MHT CET · Physics · Alternating Current
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
- A \(\frac{\mathrm{Q}}{4}\)
- B \(\frac{\mathrm{Q}}{2}\)
- C \(\frac{\mathrm{Q}}{\sqrt{2}}\)
- D \(\frac{\mathrm{Q}}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{Q}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
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 \(\mathrm{E}_2=\frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}\).
\(
\begin{aligned}
\therefore & \frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^2}{2 \mathrm{C}} \\
& \mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}} .
\end{aligned}
\)
\(
\mathrm{E}_1=\frac{\mathrm{Q}^2}{2 \mathrm{C}}
\)
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 \(\mathrm{E}_2=\frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}\).
\(
\begin{aligned}
\therefore & \frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^2}{2 \mathrm{C}} \\
& \mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}} .
\end{aligned}
\)
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