A 600,pF capacitor is charged by 200,V supply | vedclass

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Question

$Q = CV =600 	imes 10^{-12} 	imes 200=12 	imes 10^{-8} C$

$	ext { Initial energy }=\frac{1}{2} CV ^2$

$=\frac{1}{2} 	imes 600 	imes 10^{-1

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$Q = CV =600 	imes 10^{-12} 	imes 200=12 	imes 10^{-8} C$

$	ext { Initial energy }=\frac{1}{2} CV ^2$

$=\frac{1}{2} 	imes 600 	imes 10^{-12} 	imes(200)^2=12\,\mu J$

When connected to another uncharged capacitor

Charge will be equally distributed on identical capacitor

$Q ^{\prime}=\frac{ Q }{2}=6 	imes 10^{-8}$

Final energy $=2 	imes \frac{ Q ^{\prime 2}}{2 C }=\frac{ Q ^{\prime 2}}{ C }$

$\frac{\left(6 	imes 10^{-8}ight)^2}{600 	imes 10^{-12}}=6\,\mu J$

Energy lost = Initial energy - Final energy

$=(12-6)\,\mu\, J =6\, \mu\, J$

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 $.........\,\mu J$.

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