A 19-pF capacitor is charged to 6.0 kV and then removed from the battery and con

Question

A 19-pF capacitor is charged to 6.0 kV and then removed from the battery and connected to an uncharged 74-pF capacitor. (a) What is the new charge on each capacitor? Q19 = Q74 . nC nC (b) Find the energy stored in the 19-pF capacitor before it is disconnected from the battery, and the energy stored in the capacitors after they are connected to each other Uy= Does the stored energy increase or decrease when the two capacitors are connected to each other? Potential energy is gained. Potential energy is lost. Potential energy is neither gained nor lost.

Explanation / Answer

charge on capacitor,

Q = C V = (19 x 10^-12 F) (6 x 10^3 V) = 114 x 10^-9 C

Or Q = 114 nC  

now battery is disconnected so this charge will remain constant in circuit.

as two capacitors are connected in parallel so voltage drop cross them will be same say V.

Applying charge conservation,

114 nC = (74 + 19 ) V

V = 1.225 kV  

Q19 = 19 V = 23.3 nC  

Q74 = 90.7 nC  

(b) Ui = CV^2 /2 = (19 x 10^-15) (6 x 10^3)^2 / 2

Ui = 0.342 x 10^-6 J  

Or 0.342 uJ

Uf = (19 x 10^-15) (1.225 x10^3)^2 /2 = 0.014 x 10^-6 J  

Or 0.014 uJ  

-> PE is lost.

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