A capacitor of capacitance $C =900\,pF$ is charged fully by $100\,V$ battery $B$ as shown in figure $(a)$. Then it 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 $\dots\dots\times 10^{-6}\,J$
NEET 2022, Medium
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Common potential
$V _{ C } =\frac{ C _{1} V _{1}+ C _{2} V _{2}}{ C _{1}+ C _{2}}$
$=\frac{ C \times 100+ C \times 0}{ C + C }$
$=50\,Volt$
Electrostatic energy stored
$=2 \times \frac{1}{2} CV^{2}=CV^{2}$
$=900 \times 10^{-12} \times 50 \times 50$
$=225 \times 10^{-8}\,J$
$=2.25 \times 10^{-6}\,J$
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