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Question:
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A $4 \mu \mathrm{F}$ capacitor is charged by a $200 \mathrm{~V}$ supply. It is then disconnected from the supply, and is connected to another uncharged $2 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Solution:
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Verified Answer
Given, $C_1=4 \mu \mathrm{F}=4 \times 10^{-6} \mathrm{~F}$,
$$
\begin{gathered}
\mathrm{C}_2=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F} \\
\mathrm{~V}_1=200 \mathrm{~V}, \mathrm{~V}_2=0
\end{gathered}
$$
Loss of energy, $\Delta \mathrm{E}=$ ?
By formula,
$$
\begin{aligned}
\Delta \mathrm{E} &=\frac{\mathrm{C}_1 \mathrm{C}_2\left(\mathrm{~V}_1-\mathrm{V}_2\right)^2}{2\left(\mathrm{C}_1+\mathrm{C}_2\right)} \\
&=\frac{4 \times 10^{-6} \times 2 \times 10^{-6}(200-0)^2}{2(4+2) \times 10^{-6}}=\frac{8}{3} \times 10^{-2} \mathrm{~J}
\end{aligned}
$$
Hence, energy converted into heat and electromagnetic radiation $=2.67 \times 10^{-2} \mathrm{~J}$.
$$
\begin{gathered}
\mathrm{C}_2=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F} \\
\mathrm{~V}_1=200 \mathrm{~V}, \mathrm{~V}_2=0
\end{gathered}
$$
Loss of energy, $\Delta \mathrm{E}=$ ?
By formula,
$$
\begin{aligned}
\Delta \mathrm{E} &=\frac{\mathrm{C}_1 \mathrm{C}_2\left(\mathrm{~V}_1-\mathrm{V}_2\right)^2}{2\left(\mathrm{C}_1+\mathrm{C}_2\right)} \\
&=\frac{4 \times 10^{-6} \times 2 \times 10^{-6}(200-0)^2}{2(4+2) \times 10^{-6}}=\frac{8}{3} \times 10^{-2} \mathrm{~J}
\end{aligned}
$$
Hence, energy converted into heat and electromagnetic radiation $=2.67 \times 10^{-2} \mathrm{~J}$.
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