Here we are considering two parallel plates of area A, not necessarily of rectangular shape, separated by a distance d.
2. Evaluate the electric field everywhere along some path joining the two conductors.
The plates are considered to be large enough in area and close enough in distance that the electric field is uniform well away from the edges (fringing fields at the edges are ignored). In addition, the excess charge deposited on the conductors is assumed to spread out uniformly so that
Application of Gauss's law can be made to determine the dependence of the electric field to the surface charge density present; the symmetry of the charge sheet suggests a gaussian can enclosing one surface. The cross section of the appropriate gaussian can is shown below.
Considering a gaussian can with base area of a, the only contribution to the flux will be that of the end cap between the plates. The portion of the gaussian surface immersed in the conducting material will have no flux through it since the electric field is zero there. The tube wall outside the conductor is constructed so that there is no flux through it either; i.e. the surface vector at all points on the cylindrical surface is perpendicular to E.
3. Evaluate the potential difference along that same path using the relationship between V and E.
Ideally you may select a path over which the line integral will be the simplest to evaluate; i.e. one in which the line integral reduces to a simple integral. In this case, choose the shortest possible path, a straight line perpendicular to the plates, starting from the negatively charged plate (lower potential) and moving toward the positively charged plate. The direction vector of the line integral will then always be antiparallel to the electric field and the dot product in the integrand will reduce the line integral to a simple integral as shown.
4. The capacitance is the ratio of the charge deposited on one conductor to the potential difference between the two.