A capacitor of capacitance $\mathrm{C}$ and potential $\mathrm{V}$ has energy $\mathrm{E}$. It is connected to another capacitor of capacitance $2 \mathrm{C}$ and potential $2 \mathrm{~V}$. Then the loss of energy is $\frac{x}{3} \mathrm{E}$, where $x$ is _______.
Answer (integer)
2
Solution
<p>$$\begin{aligned}
& \text { Energy loss }=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2 \\
& =\frac{2}{3} \cdot E \\
& \therefore x=2
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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