Electronic – the equation for capacitance with aysmmetric plates

Intuitive approach: if the distance wouldn't be a factor then you would be able to place the plates at an infinite distance apart and still have the same capacitance. That doesn't make sense. You would expect a zero capacitance then.
If the capacitor is charged to a certain voltage the two plates hold charge carriers of opposite charge. Opposite charges attract each other, creating an electric field,

and the attraction is stronger the closer they are. If the distance becomes too large the charges don't feel each other's presence anymore; the electric field is too weak.

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.