$V=\frac{1}{C} Q$
Straight line with slope $=\frac{1}{C}$
Slope $=\frac{1}{C}=\frac{1}{2 \times 10^{-6}}=5 \times 10^{5}$
$(A)$ $I =\frac{V_0 t}{\pi \rho} \ln \left(\frac{R_2}{R_1}\right)$
$(B)$ the outer surface is at a higher voltage than the inner surface
$(C)$ the outer surface is at a lower voltage than the inner surface
$(D)$ $\Delta V \propto I ^2$
Reason: An electron has a negative charge.