 Solution

The highest-field plateau is labelled $R_H/(h/e^2)=1$, so it is the $\nu =1$ plateau in the filling factor, with the longitudinal resistivity returning to approximately zero after the transition near 10 T. The odd integer indicates a resolved spin branch at these fields; assuming a permanent twofold spin degeneracy for every orbital Landau level would misidentify the plotted sequence.

For a nominal exactly filled $\nu =1$ state at a chosen field $B_{\ast }$, the density target is $n_{\ast }=e{B}_{\ast }/h$. Combining this with the same gate capacitor model gives the required gate-voltage change relative to the original ungated-density reference:Taking $B_{\ast }\simeq 10\mathrm{T}$ as the graph's onset-scale estimate gives $n_{\ast }=2.418\times 10^{15}{\mathrm{m}}^{-2}$ andRelative to the positive 0.2 V setting in part (v), this is an additional change of approximately $-0.281\mathrm{V}$. An absolute laboratory gate voltage cannot be obtained without its initial offset or threshold; the calculated quantities are voltage changes.

There is a graphical/model qualification: a quantum Hall plateau occupies a finite field interval, and its onset is not an exact statement that $n_s=eB/h$ at that edge. The plot identifies $\nu =1$ from about 10 T onward but does not specify a unique plateau-centre field or the gate setting at which the trace was recorded. The boxed formula gives the result for any selected target field. For comparison, the nominal exact-filling fields are 12.39 T at the original density and 18.34 T at the unchanged positive 0.2 V density. Thus the graph must not be combined silently with the latter density as if its $\nu =1$ onset uniquely fixed the same gate bias.