4. A 4.0 103-nF parallel plate capacitor is contested to a 14.0 volt batter and

Question

4. A 4.0 103-nF parallel plate capacitor is contested to a 14.0 volt batter and charged a. what is the charge Q on the positive plate on the capacitor b. what is the electric potential energy stored in the capacitor
The 4.00x103-nF capacitor ha been disconnected 12 V better and used to charge two uncharged capacitors a 100.-nF capacitor. a 200.-nF capacitor connected in series
4. (15 points) A 4.00 103-nF parallel plate capacitor is connected to a 14.0-V battery and charged. a) What is the charge O on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The 4.00 103-nF capacitor is then disconnected from the 12.0-V battery and used to charge two uncharged capacitors, a 100.-nF capacitor, a 200-nF capacitor, connected in series. e) After charging, what is the potential difference across each of the three capacitors?

Explanation / Answer

a) charge = c*v = 4000*10-9*14 = 56*10-6 columb

b) energy = 1/2*c*v2 = 392*10-6 joules

c) all the three capacitor are in series so they will have equal charge = 1/3(12*4000*10-9) = 16*10-6

hence potential difference across

4000 capacitor = Q/C = 4 volt

across 100 = 160volt

across 200 = 80 volt

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