A 2 µF capacitor charged initially by a 30V source is disconnected from the source and then connected?
A 2 µF capacitor charged initially by a 30V source is disconnected from the source and then connected to an uncharged 1µF capacitor.
What is the final potential difference across the capacitors?
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I agree with the 20 V in your first answer but the explanation had an error.
It says "electrons will flow from the 2 µF capacitor to the 1 µF capacitor, until the charge is the same on both capacitors."
Sould have been "electrons will flow from the 2 µF capacitor to the 1 µF capacitor, until the VOLTAGE is the same on both capacitors."
Correction in the 1st sentence, thanks to sojsail !
When the 2 µF capacitor is attached to the 1 µF capacitor, electrons will flow from the 2 µF capacitor to the 1 µF capacitor, until the “voltage” is the same on both capacitors.
Determine the initial amount of charge on the 2 µF capacitor.
Capacitance = Charge / Voltage
Charge = Capacitance * Voltage = 2 * 10^-6 * 30 = 6 * 10^-5
The 2 µF capacitor can hold twice as much charge as the 1 µF capacitor. So â of the charge will be on the 2 µF capacitor and â of the charge will be on the 1 µF capacitor.
Now determine the voltage.
V = Q / C
For 2 µF capacitor, Q = 4 * 10^-5
V = 4 * 10^-5 / 2 * 10^-6 = 20 volts
For 1 µF capacitor, Q = 2 * 10^-5
V = 2 * 10^-5 / 1 * 10^-6 = 20 volts
The final potential difference across the capacitors = 20 volts