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Physics Question on Electric Potential And Potential Difference

A capacitor of capacitance C=900 pF is charged rally by 100 V battery B as shown in figure (a). Then B 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:

A

4.5×10−6J4.5 \times 10^{-6} J

B

3.25×10−6J3.25 \times 10^{-6} J

C

2.25×10−6J2.25 \times 10^{-6} J

D

1.5×10−6J1.5 \times 10^{-6} J

Answer

2.25×10−6J2.25 \times 10^{-6} J

Explanation

Solution

The correct option is (C): 2.25×10−6J2.25 \times 10^{-6} J