Calculate the resulting charge on each capacitor

[Calcguy]

Capacitors C1 = 6.35 µF and C2 = 2.00 µF are charged as a parallel combination across a 250 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.
q'1 = µC
q'2 = µC

Can someone please help me with this, I don't know what to do.

 

[Defennder]

First you have to find the charge on each capacitor after it has been charged. Then note from the statement of the problem whether the new connection yields a series or parallel connection.

From here, note that if the caps are in series, the resulting charge (if you let the circuit stabilise) on each capacitor will be the same. If they are in parallel, then their final potential differences will be the same. From these 2 conditions you can calculate the final charge on each cap.

 

[baba89]

To calculate the resulting charge on each capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

First, let's calculate the total capacitance of the parallel combination of C1 and C2. Since they are connected in parallel, the total capacitance (Cp) is given by Cp = C1 + C2 = 6.35 µF + 2.00 µF = 8.35 µF.

Next, we can calculate the charge on each capacitor when they are connected in parallel to the 250 V battery. Using the formula Q = CV, we get Q1 = (6.35 µF)(250 V) = 1587.5 µC and Q2 = (2.00 µF)(250 V) = 500 µC.

When the capacitors are disconnected from the battery and from each other, the charge on each capacitor remains the same. Therefore, the charge on C1 is still 1587.5 µC and the charge on C2 is still 500 µC.

Finally, when the capacitors are connected positive plate to negative plate and negative plate to positive plate, they essentially form a series combination. The total capacitance of a series combination (Cs) is given by Cs = 1/(1/C1 + 1/C2) = 1/(1/6.35 µF + 1/2.00 µF) = 1/(0.1575 µF + 0.5 µF) = 1/(0.6575 µF) = 1.521 µF.

Now, we can use the formula Q = CV to calculate the charge on each capacitor when they are connected in series. We get Q'1 = (1.521 µF)(250 V) = 380.25 µC and Q'2 = (1.521 µF)(250 V) = 380.25 µC.

Therefore, the resulting charge on each capacitor is Q'1 = 380.25 µC and Q'2 = 380.25 µC.