Final answer:
The original charge on the 40-pF capacitor is 20 nC. Upon connecting to the 10-pF capacitor, charge redistributes according to the capacitances. Calculate individual charges using the total charge and the ratio of the capacitances after finding the common voltage.
Step-by-step explanation:
To answer the student's question, first, we need to calculate the original charge on the 40-pF capacitor. We can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference (voltage).
Solution for (a):
The original charge on the 40-pF capacitor is calculated by multiplying its capacitance by the potential difference (voltage). Therefore, with C = 40 pF (40 x 10-12 F) and V = 500 V, we get:
Q = CV = (40 x 10-12 F)(500 V) = 20 x 10-9 C = 20 nC
This is the original charge on the 40-pF capacitor before it is connected to the uncharged 10-pF capacitor.
Solution for (b):
When the 40-pF capacitor is connected to the uncharged 10-pF capacitor, charge will redistribute between the two capacitors until they reach the same potential. Since they are in parallel, the total capacitance Ctotal is the sum of the individual capacitances, which is 40 pF + 10 pF = 50 pF. The total charge remains the same, being 20 nC, and it is going to be distributed in accordance with the ratio of their capacitances.
The voltage across both capacitors will then be V = Q/Ctotal, and the charge on each capacitor can be found by Qcapacitor = Ccapacitor x V.