A parallel plate capacitor of capacitance $500 \mathrm{pF}$ charged with $100 \mathrm{~V}$ supply: It is…
A parallel plate capacitor of capacitance $500 \mathrm{pF}$ charged with $100 \mathrm{~V}$ supply: It is then disconnected from the supply and connected to another uncharged $500 \mathrm{pF}$ capacitor. The electrostatic energy lost in this process is
$1.25 \mu \mathrm{J}$
$0.175 \mu \mathrm{J}$
$0.225 \mu \mathrm{J}$
$0.275 \mu \mathrm{J}$
Solution
Initial stored energy = Energy of capacitor 1
$=\frac{1}{2} C_1 V_1^2$
$\begin{aligned} & =\frac{1}{2} \times 500 \times 10^{-12} \times(100)^2\left(\because C_1=500 \mathrm{pF} V_1=100 \mathrm{~V}\right) \\ & =250 \times 10^{-12} \times 10^4=250 \times 10^{-2} \times 10^{-6} \\ & =2.5 \mu \mathrm{J}\end{aligned}$
When both capacitors are joined, common potential of combination is
$V_{\text {common }}=V=\frac{Q_1+Q_2}{C_1+C_2}$ or $V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$
or $\quad V=\frac{500 \times 10^{-12} \times 100}{2 \times 500 \times 10^{-12}}=50 \mathrm{~V}$
Final stored energy $=\frac{1}{2}\left(C_1+C_2\right) V^2$
$\begin{aligned} & =\frac{1}{2} \times 2 \times 500 \times 10^{-12} \times(50)^2 \\ & =500 \times 10^{-12} \times 2500 \\ & =125 \times 10^{-8} \\ & =1.25 \times 10^{-6} \mathrm{~J} \\ & =1.25 \mu \mathrm{J}\end{aligned}$
Loss of energy, $\Delta U=$ Final energy - Initial energy
$=25-1.25 \mu \mathrm{j}=\mathrm{I} .25 \mu \mathrm{j}$