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Q.

A capacitor of capacitance C=900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C=900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is:
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a

3.25×10−6J

b

4.5×10−6J

c

2.25×10−6J

d

1.5×10−6J

answer is B.

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Detailed Solution

after joining their common potential will be  c1v1+c20c1+c2   

9×10018=50 V 

and combine capacitance=2c=18pF 

energy =1218×10-10502 

=9×25×10-8 =2.25×10-6 J

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