The electrostatic energy of the capacitor e = Cu 2/2 = uq/2. After the electrolyte is pumped out, the capacitance c decreases. In the first case, because it is still connected to the power supply, the voltage u remains unchanged, the capacitance decreases, and e decreases. In the second case, due to the disconnection from the power supply, the power Q remains unchanged and E increases, because C=Q/U, C decreases and U increases. Hope to adopt!