Problem Set 5. Originally appeared at: http://sites.google.com/site/peeterjoot/math2011/relElectroDynProblemSet5.pdf Peeter Joot — peeter.joot@gmail.com

Problem Set 5.

Originally appeared at:

http://sites.google.com/site/peeterjoot/math2011/relElectroDynProblemSet5.pdf Peeter Joot — peeter.joot@gmail.com

Mar 14, 2011 relElectroDynProblemSet5.tex

Contents

1 Problem 1. Sinusoidal current density on an infinite flat conducting sheet. 1

1.1 Statement. . . 1

1.2 1-2. Determining the electromagnetic potentials. . . 2

1.3 3. Find the electric and magnetic fields outside the plane. . . 3

1.4 4. Give a physical interpretation of the results of the previous section. . . 4

1.5 5. Find the direction and magnitude of the energy flux outside the plane. . . 4

1.6 6. Sketch the intensity of the electromagnetic field far from the plane. . . 4

1.7 7. Continuity across the plane? . . . 5

2 Problem 2. Fields generated by an arbitrarily moving charge. 5 2.1 Statement. . . 5

2.2 0. Solution. Gradient and time derivatives of the retarded time function. . . 6

2.3 1. Solution. Computing the EM fields from the Lienard-Wiechert potentials. . . 8

2.4 2. Solution. EM fields from a uniformly moving source. . . 11

2.5 2. Solution. EM fields from a uniformly moving source along x axis.. . . 13

3 Problem 3. 15

4 Grading notes. 15

1. Problem 1. Sinusoidal current density on an infinite flat conducting sheet.

1.1. Statement

An infinitely thin flat conducting surface lying in the x−z plane carries a surface current density:

κ=e₃θ(t)κ₀sin ωt (1)

Here e3is a unit vector in the z direction, κ0is the peak value of the current density, and θ(t)is the theta function: θ(t <0) −0, θ(t >0) =1.

1. Write down the equations determining the electromagnetic potentials. Specify which gauge you choose to work in.

2. Find the electromagnetic potentials outside the plane.

3. Find the electric and magnetic fields outside the plane.

4. Give a physical interpretation of the results of the previous section. Do they agree with your qualitative expectations?

5. Find the direction and magnitude of the energy flux outside the plane.

6. Consider a point at some distance from the plane. Sketch the intensity of the electromagnetic field near this point as a function of time. Explain physically.

7. Consider now a point near the plane. Are the electric and magnetic fields you found contin- uous across the conducting plane? Explain.

1.2. 1-2. Determining the electromagnetic potentials.

Augmenting the surface current density with a delta function we can form the current density for the system

J=δ(y)κ=e₃θ(t)δ(y)κ₀sin ωt. (2) With only a current distribution specified use of the Coulomb gauge allows for setting the scalar potential on the surface equal to zero, so that we have

^A= ^4πJ

c (3)

E= −¹ c

∂A

∂t (4)

B= ∇ ×B (5)

Utilizing our Green’s function

G(_{x, t}) = ^δ(t− |_x|/c)

4π|x| =δ³(_x)δ(t)_{, (6)} we can invert our vector potential equation, solving for A

A(x, t) =

Z

d³x⁰dt⁰x⁰,t⁰G(x−x⁰, t−t⁰)A(x⁰, t⁰)

=

Z

d³x⁰dt⁰G(x−x⁰, t−t⁰)^4πJ(x⁰, t⁰) c

=

Z

d³x⁰dt⁰δ(t−t⁰− |x−x⁰|/c) 4π|x−_x⁰|

4πJ(x⁰, t⁰) c

=

Z

d³x⁰J(x⁰, t− |x−x⁰|/c c|x−x⁰|

= ¹ c

Z

dx⁰dy⁰dz⁰e₃θ(t−x−x⁰

/c)δ(y)κ₀sin(ω(t−x−x⁰

/c)) ¹

|x−x⁰|

= ^e3κ0 c

Z

dx⁰dz⁰θ(t−x− (x⁰, 0, z⁰)/c)sin(ω(t−x− (x⁰, 0, z⁰)/c)) ¹

|x− (x⁰, 0, z⁰)|

= ^e3κ0 c

Z

dx⁰dz⁰θ

 t− ¹

c q

(x−x⁰)²+y²+ (z−z⁰)²

sin ω



t−¹_c^p(x−x⁰)²+y²+ (z−z⁰)^2

p(x−x⁰)²+y²+ (z−z⁰)²

Now a switch to polar coordinates makes sense. Let’s use

x⁰−x=r cos α (7)

z⁰−z=r sin α (8)

