Q193 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
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 is q. The relation between Q and q is
- A$Q = q$
- Correct answerB$Q = \sqrt{2}\, q$
- C$Q = \sqrt{3}\, q$
- D$Q = 2q$
Solution
Total energy of the LC circuit is given by the maximum energy stored in the capacitor:
$U_{total} = \dfrac{Q^2}{2C}$
When the energy is stored equally between the electric and magnetic fields, the electric energy $U_E$ is half of the total energy:
$U_E = \dfrac{1}{2} U_{total}$
The electric energy at this instant is given by:
$U_E = \dfrac{q^2}{2C}$
Equating the expressions for $U_E$:
$\dfrac{q^2}{2C} = \dfrac{1}{2} \left( \dfrac{Q^2}{2C} \right)$
$q^2 = \dfrac{Q^2}{2}$
$Q = \sqrt{2} q$
Answer: $Q = \sqrt{2}\, q$
$U_{total} = \dfrac{Q^2}{2C}$
When the energy is stored equally between the electric and magnetic fields, the electric energy $U_E$ is half of the total energy:
$U_E = \dfrac{1}{2} U_{total}$
The electric energy at this instant is given by:
$U_E = \dfrac{q^2}{2C}$
Equating the expressions for $U_E$:
$\dfrac{q^2}{2C} = \dfrac{1}{2} \left( \dfrac{Q^2}{2C} \right)$
$q^2 = \dfrac{Q^2}{2}$
$Q = \sqrt{2} q$
Answer: $Q = \sqrt{2}\, q$
Hence, the correct answer is option: 'B'.