Final answer:
The charge on each capacitor after connecting a charged 25 pF capacitor to an uncharged 75 pF capacitor remains 600 pC. The total energy stored is reduced after redistribution of charge due to the nature of energy being proportional to the square of voltage, which decreases in the larger capacitor.
Step-by-step explanation:
When a capacitor of capacitance 25 pF is connected to a 24 V battery for a long time and then removed, the capacitor would have stored a charge given by Q = CV, where C is capacitance and V is the voltage. Therefore, the charge on the 25 pF capacitor is Q = 25 pF × 24 V = 600 pC (picoCoulombs).
When this charged capacitor is connected to an uncharged capacitor of capacitance 75 pF, charge will redistribute until the voltage across both capacitors is equal, because they are in parallel. Since the total charge is conserved, the charge on each capacitor (Q1 on the 25 pF and Q2 on the 75 pF) will be Q1 = Q2 = 600 pC.
The change in total energy stored before and after the battery was disconnected can be found by calculating the energy before redistribution, which is E = (1/2)CV^2, and the energy after, which would be the sum of energies in each capacitor, E_total = (1/2)C1V1^2 + (1/2)C2V2^2. There will be a loss in energy after redistribution due to the difference in capacitances.
The reason why the total energy is less than before is due to the redistribution of charge. When the capacitors achieve equal voltage, the larger capacitor will have a lower voltage than what was initially on the smaller capacitor, due to its bigger capacitance. As the energy stored in a capacitor is proportional to the square of the voltage, the total energy decreases.