Physics olympiad question about a charge capacitor
This is problem 1463 from the Russian magazine Quant. Unfortunately my Russian is not good enough, so I apologize if the translation of the problem contains errors.
Problem. Inside a flat charge capacitor with distance between the plates $d$ a weakly conducting slab of thickness $h<d$ and resistivity $\rho$ is inserted. The area of the plates and the slab is $S$ (as always it is assumed $d\ll\sqrt{S}$). The capacitor is charged to potential $U_0$. Then the plates of the capacitor are short circuited. Find the maximal current across the slab.
Answer. $I_{max}=U_0 S/\rho h$
I obtain a completely different answer $I_{max}=U_0 S/\rho d$ and think that their solution is flawed.
Let's show that their solution doesn't make any sense. In the official solution they draw the picture
where $C=\varepsilon_0S/(d-h)$ and $R=\rho h/S$, claiming that it is equivalent to the initial system. Then they say that after the short circuit the capacitor $C$ (which initially has been charged to voltage $U_0$) will discharge through resistance $R$. 903
This means that it will take approximately time $RC$ for the capacitor to discharge.
Let's return to the initial system. We are free to assume that $\rho\to\infty$, $h\to 0$, so that $R$ is finite. Technically, this means that the slab is a very thin insulator placed inside the capacitor. So when the capacitor is short circuited, plate charges will disappear instantly due to current spike in the short circuit. This contradicts the official solution, where it takes time $RC$ to discharge the capacitor.
Q: So which of the answers is correct?
My solution is as follows:
The charges on the faces of the slab can not change abruptly, but the charges on the plates of the capacitor will change abruptly as a result of the short circuit. So the words "right after the short circuit" here means that we are considering the system when this spike of current across the wire connecting the plates has quickly adjusted the plate charges.
Let the charges on the plates and the faces of the slab (right after short circuit) be $q_1,Q,-Q,q_2$ (see the picture). Then we calculate initial charges on the surfaces of the slab $\pm Q$, where $$ Q=\frac{U_0\varepsilon_0 S}{d-h}. $$ Then calculate charges on the plates (right after short circuit) so that voltage drop between the plates is $0$: $$ \frac{q_1}{2\varepsilon_0 S}d+\frac{Q}{2\varepsilon_0 S}h-\frac{-Q}{2\varepsilon_0 S}h-\frac{q_2}{2\varepsilon_0 S}d=0, $$ and the electroneutrality is satisfied $$ q_1+q_2=0. $$ Thus $$ q_{1,2}=\mp Q\frac{h}{d}. $$ Then we calculate the electric field strength inside the slab, $$ E=\frac{q_1+Q+Q-q_2}{2\varepsilon_0 S}=\frac{U_0}{d}. $$ Thus current density is $j=E/\rho$ and the current $I=jS=\frac{U_0 S}{\rho d}$. 903