- #1
The Blind Watchmaker said:
They are in series thus Q1=Q2. I do not get why they have the same potential difference as I assume only 1 battery is usedgneill said:
There are two issues in your understanding. When we connect two capacitors in series using only one battery q for each is taken to be the same. Just make such a connection with two different capacitors and then argue why we take charge stored by these capacitors to be the same. But in this problem they have been charged separately to the same potential difference.The Blind Watchmaker said:
Read the problem statement carefully as to how the capacitors are prepared *before* they are connected to each other.The Blind Watchmaker said:
To calculate the final potential difference, you need to add the charges of the capacitors together and divide by the total capacitance. This will give you the final voltage across the capacitors.
The formula for calculating the final potential difference in a series circuit is Vf = Q/C, where Vf is the final potential difference, Q is the total charge of the capacitors, and C is the total capacitance.
Yes, charged capacitors can be connected together in parallel. When connected in parallel, the final potential difference will be the same as the potential difference of the individual capacitors.
When charged capacitors are connected together in series, the potential difference across each capacitor is divided among them. This means that the final potential difference will be less than the potential difference of the individual capacitors.
The capacitance affects the final potential difference by determining how much charge can be stored on the capacitors. A higher capacitance will result in a larger final potential difference, while a lower capacitance will result in a smaller final potential difference.
