A capacitor with capacitance `6 xx 10^(-5) F` is charged by connecting it to a `12 -V` battery. The capacitor is disconnected from the battery and connected across an inductor with `L = 1.50 H`.
a. What are the angular frequency `omega` of the electrical oscillations and the period of these oscillations (the time for one oscillation)?
b. What is the intial charge on the capacitor?
c. How much energy is intially stored in the capacitor?
d. What is the charge on the capacitor `0.0230` s after the connecting to the inductor is made? Interpret the sign of the your answer.
e. At the times given in part (d), what is the current in the inductor? Interpret the sign of your answer.
f. At the time given in part (d), how much electrical energy is stored in the capactior and how much is stored in the inductor?

Explanation

a. `T = (2 pi)/(omega) = 2 pi sqrt(LC) = 2 pi sqrt((1.50 H)(6.00 xx 10^(-5) F)`
`= 0.0596 s`, `omega = 105 red//s`
b. `Q = CV = (6.00 xx 10^(-5) F)(12.0 V)= 7.20 xx 10^(-4) C`
c. `U_(0) = (1)/(2) CV^(2) = (1)/(2) (6.00 xx 10^(-5) F)(12.0 V)^(2) = 4.32 xx 10^(-3) J`
d. `t = 0.230 s`,
`q = Q cos (omega t)`
`= (7.20 xx 10^(-4) C) xx cos((0.0230 _(S))/sqrt((1.50 H)(6.00 xx 10^(-5) F)))`
`= - 5.43 xx 10^(-4) C`
Negative sign indicates that signs on plates are opposite to those at initials
e. `t = 0.0230 s`, `i = (dq)/(dt) = - mu Q sin (omega t)`
`i = - (7.20 xx 10^(-4)C)/sqrt((1.50 H)(6.00 xx 10^(-5) H))`
`xx sin ((0.0230_(S))/sqrt((1.50 H)(6.00 xx 10^(-5) H))) = - 0.05 A`
Negative sign indicates that positive charge flowing away from plate which had positive charge at the given time.
f. Capacitor : `U_(C) = (q^(2))/(2C) = ((5.43 xx 10^(-4) C)^(2))/(2(6.00 xx 10^(-5) F)) = 2.46 xx 10^(-3)J`
Inductor : `U_(L) = (1)/(2) Li^(2) = (1)/(5) (1.50 H)(0.05 A)^(2)`