 Solution

Treat the metal gate and conducting electron sheet as a parallel-plate capacitor separated by a uniform GaAs-like dielectric. Its capacitance per area isA gate-voltage change gives an electron-number change of magnitudeFor the ordinary accumulation convention, a more positive gate voltage attracts more electrons, while a more negative gate depletes them. The magnitude is about $13\%$ of the preceding sheet density.

This neglects depletion, trapped charge, finite well width and quantum capacitance. The occupied-subband quantum capacitance is $C_q/A=e^2{D}_{2D}\simeq 0.0449\mathrm{F}{\mathrm{m}}^{-2}$, much larger than $C_g/A$ here; putting the two in series would reduce the estimate by only about $1.3\%$ within this simple model.