Final answer:
After a 40 pF capacitor charged to 5 kV is connected in parallel to an uncharged 55 pF capacitor, the charge on the second capacitor becomes 110 µC.
Step-by-step explanation:
The initial charge on the first capacitor (40 pF) can be calculated by the formula Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage. So for the first capacitor charged to 5 kV, the charge will be Q = 40 pF × 5000 V = 200 µC. When this charged capacitor is connected in parallel with the second uncharged capacitor (55 pF), the total charge will be conserved as no charge is lost in the process.
Upon connection, since both capacitors are in parallel, the voltage across both becomes the same and the charge is distributed according to their capacitances. The total charge Q (200 µC) is shared, resulting in the charge on the 55 pF capacitor (Q2) being Q2 = (C2 / (C1 + C2)) × Q = (55 pF / (40 pF + 55 pF)) × 200 µC = 110 µC. Therefore, the new charge on the second capacitor is 110 µC.