Motional emf across $PQ$

$V=Blv=4\left(1\right) \left(2\right)= 8 \mathrm{V}$

This is the potential to which the capacitor is charged.

As $q=CV$

$\therefore q=\left(10\times {10}^{-6}\right)8={10}^{-5}C=80 \mathrm{\mu C}$

As the magnetic force on an electron in the conducting rod $PQ$ is towards $Q,$ therefore, $\mathrm{A}$ is positively charged and $B$ is negatively charged

$ie, q_A=+80 \mathrm{\mu C} \mathrm{a}\mathrm{n}\mathrm{d} q_B=-80\mathrm{ }\mathrm{\mu C}\mathrm{ }$