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Unformatted text preview: 92. (a) Eqs. 314 and 3114 lead to Q = 1 ω = I LC = 1.27 × 10−6 C . (b) We choose the phase constant in Eq. 3112 to be φ = − π / 2 , so that i0 = I in Eq.
3115). Thus, the energy in the capacitor is
UE = q 2 Q2
=
(sin ωt ) 2 .
2C 2C Differentiating and using the fact that 2 sin θ cos θ = sin 2θ, we obtain
dU E Q 2
=
ω sin 2ωt .
dt
2C
We find the maximum value occurs whenever sin 2ωt = 1 , which leads (with n = odd
integer) to
1 nπ nπ nπ
t=
=
=
LC = 8.31 × 10−5 s, 2.49 × 10−4 s,… .
2ω 2 4ω
4 The earliest time is t = 8.31× 10−5 s.
(c) Returning to the above expression for dU E / dt with the requirement that sin 2ωt = 1 ,
we obtain FG dU IJ
H dt K d E max I LC
Q2
=
ω=
2C
2C i 2 I
I2
=
2
LC L
= 5.44 × 10−3 J / s .
C ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy

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