This gives us

A(x, t) = ^e3κ0 c

Z _∞

r=0

Z _2π

α=0rdrdαθ

 t−¹

c q

r²+y²

sin ω



t−¹_cpr²+y²

pr²+y²

= ^2πe3κ0 c

Z _∞

r=₀rdrθ

 t−¹

c q

r²+y²

sin

 ω



t−¹_cpr²+y²

pr²+y²

Since the theta function imposes a t− ¹

c q

r²+y² >0 (9)

constraint, equivalent to

c²t²>r²+y², (10)

we can reduce the upper range of the integral and drop the theta function explicitly

A(x, t) = ^2πe3κ0 c

Z

√

c²t²−y²

r=0 rdr

sin ω



t−¹_cpr²+y²

pr²+y² (11)

Here I got lazy and used Mathematica to help evaluate this integral, for an end result of A(x, t) = ^2πκ0ω

c² e₃(1−cos(ω(t− |y|/c))). (12) 1.3. 3. Find the electric and magnetic fields outside the plane.

Our electric field can be calculated by inspection E= −¹

c

∂A

∂t

= −^2πκ0ω

2

c³ e3sin(ω(t− |y|/c))_{. (13)} For the magnetic field we have

B= ∇ ×A

= −^2πκ0ω

c² e3× ∇(1−cos(ω(_t− |y|/c))))

= ^2πκ0ω

c² (−sin(ω(t− |y|/c))e₃× ∇ω(t− |y|/c)

= ^2πκ0ω

2

c³ sin(ω(t− |y|/c))e3× ∇|y|

= ^2πκ0ω

2

c³ sin(ω(t− |y|/c))e₃×e₂,

which gives us

B= −^2πκ0ω

2

c³ e₁sin(ω(t− |y|/c) (14) 1.4. 4. Give a physical interpretation of the results of the previous section.

It was expected that the lack of boundary on the conducting sheet would make the potential away from the plane only depend on the y components of the spatial distance, and this is precisely what we find performing the grunt work of the integration.

Given that we had a sinusoidal forcing function for our wave equation, it seems logical that we also find our non-homogeneous solution to the wave equation has sinusoidal dependence. We find that the sinusoidal current results in sinusoidal potentials and fields very much like one has in the electric circuits problem that we solve with phasors in engineering applications.

We find that the electric and magnetic fields are oriented parallel to the plane containing the surface current density, with the electric field in the direction of the current, and the magnetic field perpendicular to that, but having energy propagate outwards from the plane.

It’s kind of curious that despite introducing a step function in time for the current, that the end result appears to have no constraints on time t > 0, and that we have fields at all points in space and time from this current distribution, even the t < 0 case (ie: we have a non-causal system). I’d have expected a transient response to switching on the current. Perhaps this is because instantaneously inducing a current on all points in an infinite sheet is not physically realizable?

1.5. 5. Find the direction and magnitude of the energy flux outside the plane.

Our energy flux, the Poynting vector, is

S= ^c 4π

 2πκ₀ω² c³

2

sin²(ω(t− |y|/c)e₃×e₁. (15) This is

S= ^πκ

20ω⁴

c⁵ sin²(ω(t− |y|/c)e₂ = ^πκ

20ω⁴

2c⁵ (1−cos(2ω(t− |y|/c)))e₂. (16) This energy flux is directed outwards along the y axis, with magnitude oscillating around an average value of

|hSi| = ^πκ

20ω⁴

2c⁵ . (17)

1.6. 6. Sketch the intensity of the electromagnetic field far from the plane.

I’m assuming here that this question does not refer to the flux intensity hSi, since that is con- stant, and boring to sketch.

The time varying portion of either the electric or magnetic field is proportional to

sin(ωt−ω|y|/c) (18)

We have a sinusoid as a function of time, of period T = 2π/ω where the phase is adjusted at each position by the factor ω|y|/c. Every increase of∆y=2πc/ω shifts the waveform back.

A sketch is attached.

1.7. 7. Continuity across the plane?

It is sufficient to consider either the electric or magnetic field for the continuity question since the continuity is dictated by the sinusoidal term for both fields.

