A capacitor of C1 = 10 uF is connected to a battery, and a charge of 100 uC builds up on the capacitor. Then, the battery is disconnected, the capacitor is discharged, and a second capacitor of capacity C2 = 20 uF is connected to it in parallel. Then, the two capacitors are connected to the same battery. What is the charge on the plates (in uC) of the second capacitor at that time?

Can you walk me through this? I need to understand :)

Question

A capacitor of C1 = 10 uF is connected to a battery, and a charge of 100 uC builds up on the capacitor. Then, the battery is disconnected, the capacitor is discharged, and a second capacitor of capacity C2 = 20 uF is connected to it in parallel. Then, the two capacitors are connected to the same battery. What is the charge on the plates (in uC) of the second capacitor at that time?

Can you walk me through this? I need to understand :)

radar · Answer

The formula you need is: C=Q/V  Solve for V
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