A capacitor with a capacitance of 100 is charged to a potential...

The initial charge on the capacitor is given by $Q_{i} = C V_{1}$.

Substituting the given values, $Q_{i} = 100 \mu\text{F} \times 24 \text{ V} = 2400 \mu\text{C}$.

When the capacitor is connected to a new battery of emf $12 \text{ V}$, the potential difference across the capacitor in steady state becomes equal to the emf of the new battery.

The final charge on the capacitor is given by $Q_{f} = C V_{2}$.

Substituting the values, $Q_{f} = 100 \mu\text{F} \times 12 \text{ V} = 1200 \mu\text{C}$.

The charges before and after reconnection are $2400 \mu\text{C}$ and $1200 \mu\text{C}$ respectively.