Let's revisit the calculation of the capacitance of parallel plates. The capacitance is defined as \(C={Q\over\Dl V}\), and we found the potential difference between the plates by integrating the electric field. So $$C=\frac{Q}{\int E_z\,dz}$$ Notice that the capacitance is inversely proportional to the electric field between the plates. If I can make that field smaller, then the capacitance would be bigger.

Now remember what happens when you place an insulator in an electric field? The insulator polarizes, creating a counterfield which partially cancels out the original field, so that the net electric field inside the insulator is smaller than the field outside, by a factor of the dielectric constant \(\kappa\).

Repeat the calculation in Example 9.5.1 to find the capacitance of a
parallel-plate capacaitor, where each plate has area \(A\), the plates
are a distance \(d\) apart, but the space between the plates is filled
with paper, which has a dielectric constant of \(\kappa=3.5\).

Suppose that charge \(+Q\) is on the top plate and charge \(-Q\) is
on the bottom plate. As in Example 9.5.1, the plates would
create an electric field of \(4\pi kQ/A\) between them if the space
wasn't filled with paper. With the paper, the electric field is
smaller by a factor of \(\kappa\):
\(E=\frac{4\pi kQ}{\kappa A}\).
Continuing on with the original calculation, we have that the
potential difference between the plates is
\(\Dl V=\frac{4\pi kQd}{\kappa A}\), and the capacitance is
$$C={Q\over\Dl V}={Q\over 4\pi kQd/\kappa A}=\kappa\frc{4\pi k}{A\over
d}$$

Consider, however, that the materials with the highest dielectric
constants are the metals: a pure conductor has \(\kappa=\infty\).
Of course, we can't insert metal in between the plates of a
capacitor: it would immediately discharge. Even if you use an
insulator, you must look out for the phenomenon of *electric
breakdown* mentioned in . If you place too much charge
on a capacitor, so that the electric field between the plates
exceeds the breakdown threshold of the insulator between them (and
air counts as an insulator in this case!), then the insulator will
become a temporary conductor, and the capacitor will discharge.

To construct a high-capacitance capacitor, then, we would like to
find materials which have high dielectric constants *and* high
breakdown thresholds. There are other tricks one can use to
construct these ultracapacitors, including creating
battery-capacitor hybrids which have the benefits of both.