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Phys436: Homework 7 Solution
Grader: Hao-Chien Wang
October 29, 2024
- (a) The energy of electromagnetic field is given by:
W =
ϵ 0 E^2 +
μ 0
B^2
dτ ,
and, for |x| < ct, the fields are given by :
E = −
μ 0 k 2 (ct − |x|) zˆ
B = ± μ 0 k 2 c
(ct − |x|) yˆ, “+” for x > 0 and “−” for x < 0 ,
and both fields vanish for |x| > ct. At t 1 = dc , x ≥ ct 1 within the box, so E = 0, B = 0, and hence W (t 1 ) = 0 (0.1 pt). At t 2 = (d+ c h), ct 2 = d + h > x at all points in the box, and the fields are:
E = − μ 0 k 2
(d + h − x) zˆ
B = μ 0 k 2 c (d + h − x) yˆ,
0.1 pt for writing down the field, but 0 pt if not distinguishing the case which |x| ≶ ct and x ≶ 0. We have B^2 = E^2 /c^2 , and the energy density is:
1 2
ϵ 0 E^2 +
μ 0
B^2
(0.1 pt)
ϵ 0 2
E^2 +
μ 0 ϵ 0
c^2
E^2
= ϵ 0 E^2
= ϵ 0 μ^20 k^2 4
(d + h − x)^2 (0.1 pt)
Therefore,
W (t 2 ) = ϵ 0 μ^20 k^2 4
∫ (^) d+h
d
(d + h − x)^2 lw dx (0.1 pt)
ϵ 0 μ^20 k^2 lw 4
[
(d + h − x)^3 3
]d+h
d
ϵ 0 μ^20 k^2 lwh^3 12
μ 0 k^2 lwh^3 12 c^2
. (0.2 pt)
(b) The Poynting vector is given by:
S(x) =
μ 0 (E × B) (0.1 pt)
=
μ 0 c
E^2 [−zˆ × (±yˆ)]
= ±
μ 0 c
E^2 xˆ = ± μ 0 k^2 4 c
(ct − |x|)^2 xˆ (0.3 pt),
(“+” for ct > x > 0 and “−” for −ct < dx < 0 , as before). For |x| > ct, S = 0. You need to check that S = 0 at the upper surface (using the fact that S = 0 if |x| > ct), in order to deduce only the integral over the lower surface contributes to the energy influx. Therefore, just calculating S at the lower surface and not distinguishing the case which |x| ≶ ct and x ≶ 0 is not enough. There will be − 0. 1 pt deduction for this. At t 1 < t < t 2 , the only surface that has non-zero energy flux is the lower face of the box at x = d, so the energy per unit time entering the box in this time interval is: dW dt
= P =
S(d) · da =
μ 0 k^2 lw 4 c (ct − d)^2. (0.3 pt)
Note that no energy flows out the top, since S(d + h) = 0. (c) The change in total energy in the box is:
W =
∫ (^) t 2
t 1
P dt = μ 0 k^2 lw 4 c
∫ (^) (d+h)/c
d/c
(ct − d)^2 dt (0.2 pt)
μ 0 k^2 lw 4 c
[
(ct − d)^3 3 c
](d+h)/c
d/c
μ 0 k^2 lwh^3 12 c^2 (0.4 pt),
which agrees with the answer to (a).
- The fields are:
E = −∇V −
∂A
∂t
4 πϵ 0
2 q r^2
rˆ (0.2 pt) B = ∇ × A = 0 (0.2 pt),
which are just the fields of a point charge 2 q at the origin. Therefore, using Eq. (1.99), ρ = ϵ 0 ∇ · E = 2q δ(3)(r) (0.3 pt) J = 0 (0.3 pt).
- (a) Ex. 10.1: ∇ · A = 0; ∂V∂t = 0. [Both Coulomb and Lorenz.]
(b) Prob. 10.3: ∇ · A = − (^4) πϵqt 0 ∇
( (^) rˆ r^2
= − (^) ϵqt 0 δ^3 (r); ∂V∂t = 0. [Neither.] (c) Prob. 10.4: ∇ · A = 0; ∂V∂t = 0. [Both.] 1/6 pt for each answer (Coulomb gauge and Lorenz gauge count as individual answers)
- To prove this, we can adopt the approach hinted by Sec. 5.4.1 in the textbook. We can start with any potentials A and V that do not satisfy the condition of Lorenz gauge,
Φ ≡ ∇ · A + μ 0 ϵ 0
∂V
∂t
