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homework solution for hw6, Assignments of Physics

homework solution of HW6 in spring 2025

Typology: Assignments

2024/2025

Uploaded on 05/09/2025

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Phys436: Homework 7 Solution
Grader: Hao-Chien Wang
October 29, 2024
1. (a) The energy of electromagnetic field is given by:
W=1
2(ϵ0E2+1
µ0
B2)dτ ,
and, for |x|< ct, the fields are given by :
E=µ0k
2(ct |x|)ˆz
B=±µ0k
2c(ct |x|)ˆy,+ for x > 0and for x < 0,
and both fields vanish for |x|> ct. At t1=d
c, x ct1within the box, so E= 0,B= 0, and
hence W(t1) = 0 (0.1 pt). At t2=(d+h)
c,ct2=d+h > x at all points in the box, and the
fields are:
E=µ0k
2(d+hx)ˆz
B=µ0k
2c(d+hx)ˆy,
0.1 pt for writing down the field, but 0 pt if not distinguishing the case which |x|ct and
x0.We have B2=E2/c2, and the energy density is:
1
2(ϵ0E2+1
µ0
B2)(0.1 pt)
=ϵ0
2(E2+1
µ0ϵ0
1
c2E2)=ϵ0E2
=ϵ0µ2
0k2
4(d+hx)2(0.1 pt)
Therefore,
W(t2) = ϵ0µ2
0k2
4d+h
d
(d+hx)2lw dx(0.1 pt)
=ϵ0µ2
0k2lw
4[(d+hx)3
3]d+h
d
=ϵ0µ2
0k2lwh3
12 =µ0k2lwh3
12c2.(0.2 pt)
1
pf3

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Phys436: Homework 7 Solution

Grader: Hao-Chien Wang

October 29, 2024

  1. (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).

  1. 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).

  1. (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)

  1. 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