 Solution

Treat the top gate as a parallel-plate capacitor, neglecting fringing and assuming its voltage change couples to the 2DEG without a dominant trap or series-capacitance correction. Its capacitance per area and induced electron density areIn this top-gate density tuning, a positive gate change increases the electron density in this model, giving $n_s^{\prime }=4.43371\times 10^{15}{\mathrm{m}}^{-2}$. Since $B_j\propto k_F\propto \sqrt{n_s}$,The fields rise by a factor $\sqrt{n_s^{\prime }/n_s}=1.2163$, not by the density ratio itself. Finite quantum capacitance is a small correction here: the one-subband value $e^2{m}^{\ast }/(\pi {\hslash }^2)$ is much larger than the stated geometric capacitance per area.