What's the dual of Faraday's law of induction?
The dual of Faraday's Law is Ampere's Law but, while Faraday's Law is fundamental to the physics of an inductor, Ampere's Law is not fundamental to the physics of a capacitor.

Now, it is true that, in circuit theory, the capacitor and inductor are duals:

$$i_C = C\frac{dv_C}{dt} \leftrightarrow v_L = L \frac{di_L}{dt}$$

However, we have to be more careful outside the context of circuit theory.

In physics, the fundamental relationship

$$Q = CV$$

clearly requires the existence of electric charge and an electric scalar potential due to a conservative electric field.  This equation relates electric charge and electric scalar potential.

The closest we can get to a dual of this is

$$\Phi = LI $$

which relates magnetic flux and electric current.  But magnetic flux is not the dual of electric charge.

The missing ingredient here is the hypothetical magnetic charge (magnetic monopole) which is the dual of electric charge.*  Were magnetic charge \$Q_m\$ (measured in webers) to exist, it would be a source or sink of a conservative magnetic field (measured in amperes per meter) and there would be an associated scalar magnetic potential (measured in amperes).

We could thus relate magnetic charge and magnetic scalar potential with a magnetic "capacitance" measured in henrys.

Further, we could relate electric flux to magnetic current (measured in volts) with an electric "inductance" measured in farads.

To summarize, while electric flux and magnetic flux are duals, and changing magnetic flux is fundamental to the physics of an inductor, changing electric flux is not fundamental to the physics of a capacitor.  Indeed, it is the electric field itself, not the electric flux, that is fundamental.



*Assuming magnetic charge exists, Maxwell's equations become

$$\nabla \cdot \vec D = \rho_e$$

$$\nabla \cdot \vec B = \rho_m$$

$$\nabla \times \vec E = - (\vec J_m + \frac{\partial \vec B}{\partial t})$$

$$\nabla \times \vec H = \vec J_e + \frac{\partial \vec D}{\partial t}$$