What is the "positive charge" on a capacitor if only electrons and not protons flow through a circuit?

In metals, where the only mobile charge carriers are electrons (other materials use other charge carriers), the positive charge is the excess of nuclear protons, whose charge is now no longer being neutralised by the electrons in the shell.

To put some numbers on this, consider a capacitor whose plates are 1 square metre, and with a separation of 10 um. The standard capacitance formula of C=Ae/d gives a capacitance of about 1 uF, assuming a vacuum dielectric.

Let's say the metalisation is aluminium, with a unit cell dimension of 0.4 nm. If we count the number of atoms on the surface layer of the electrodes, it comes to 1/(0.4 n)² = 6.3e18. There are many more atoms in the film, but let's only consider the surface monolayer. As each atom has 13 electrons and 13 neutrons, that's a total charge of (surprisingly, I picked 1 m² and Al first) 13 Coulombs each of proton and electron charge. However, only the outermost electron is freely available to move in the conduction band, so that's 1 C of mobile charge.

Let's charge the capacitor to 1 kV. As Q = CV, that's 1 mC. So 0.1% of the mobile electronic charge has left the positive plate, which stops it neutralising 1 mC of the total 13 C nuclear charge. Meanwhile, 1 mC of electrons have arrived on the negative plate, leaving the capacitor overall neutral.