Final answer:
The potential difference across the 10µF capacitor is 133.4V, and across the 20µF capacitor is 66.7V when they are reconnected with their positive plates together and negative plates together after being charged in series from a 200V supply.
Step-by-step explanation:
When a 10µF capacitor and a 20µF capacitor are connected in series across a 200 V supply, the same charge accumulates on both because in series connection, the charge is the same. The total voltage is distributed across the capacitors in inverse proportion to their capacitances.
First, we calculate the equivalent capacitance (Ceq) for the series connection:
Ceq = (C1 * C2) / (C1 + C2) = (10µF * 20µF) / (10µF + 20µF) = 200 / 30 = 6.67µF
The charge (Q) on the series capacitors is found by Q = Ceq * V = 6.67µF * 200V = 1334µC
When disconnected and reconnected with like terminals together, they share the same charge. The voltage across each capacitor (V) is now different because of their capacitances and is given by V = Q / C:
For 10µF capacitor: V = Q / C = 1334µC / 10µF = 133.4V
For 20µF capacitor: V = Q / C = 1334µC / 20µF = 66.7V
Thus, the potential difference across the 10µF capacitor is 133.4V and across the 20µF capacitor is 66.7V after they are reconnected with their positive plates together and negative plates together with no external voltage applied.