A $600\,pF$ capacitor is charged by $200\,V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\,pF$ capacitor. Electrostatic energy lost in the process is $.........\,\mu J$.
JEE MAIN 2023, Diffcult
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$Q = CV =600 \times 10^{-12} \times 200=12 \times 10^{-8} C$
$\text { Initial energy }=\frac{1}{2} CV ^2$
$=\frac{1}{2} \times 600 \times 10^{-12} \times(200)^2=12\,\mu J$
When connected to another uncharged capacitor
Charge will be equally distributed on identical capacitor
$Q ^{\prime}=\frac{ Q }{2}=6 \times 10^{-8}$
Final energy $=2 \times \frac{ Q ^{\prime 2}}{2 C }=\frac{ Q ^{\prime 2}}{ C }$
$\frac{\left(6 \times 10^{-8}\right)^2}{600 \times 10^{-12}}=6\,\mu J$
Energy lost = Initial energy - Final energy
$=(12-6)\,\mu\, J =6\, \mu\, J$
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