- #1
binbagsss
- 1,291
- 12
A capacitor is connected to a cell of emf 200V and fully charged.
Another capacitor is then connected in parallel, which has 1/100th of the capacitance of the charged capacitor.
The total charge is instantly shared between the two capacitors, and so the capacitor connected in parallel to the original capacitor will get 1/100th of the charge as components in parallel have the same voltage and so Q is proportional to C.
However, what I don't fully understand is why the charge is instantly shared, and, eventually will both capacitors obtain a voltage equal to the emf of the cell?
- Is it because intially the emf is not able to do any work, because the capacitor which is fully charged has an equal voltage across it, acting in the opposite direction. However charge is able to flow from the charged capacitor to the capacitor in parallel as this has no pd across it, and so the charge will be distributed until both capacitors reach the same voltage and no current flows.
- But, as the charge flows from the charged capacitor to the uncharged capacitor, the p.d across the charged capacitor decreases, and so the cell has the greater voltage again and is able to do work, so eventually will both capacitors reach 200v?
Another capacitor is then connected in parallel, which has 1/100th of the capacitance of the charged capacitor.
The total charge is instantly shared between the two capacitors, and so the capacitor connected in parallel to the original capacitor will get 1/100th of the charge as components in parallel have the same voltage and so Q is proportional to C.
However, what I don't fully understand is why the charge is instantly shared, and, eventually will both capacitors obtain a voltage equal to the emf of the cell?
- Is it because intially the emf is not able to do any work, because the capacitor which is fully charged has an equal voltage across it, acting in the opposite direction. However charge is able to flow from the charged capacitor to the capacitor in parallel as this has no pd across it, and so the charge will be distributed until both capacitors reach the same voltage and no current flows.
- But, as the charge flows from the charged capacitor to the uncharged capacitor, the p.d across the charged capacitor decreases, and so the cell has the greater voltage again and is able to do work, so eventually will both capacitors reach 200v?
