# Two-Piece Capacitors

## How to improve the capacitance

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.

Suppose we want to calculate the capacitance of two spherical conducting shells, the smaller one with radius $$R_1$$ and the larger with radius $$R_2$$.
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 $$C=\frac{Q}{\Delta V}=\frac{Q}{kQ\frac{R_2-R_1}{R_1R_2}}=\boxed{\displaystyle\frc{k}\frac{R_1R_2}{R_2-R_1}}$$

In case you didn't click through, the capacitance of the two spherical shells in the example is $$C=\frc{k}\frac{R_1R_2}{R_2-R_1}$$ If we compare that with the capacitance of a single sphere with radius $$R$$ $$C=\frc{k}R,$$ the obvious difference is the denominator in the first expression. The capacitance of the two shells is inversely proportional to the spacing between them. Theoretically, we could make the capacitance of this combination as large as we pleased, simply by making the gap between the spheres as small as possible.
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.