Final answer:
The potential difference across capacitor C2, which is connected in parallel with capacitor C3, is 8.0 volts, resulting in a charge of 16.0 µC on capacitor C2.
Step-by-step explanation:
The potential difference, in volts, across capacitor C2 would be equivalent to the potential difference across any capacitors connected in parallel to it. In the provided scenario, since capacitors C2 and C3 are connected in parallel, and it's given that they experience a potential difference of 8.0 V, we can conclude that the potential difference across capacitor C2 is also 8.0 V. This is because in a parallel circuit, the voltage across each component is the same, hence V2 is equal to 8.0 volts. The charge on capacitor C2 can be calculated using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference, which yields Q2 = (2.0 µF)(8.0 V) = 16.0 µC.