The point in time only changes the phase, so let’s consider the electric field at t = 0, and an infinitesimal distance y= ±ec/ω. At either point we have

E(_0,±ec/ω, 0, 0) = ^2πκ0ω

2

c³ e3e (19)

In the limit as e → 0 the field strength matches on either side of the plane (and happens to equal zero for this t=0 case).

We have a discontinuity in the spatial derivative of either field near the plate, but not for the fields themselves. A plot illustrates this nicely

Figure 1: sin(t− |y|)

2. Problem 2. Fields generated by an arbitrarily moving charge.

2.1. Statement

Show that for a particle moving on a worldline parametrized by(ct, x_c(t)), the retarded time t_rwith respect to an arbitrary space time point(ct, x), defined in class as:

|x−x_c(t_r)| =c(t−t_r) (20) obeys

∇t_r= − ^x−x_c(t_r)

c|x−_xc(t_r)| =c(t−t_r) −_vc(t_r)_cot(_x−_xc(t_r)) ⁽²¹⁾

and

∂t_r

∂t = ^c|x−x_c(t_r)|

c|x−xc(tr)| =c(t−tr) −vc(tr)cot(x−xc(tr)) ⁽²²⁾ 1. Then, use these to derive the expressions for E and B given in the book (and in the class

notes).

2. Finally, re-derive the already familiar expressions for the EM fields of a particle moving with uniform velocity.

2.2. 0. Solution. Gradient and time derivatives of the retarded time function.

Let’s use notation something like our text [1], where the solution to this problem is outlined in

§63, and write

R(t_r) =x−x_c(t_r) (23)

R=|R| (24)

where

∂R

∂tr

= −vc. (25)

From R²=R·Rwe also have

2R∂R

∂tr

=2R·^∂R

∂tr, (26)

so if we write

Rˆ = ^R

R, (27)

we have

R⁰(tr) = −R^ˆ ·vc. (28) Proceeding in the manner of the text, we have

∂R

∂t

= ^∂R

∂tr

∂t_r

∂t

= −_R^ˆ ·_vc^∂tr

∂t. (29)

From20we also have

R=|x−x_c(t_r)| =c(t−t_r), (30) so

∂R

∂t =c

 1− ^∂tr

∂t



. (31)

This and29gives us

∂t_r

∂t = ¹

1−R^ˆ ·^v_c^{c (32)}

For the gradient we operate on the implicit equation30again. This gives us

∇R= ∇(ct−ct_r) = −c∇t_r. (33) However, we can also use the spatial definition of R = |x−x_c(t⁰)|. Note that this distance R= R(t_r)is a function of space and time, since tr = t_r(_{x, t})is implicitly a function of the spatial and time positions at which the retarded time is to be measured.

∇R= ∇ q

(x−xc(tr))²

= ¹

2R∇(x−x_c(t_r))²

= ¹

R(x^β−x^β_c)eα∂α(x^β−x_c^β(t_r))

= ¹

R(R)_βeα(δαβ−∂αx_c^β(tr))

We have only this bit ∂_αx^β_c(t_r)to expand, but that’s just going to require a chain rule expansion.

This is easier to see in a more generic form

∂ f(g)

∂x^α = ^{∂ f}

∂g

∂g

∂x^α, (34)

so we have

∂x_c^β(t_r)

∂x^α = ^∂x

β c(t_r)

∂t_r

∂t_r

∂x^α, (35)

which gets us close to where we want to be

∇R= ¹

R R− (R)_β^∂x

βc(t_r)

∂tr eα

∂t_r

∂x^α

!

= ¹

R R−R·^∂x

βc(t_r)

∂tr

∇tr

!

Putting the pieces together we have only minor algebra left since we can now equate the two expansions of∇R

−c∇t_r= _R^ˆ −_R^ˆ ·_vc(t_r)∇t_r. (36) This is given in the text, but these in between steps are left for us and for our homework assignments! From this point we can rearrange to find the desired result

∇t_r = −¹ c

Rˆ

1−R^ˆ ·^v_c^c = −^Rˆ c

∂t_r

∂t (37)

2.3. 1. Solution. Computing the EM fields from the Lienard-Wiechert potentials.

Now we are ready to derive the values of E and B that arise from the Lienard-Wiechert poten- tials. We have for the electric field.

