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In an oscillating LC circuit, the maximum charge on the capacitor is 'Q'. 'When the energy is stored equally between the electric and magnetic fields, the charge on the capacitor becomes
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Verified Answer
The correct answer is:
$\frac{\mathrm{Q}}{\sqrt{2}}$
Maximum energy stored in a capacitor,
$$
\mathrm{E}_1=\frac{\mathrm{Q}^2}{2 \mathrm{C}}
$$
When energy is stored equally between the electric and magnetic fields, then energy in the capacitor is $\mathrm{E}_2=\frac{1}{2} \mathrm{E}_1$
If $\mathrm{Q}^{\prime}$ is the charge on the capacitor in this case, then $\mathrm{E}_2=\frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}$.
$$
\begin{aligned}
\therefore \quad & \frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^2}{2 \mathrm{C}} \\
& \mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}} .
\end{aligned}
$$
$$
\mathrm{E}_1=\frac{\mathrm{Q}^2}{2 \mathrm{C}}
$$
When energy is stored equally between the electric and magnetic fields, then energy in the capacitor is $\mathrm{E}_2=\frac{1}{2} \mathrm{E}_1$
If $\mathrm{Q}^{\prime}$ is the charge on the capacitor in this case, then $\mathrm{E}_2=\frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}$.
$$
\begin{aligned}
\therefore \quad & \frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^2}{2 \mathrm{C}} \\
& \mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}} .
\end{aligned}
$$
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