Look at the unit analysis: Resistivity is measured in Ω-m. When you mutliply by length (m) and divide by area (m2), you're left with just Ω.
Permittivity is measured in Farads/m. When you multiply by area (m2) and divide by spacing (m), you're left with just Farads, or capacitance.
Capacitance of a parallel place capacitor is related to the dielectric constant, area, and distance between plates by a very simple equation:
$$C = \frac{\epsilon_r \epsilon_0 A}{d}$$
Where \$C\$ is capacitance, \$A\$ is plate area, and \$d\$ is the separation between plates. \$\epsilon_0\$ is the permittivity of free space and \$\epsilon_r\$ is the relative permittivity.
The voltage across a capacitor at any given instant is given by:
$$V = \frac{Q}{C}$$
Where \$V\$ is voltage, \$Q\$ is the charge on the plates, and \$C\$ is the capacitance.
So clearly as capacitance increases, the amount of charge required for a given voltage across the plates will also increase proportionally.
If you require more charge, it will take longer to reach a given voltage if you have the same initial current - which in an R-C circuit is governed by the resistor. In essence at time \$t=0\$, if there is no charge on the capacitor, all of the voltage is across the resistor, so this sets the maximum current. As the capacitor charges, the voltage across the resistor will decrease (it increases over the capacitance), so current decreases (Ohms law).
Given that current is the rate of change of charge, then with the same limitation on current, it will take longer to charge the larger capacitor to a given voltage.
Best Answer
Oh, the formula is easy: it's $$C = Q/V$$ and you find V by solving for the E field with a test charge +Q on one plate, -Q on the other, and integrating (line integral) $$V = \int \overline E \cdot d\overline L $$ from any convenient point on one electrode, to the other.
The charge isn't equally distributed over the area of the plates, its location IS on the outer surface, with E inside the outer surface set to zero.
It gets more complicated because of the dielectric and/or nearby metal structures, and the geometry is not self-shielding, so technically you need to know all the surroundings out to infinity to complete the E-field calculation.
The electrode shapes are boundaries, with boundary conditions that amount to the E-field component parallel to the surface vanishing at zero distance from the surface. As boundaries go, these are not simple shapes. The field solution, therefore, is not going to be simple.
If the problem can be made two-dimensional (i.e. very long device, with constant cross-section), there are conformal mapping solutions that can create symmetry and make the on-paper problem tractable. Otherwise, the use of finite-element modeling would give good results, but not a general formula. Practical capacitors have uniform field in the dielectric, and negligible field outside the dielectric, so approximate solutions are very simple. This, however, is not a practical capacitor in that sense.