A 12pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ? If another capacitor of 6pF is connected across the combination, find the charge stored and potential difference across each capacitor.
Here, C1=12pF=12×10−12F.
V = 50 volt.
Electrostaic energy stored U=12C1V2
=12×12×10−12(50)2
U=1.5×10−8J
Now, C2=6pF=6×10−12F
1Cs=1C1+1C2=112+16=312=14
Cs=4pF=4×10−12F
Charge on each condenser,
q1=q2=CsV=4×10−12×50
=2×10−10C
V1=q1C1=2×10−1012×10−12=503 volt
V2=q2C2=2×10−106×10−12=1003 volt
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