Solution

During a short time $\Delta t$, the additional charge of $I\Delta t$ is delivered to the capacitor. Since the energy stored in the capacitor is $QV/2$, where $Q$ is the charge on the capacitor and $V=q/C$ is the constant voltage drop across the capacitor (and the generator), the change in the energy stored in the capacitor is $I\Delta tV/2$. Thus, the power absorbed by the capacitor is $IV/2=Iq/(2C)$, which is half of the power $IV=Iq/C$ consumed from the generator. The difference arises from the fact that we need to do negative work to bring the plates of the capacitor (which have a tendency to attract each other) closer to increase its capacitance while keeping the voltage drop across it constant.