We’ll evaluate

E= −¹ c

∂A

∂t − ∇φ B= ∇ ×B

For the electric field we’ll use the chain rule on the vector potential

∂A

∂t = ^∂tr

∂t

∂A

∂t_r. (38)

Similarly for the gradient of the scalar potential we have

∇φ=eα

∂φ

∂x^α

=eα

∂φ

∂t_r

∂t_r

∂x^α

= ^∂φ

∂t_r∇tr

= −^Rˆ c

∂t_r

∂t. Our electric field is thus

E= −^∂tr

∂t

 1 c

∂A

∂tr

−^Rˆ c

∂φ

∂tr



(39) For the magnetic field we have

∇ ×A=eα× ^∂A

∂x^α

=e_α× ^∂A

∂t_r

∂t_r

∂x^α. The magnetic field will therefore be found by evaluating

B= (∇t_r) × ^∂A

∂t_r = −^∂tr

∂t Rˆ

c × ^∂A

∂t_r (40)

Let’s compare this to ˆR×_E

Rˆ ×E=R^ˆ ×



−^∂tr

∂t

 1 c

∂A

∂t_r − ^Rˆ c

∂φ

∂t_r



=R^ˆ ×



−^∂tr

∂t 1 c

∂A

∂t_r



This equals40, verifying that we have

B=_R^ˆ ×E, (41)

something that we can determine even without fully evaluating E.

We are now left to evaluate the retarded time derivatives found in39. Our potentials are

φ(x, t) = ^e R(t_r)

∂t_r

∂t (42)

A(x, t) = ^evc(t_r) cR(tr)

∂t_r

∂t (43)

It’s clear that the quantity ∂tr/∂t is going to show up all over the place, so let’s label it γtr. This is justified by comparing to a particle’s boosted rest frame worldline

ct⁰ x⁰



=γ

 1 −β

−β 1

 ct 0



=

 γct

−γβct



, (44)

where we have ∂t⁰/∂t=γ, so for the remainder of this part of this problem we’ll write γ_tr ≡ ^∂tr

∂t

= ¹

1−R^ˆ ·^v_c^{c. (45)}

Using primes to denote partial derivatives with respect to the retarded time t_rwe have

φ⁰ =e



−^R

0

R²γ_tr +^γ

t0r

R



(46) A⁰ =evc

c



−^R

0

R²γ_tr +^γ

t0r

R

 +eac

c γtr

R , (47)

so the electric field is

E= −γ_tr 1

cA⁰−^Rˆ cφ⁰



= −^eγtr c

 v_c c



−^R

0

R²γ_tr+ ^γ

0tr

R

 + ^ac

c γ_tr

R −_R^ˆ



−^R

0

R²γ_tr + ^γ

0tr

R



= −^eγtr c

 v_c c

 c

R²γ_tr+ ^γ

0tr

R

 + ^ac

c γ_tr

R −_R^ˆ

 c

R²γ_tr +^γ

t0r

R



= −^eγtr cR



γ_tr ac

c +^vc R −^Rcˆ

R

 +γ⁰_tr

vc

c −_R^ˆ

 . Here’s where things get slightly messy.

γ⁰_tr = ^∂

∂t_r 1 1−^v_c^c ·R^ˆ

= −γ²_tr ∂

∂tr

 1− ^vc

c ·R^ˆ

=γ²_tr

a_c

c ·R^ˆ +^vc c ·R^ˆ0

,

and messier

Rˆ⁰ = ^∂

∂t_r R R

= ^R

0

R − ^RR

0

R²

= −^vc

R − ^Rˆ(−c) R

= ¹

R −vc+c ˆR
, then a bit unmessier

γ⁰_tr =γ²_tr

ac

c ·R^ˆ +^vc c ·R^ˆ0

=γ²_tr

ac

c ·R^ˆ + ^vc

cR · (−vc+c ˆR)^

=γ²_tr

 Rˆ ·^ac

c + ^vc R

− ^v2c cR

 .

Now we are set to plug this back into our electric field expression and start grouping terms

