Capacitors C1 = 5.0 µF and C2 = 2.0 µF are charged as a parallel combination across a 150 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.
C1 acquires C1*V = 7.5*10^-4 C and
C2 acquires C2*V = 3.0*10^-4 C, initially. When rewired, there is a total of 4.5*10^-4 C on the positive-plate sides and -4.5*10^-4 C on the negative-plate sides. The charges redistribute to maintain the same potential on each capacitors.
(2/7)*4.5*10^-4 = 1.29*10^-4C will be on the smaller capacitor and
(5/7)*4.5*10^-4 = 3.21*10^-4 C will be on the larger capacitor.
To calculate the resulting charge on each capacitor when they are connected positive plate to negative plate and negative plate to positive plate, you can use the principle of conservation of charge. The total charge on the capacitors before and after the connection remains the same.
Before the connection, the charges on the capacitors can be calculated using the formula Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage.
For C1:
Q1 = C1 * V = (5.0 µF) * (150 V) = 750 µC
For C2:
Q2 = C2 * V = (2.0 µF) * (150 V) = 300 µC
Since the total charge is conserved, the resulting charges on the capacitors after the connection will also be the same.
Therefore, after the connection:
Q1 = Q2 = 750 µC.
To find the resulting charge on each capacitor, we can use the principle of charge conservation. According to this principle, the total charge before and after the capacitors are connected in series will remain the same.
1. Calculate the initial charge on each capacitor before they are connected in series:
Since the capacitors are connected in parallel across the battery, the charge on each capacitor will be the same as the charge on the combination.
Q1 = Q2 = Q (let's assume the charge on each capacitor is Q)
The charge on a capacitor can be calculated using the formula:
Q = C * V
where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.
For capacitor C1:
Q1 = C1 * V
= 5.0 µF * 150 V
= 750 µC (microcoulombs)
For capacitor C2:
Q2 = C2 * V
= 2.0 µF * 150 V
= 300 µC (microcoulombs)
2. When the capacitors are disconnected from the battery, the charge remains the same. Therefore, the charge on each capacitor will still be Q1 and Q2.
3. When the capacitors are connected positive plate to negative plate and negative plate to positive plate, they are connected in series. In a series connection, the charges add up.
Q_total = Q1 + Q2
Q_total = 750 µC + 300 µC
= 1050 µC (microcoulombs)
Now we need to find the charge on each capacitor when they are connected in series.
4. Since the series combination of capacitors has the same charge (Q_total) on each capacitor, we can say:
Q1 = Q_total
Q2 = Q_total
Therefore, the resulting charge on each capacitor will be:
Q1 = Q2 = Q_total = 1050 µC
So, the resulting charge on each capacitor when they are connected positive plate to negative plate and negative plate to positive plate is 1050 µC.