Charging two capacitors then putting them in parallel

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SUMMARY

In the discussion, capacitors C_1 (13 µF) and C_2 (26 µF) are charged to 20 V and then connected in parallel with opposite polarities. The initial charges on the capacitors are calculated as Q1 = 87 µC and Q2 = 170 µC. After connecting them in parallel, the equivalent capacitance is determined to be 39 µF. The potential difference across each capacitor after connection is a key point of inquiry.

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Linus Pauling
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1. Capacitors C_1 = 13 uF and C_2 = 26 uF are each charged to 20 V, then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C_1 connected to the negative plate of C_2 and vice versa.

Afterward, what is the charge on each capacitor?




2. Q = VC, equivalent C for parallel capacitors add linearly



3. I know Q1 and Q2 are 87 and 170 uC, respectively. I know that the equivalent capacitance is simple 39 uF. But I still don't see how these charges were obtained.
 
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Linus Pauling said:
The two capacitors are then connected in parallel, with the positive plate of C_1 connected to the negative plate of C_2 and vice versa.

If they are connected in parallel, what is the potential difference across each capacitor?
 

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