Final answer:
The new charge on the second capacitor, when a 35pF capacitor charged to 5kV is connected in parallel with an uncharged 55pF capacitor, is 275nC.
Step-by-step explanation:
The question involves a physics concept called charge conservation in capacitors. When two capacitors are connected in parallel, the total charge is conserved. To find the new charge on the second capacitor when a 35pF capacitor charged to 5kV is connected in parallel with an uncharged 55pF capacitor, we use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Initially, the charge on the 35pF capacitor is Q = C1 V1 = 35pF * 5000V = 175µC. After the connection, this charge redistributes over the total capacitance (Ctotal = C1 + C2).
The voltage across both capacitors after connection is Vnew = Q / Ctotal. Therefore, the charge on the second capacitor (55pF) at this new voltage is Q2 = C2 Vnew. Here, Vnew is the same as before since they are in parallel, so the new charge Q2 would be 55pF * 5000V = 275µC. To convert this into nanocoulombs (nC), we multiply by 1000, getting Q2 = 275nC.