14.3 Solutions
657
14.19 If U is the total field energy then
U = U B + U E =
1
2
Li
2
+
1
2
Q 2
C
(1)
which shows that at any time the energy is stored partly in the magnetic
field in the conductor and partly in the electric field in the capacitor. In the
presence of the resistance R the energy is transferred to Joule heat, being
given by
dU
dt
= −i
2 R
(2)
the minus sign signifying that the stored energy U decreases with time. Differentiating (1) with respect to time and equating the result with (2) gives
Li
di
dt
+
Q
C
dQ
dt
= −i
2 R
(3)
Substituting i = dQ/dt and di/dt = d 2 Q/dt 2 , (3) becomes
d 2 Q
dt 2 +
R
L
dQ
dt
+
Q
LC
= 0
( 4 )
Writing R/L = 2γ and 1/LC = ω 2
0 , (4) takes the required form
d 2 Q
dt 2 + 2γ
dQ
dt
+ ω
2
0 Q = 0
( 5 )
(a) f 0 =
1
2π
√
LC
=
1
2π
√
80 × 10 −3 × 700 × 10 −12
= 2.128 × 10 4 Hz
(b) τ =
L
R
=
80 × 10 −3
100
= 8 × 10 −4 s
14.20
d 2 Q
dt 2 + 2γ
dQ
dt
+ ω
2
0 Q = 0
( 1 )
Let Q = e λt so that dQ/dt = λe λt and d 2 Q/dt 2 = λ 2 e λt .
The characteristic equation is then
λ
2
+ 2γ λ + ω
2
0 = 0
whose roots are λ = −γ ±
γ 2 − ω 2
0
(2)
Calling α =
γ 2 − ω 2
0
λ 1 = −γ + α, λ 2 = −γ − α