In general, capacitance is   C

=

Q DV

Consider a parallel-plate capacitor made up of two conducting plates of area A separated by a distance d. One plate has a charge +Q, the other a charge -Q. We'll assume the plates are large and close together, so the charge is distributed uniformly over each plate. The charge per unit area has a magnitude of:

s

=

Q A

It follows from our assumption above that the electric field between the plates is uniform, so the potential difference is:

DV = Ed

We showed previously using Gauss' Law that the field between equal-and-oppositely charged plates is:

E

=

s eo

Combining all this information, we can derive the fact that the capacitance depends solely on the geometry of the capacitor (and what is between the plates, but we'll get to that next time):

C

=

Q DV

=

Q Ed

=

sAeo sd

For a parallel-plate capacitor   C

=

eoA d

Does this result make any sense? A larger capacitance means that it is easier to store charge. If we increase the plate area the charge can spread out more, reducing the repulsive forces between the like charges - that's consistent with making it easier to store charge.

If we put the plates closer together the repulsive forces between the like charges on each plate is offset by attractive forces between the plates - that also makes it easier to store charge.