Here is the question: A capacitor, C1 = 12.0 µF is charged to Vi = 15.0 V. It is
ID: 1658768 • Letter: H
Question
Here is the question: A capacitor, C1 = 12.0 µF is charged to Vi = 15.0 V. It is next connected in series with an uncharged capacitor, C2 = 7.00 µF. The series combination is finally connected across a battery, Vf = 52.0 V as diagrammed in the figure below. Find the new potential differences across the 7.00 µF and 12.0 µF capacitors.
I don't need the answer or the solution, I need an explanation! My understanding is that capacitors in series have the same charge, but these two capacitors in this series circuit have different charges. According to the solution, after the two capacitors are connected, C1 has the initial charge, equal to C1 times Vi, plus an additional charge q, which is the charge on C2. What is going on?
Explanation / Answer
Here Law of conservation of charge has been followed.
When C1 is connected to C2, it was intially charged and has a magnitude of Q1 = C1Vi this is stored in the capacitor and its now connected to the new uncharges capacitor.
This system will have an equivalent capacitance and the charge that the circuit has can be calculated by Q = CV.
Now, Its a fact that the same charge flows through the capacitors connected in series, Here also it will be followed but in accordance with the law of conservation of charge.