A 6.50-nF capacitor is charged to 23.0 V then disconnected from the battery in t

Question

A 6.50-nF capacitor is charged to 23.0 V then disconnected from the battery in the circuit and connected in series with a coil that has L = 0.0650 H and negligible resistance. After the circuit has been completed, there are current oscillations.

part A) At an instant when the charge on the capacitor is 0.0500 C, what is the current in the inductor? answer 6.85mA

part B) How much energy is stored in the capacitor?

part C) How much energy is stored in the inductor?

part D) At an instant when the charge on the capacitor is 0.0500 C, what is the rate at which current in the inductor is changing?

part E) What is the voltages across the capacitor?

part F) What is the voltages across the inductor?

Please show work. Will rate. Thanks!

Explanation / Answer

a)

Inital total energy stored in Capacitor

Uinitial=(1/2)CV2 =(1/2)*(6.5*10-9)*232 =1.72*10-6 J

Final energy stored in LC circuit

Ufinal=(1/2)LI2+(1/2)Q2/C

=>Uinitial =Ufinal

1.72*10-6 =(1/2)*0.065*I2+ (1/2)(0.05*10-6)2/(6.5*10-9)

I=6.85 mA

b)

Energy stored in capacitor

UC=(1/2)(Q2/C) =(1/2)(0.05*10-6)2/(6.5*10-9)

UC=1.923*10-7 J

c)

Energy stored in inductor

UL=(1/2)LI2 =(1/2)(0.065)(6.85*10-3)2

UL=1.525*10-6 J

d)

Rate of change of current in inductor

dI/dt =Q/LC = (0.05*10-6)/(6.5*10-9)*0.065

dI/dt =118.34 A/s

e)

Vc=Q/C=(0.05*10-6)/(6.5*10-9)=7.7 Volts

f)

VL=L(dI/dt)=0.065*118.34=7.7 Volts

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