Parallel-Plate Capacitors
From the definition of C = Q/∆V, we can calculate C for a given pair of conductors, if we
imagine imposing a potential difference between the two plates and calculate the charge on either
plate (since they will be equal and opposite). Obviously, this will be harder or easier depending
on the geometry of the conducting surfaces.
The most important case is the situation with two parallel conducting plates of equal area
A and separated by a distance d. If we put equal and opposite charges of magnitude Q on the
plates, we can calculate the potential difference between the plates and use that to find C for the
plates. But, before doing this calculation, let’s try to use intuition to see what parameters C
should depend on.
We take our parallel plates and connect them to a battery. As electrons flow onto the
negative plate of the capacitor, they repel electrons on the other plate, leaving it positive. The
electrons on the negative plate repel one another and try to get as far away from each other as
possible. The larger the area of the plates, the more electric charge that can be put on the plates
for a given potential difference supplied by the battery. So our intuition tells us that we should
expect that C is proportional to A.
The only other geometric parameter of the problem is the separation distance d between
the plates. How should we expect C to depend on d? We know that the electric field between
the plates is uniform (we learned this using Gauss’s Law, where we found that E = σ/εo) and
therefore from the definition of ∆V, we have (in this case of a uniform electric field only) that
∆V = Ed or ∆V = σd/εo. Therefore we expect that, since in the definition of C, ∆V appears in the
denominator, C should depend inversely on d so that the smaller d, the larger C.
We can get an explicit expression for C in this case by putting these ideas together:
Q
Q
Q
Aε
C=
=
=
= o,
d
∆V  σ d   Qd 

 

 ε o   Aε o 
where we have used the fact that σ = Q/A. Note that C depends only on the two geometric
factors of the problem and nothing else.
Example: Cylindrical Capacitor
A solid cylindrical conducting core of radius a is coaxial with a cylindrical conducting shell of
radius b and negligible thickness. If equal and opposite charges Q are put on the two conductors
and they have length L find their capacitance C.