Final answer:
The charge on the first capacitor is 2.61nC and the charge on the second capacitor is 1.74nC.
Step-by-step explanation:
In this scenario, a 3.00pF capacitor is connected in series with a 2.00pF capacitor, and a 870V potential difference is applied across the pair. To find the charge on each capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.
For the 3.00pF capacitor: Q1 = (3.00pF)(870V) = 2.61nC
For the 2.00pF capacitor: Q2 = (2.00pF)(870V) = 1.74nC