Question

A 27-pF capacitor is charged to 5.0 kV and then removed from the battery and connected...

A 27-pF capacitor is charged to 5.0 kV and then removed from the battery and connected to an uncharged 70-pF capacitor.

(a) What is the new charge on each capacitor?

Q27 = nC

Q70 = nC

(b) Find the energy stored in the 27-pF capacitor before it is disconnected from the battery, and the energy stored in the capacitors after they are connected to each other.

Ui = µJ

Uf = µJ

c.Does the stored energy increase or decrease when the two capacitors are connected to each other?

Potential energy is lost.

Potential energy is gained.

Potential energy is neither gained nor lost.

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