Question

A capacitor of capacitance $\text{C}=900\text{pF}$ is charges fully by $100\text{V}$ battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance $\text{C}=900\text{pF}$ as shown in figure (b). The electrostatic energy stored by the system (b) is

Answer 1

Initial stored energy of capacitor is:
$U_{\text{initial}}=\frac{1}{2}C{V}^2=\frac{1}{2}C\times (100{)}^2=5000C$
Energy lost is given as:
$\text{Lost}\Delta U=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{(V_1-V_2)}^2=\frac{1}{2}\frac{C^2}{2C}(100-0{)}^2$

$=\frac{1}{4}C(100{)}^2=2500C$

$\therefore U_{\text{final}}=U_i-\Delta U=5000C-2500C=2500C$

$=2500\times 900\times {10}^{-12}$

$=225\times {10}^{-8}$

$U_{\text{final}}=2.25\times {10}^{-6}J$
Hence the correct option is $\text{(}a)$