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Sagot :
The maximum voltage across the capacitor is 2.1 * 10⁻² V and the initial energy in the capacitor is 7.042 * 10⁻⁶ J.
In the L-C oscillation, energy is transferred between capacitor and inductor with a certain periodicity.
Given that, Inductance L = 65 * 10⁻³ h
Capacitance C = 300 * 10⁻⁴ F
Initial charge q = 6.5 * 10⁻⁴ C
We know that,
1/2 q²/C + 1/2 L² = 1/2 C* v₀²
From the above equation, we can write v₀ = q/C
Substituting the values in the above equation, we get
v₀ = q/C = (6.5 * 10⁻⁴) / (300 * 10⁻⁴) = 0.021 V = 2.1 * 10⁻² V
The formula to calculate initial energy in the capacitor is 1/2* q²/C
where q is charge and C is capacitance
E = 1/2 * q²/C = (6.5 * 10⁻⁴)²/ 2 * (300 * 10⁻⁴)
⇒ (42.25 * 10⁻⁸)/ (6 * 10⁻²)
⇒ 7.042 * 10⁻⁶ J
Thus, maximum voltage across the capacitor is 2.1 * 10⁻² V and the initial energy in the capacitor is 7.042 * 10⁻⁶ J.
To know more about LC circuit:
https://brainly.com/question/13200678
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