How does Ohm's law, I=V/R, differ from the Q=V/C law?
Note that C, like R, depends only on geometry, and on the
medium (epsilon, not rho). Since epsilon is likely to depend
of conditions, such as pressure or temperature I would not
be surprised by a "nonlinear capacitor." Does it exist?
To emphasize the analogy one can say that Q stands for the
flux of field lines linking two parallel plates. Field lines in
a capacitor versus the current density lines in a wire.
Ludwik Kowalski