Question

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

• A. zero
• B. $ 4\times 10^{-6}J $
• C. $ 6\times 10^{-6}J $
• D. $ 8\times 10^{-6}J $

Answer 1

Answer: (C)

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