&=& kQ\frac{R_2-R_1}{R_1R_2}\\
\end{eqnarray}
$$

What is the capacitance?

The capacitance is
$$C=\frac{Q}{\Delta
V}=\frac{Q}{kQ\frac{R_2-R_1}{R_1R_2}}=\boxed{\displaystyle\frc{k}\frac{R_1R_2}{R_2-R_1}}$$

Find the capacitance of the two shells above, if \(R_2\gg R_1\).

If \(R_2\gg R_1\), then \(R_2-R_1\approx R_2\), and so
$$C\approx\frc{k}\frac{R_1R_2}{R_2}=\frc{k}R_1$$
which is the capacitance of a single sphere. If the outer sphere is very large, then when we transfer charge from it to the inner sphere we are basically bringing charge in from infinity, which is how we would charge a single sphere.