A 4.00*10^3 -nF parallel plate capacitor is connected to a 12.0-v battery and ch

Question

A 4.00*10^3 -nF parallel plate capacitor is connected to a 12.0-v battery and charged.

a) what is the charge Q on the positive plate of the capacitor?
b)what is the electric potential energy stored in the capacitor?

the 4.00*10^3-nF capacitor is then disconnected from the 12.0 v battery and used to charge two uncharged capacitors, a 200.nF capacitor, and a 200.nF capacitor, connected in series
c) after charging, what is the potential difference across each of the 4 capacitors?
d) how much of the electrical energy stored in the 4.00×10^3-nF capacitor was transferred to the other 3 capacitors?

Explanation / Answer

charge on capacitor=C*V

where C=capacitance
V=voltage difference

a)C=4000 nF
V=12
charge=Q=48000 nC=48 uC

b)electric potential energy=0.5*C*V^2=0.288 mJ

c)now as it is discharged, and connectred to two 200 nF capacitors,

the net charge in the combination will remain constant

as they are connected in series, the charge on each of the capacitors will be same also.

hence charge on each capacitor=48/3=16 uC

so potential difference across 4000 nF =charge/capacitance=4 volts

potential difference across 200 nF=80 volts

d) after the connection in series, the energy left in 4000 nF capacitor=charge^2/(2*capcitance)=3.2*10^(-5) J

so energy transferred=2.56*10^(-4) J

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