Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is

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(3) 200 pF. Initial charge on the first capacitor: Q=C₁V₀=100×60=6000 pC, and this total is conserved because the battery has been removed. In parallel the two capacitors share a common final voltage, and we are told it is 20 V. Charge left on the first: q₁=100×20=2000 pC, so the charge that moved to the second is q₂=6000-2000=4000 pC. C₂=(q₂)/V=(4000)/(20)=200 pF.

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is

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