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Physics Question on electrostatic potential and capacitance

A 600 pF600 \,pF capacitor is charged by 200 V200 \,V supply. It is then disconnected from the supply and is connected to another 300 pF300 \,pF capacitor. The electrostatic energy lost in the process is

A

zero

B

4×10−6J4\times 10^{-6}J

C

6×10−6J6\times 10^{-6}J

D

8×10−6J8\times 10^{-6}J

Answer

6×10−6J6\times 10^{-6}J

Explanation

Solution

Energy stored in capacitor E1=12CV2E_{1}=\frac{1}{2} C V^{2}
=12×600×10−12×(200)2=\frac{1}{2} \times 600 \times 10^{-12} \times(200)^{2}
=12×10−6 J=12 \times 10^{-6} \,J
Again E2=12(600+300)×10−12×(200)2E_{2}=\frac{1}{2}(600+300) \times 10^{-12} \times(200)^{2}
=12×900×10−12×4×104=\frac{1}{2} \times 900 \times 10^{-12} \times 4 \times 10^{4}
=18×106 J=18 \times 10^{6} \,J
Energy lost =E1=−E2=E_{1}=-E_{2}
=18×10−6−12×10−6=18 \times 10^{-6}-12 \times 10^{-6}
=6×10−6 J=6 \times 10^{-6} \,J