E= −^eγ

2tr

cR

 ac

c + ^vc R − ^Rcˆ

R +γtr

 Rˆ ·^ac

c + ^vc R

− ^v

2c

cR

vc

c −R^ˆ

= −^eγ

3tr

cR

 a_c c +^vc

R −^Rcˆ R



1−R^ˆ ·^vc c

 +

 Rˆ ·^ac

c +^vc R

− ^v

2c

cR

v_c

c −R^ˆ

= −^eγ

3tr

c²R

 ac



1−R^ˆ ·^vc c



+R^ˆ ·ac

vc

c −R^ˆ

− ^eγ

3tr

cR

 vc

R −^Rcˆ R



1−R^ˆ · ^vc c

 +



Rˆ ·^vc R

− ^v

2c

cR

vc

c −R^ˆ

Using

a× (_b×_c) =_b(_a·_c) −_c(_a·_b) ₍₄₈₎ We can verify that

−^a_c

1−R^ˆ ·^vc c



+R^ˆ ·a_cv_c

c −R^ˆ

= −a_c+a ˆR·^v

c −R^ˆ ·a_cv_c

c +R^ˆ ·a_cRˆ

=_R^ˆ × ˆR−^vc c

×_ac^, which gets us closer to the desired end result

E= ^eγ

t3r

c²RRˆ × ˆR−^vc c

×ac

−^eγ

3tr

cR²



vc−Rc^ˆ


1−R^ˆ ·^vc c

 +



Rˆ ·vc− ^v

2c

c

vc

c −R^ˆ . (49)

It is also easy to show that the remaining bit reduces nicely, since all the dot product terms conveniently cancel

−



v_c−Rc^ˆ


1−R^ˆ ·^vc c

 +



Rˆ ·v_c− ^v

2c

c

v_c

c −R^ˆ

=c

 1− ^v

2c

c²

 ˆR− ^v c



(50) This completes the exercise, leaving us with

E= ^eγ

3tr

c²RRˆ × ˆR− ^vc c

×_ac^+ ^eγ

3tr

R²

 1−^v2c

c²

 ˆR− ^vc c

 B=_R^ˆ ×E.

(51)

Looking back to45where γtr was defined, we see that this compares to (63.8-9) in the text.

2.4. 2. Solution. EM fields from a uniformly moving source.

For a uniform source moving in space at constant velocity

xc(t) =vt, (52)

our retarded time measured from the spacetime point(ct, x)is defined implicitly by

R=|x−x_c(t_r)| =c(t−t_r). (53) Squaring this we have

x²+v²t²_r−2t_rx·v= c²t²+c²t²_r −2ctt_r, (54) or

(c²−_v²)t²_r+2t_r(−ct+_x·_v) =_x²−c²t². (55) Rearranging to complete the square we have



pc²−_v²t_r−^tc2−x·v

√c²−v²

2

=_x²−c²t²+ (tc²−x·v)² c²−v²

= (_x²−c²t²)(c²−_v²) + (tc²−_x·_v)² c²−v²

= ^x

2c²−_x²_v²−_c^4t²+c²t²v²+_t^2c⁴+ (_x·_v)²−_2tc²(_x·_v) c²−v²

= ^c

2(x²+t²v²−2t(x·v)) −x²v²+ (x·v)² c²−v²

= ^c

2(x−vt)²− (x×v)² c²−_v²

Taking roots (and keeping the negative so that we have tr = t− |x|/c for the v = 0 case, we have

r 1−^v2

c²ctr = _q ¹ 1−^v_c2²

ct−x·^v c −

r

(x−vt)²−^x×^v c

2!

, (56)

or with β= v/c, this is ctr = ¹

1−β²



ct−x·β− q

(x−vt)²− (x×β)²



. (57)

What’s our retarded distance R =ct−ctr? We get

R= ^β· (x−vt) +^p(x−vt)²− (x×β)² 1−β²

. (58)

For the vector distance we get (with β· (x∧β) = (β·x)β−xβ²) R= ^x−vt+β· (x∧β) +βp

(x−vt)²− (x×β)² 1−β²

. (59)

For the unit vector ˆR=R/R we have

Rˆ = ^x−_vt+β· (_x∧β) +βp

(_x−_vt)²− (_x×β)²

β· (x−vt) +^p(x−vt)²− (x×β)^{2 . (60)} The acceleration term in the electric field is zero, so we are left with just

E= ^eγ

3tr

R²

 1− ^v

2c

c²

 ˆR− ^vc c



. (61)

Leading to γtr, we have

Rˆ ·β= ^β· (x−vt+R^∗β)

β· (_x−_vt) +R^∗ , (62) where, following §38 of the text we write

R^∗ = q

(x−vt)²− (x×β)² (63) This gives us

γtr = ^β· (x−vt) +R^∗

R^∗(1−β²) ^{. (64)}

Observe that this equals one when β=0 as expected.

We can also compute

Rˆ −β= ^x+β· (x∧β) −vt+βp

(x−vt)²− (x×β)² β· (x−vt) +^p(x−vt)²− (x×β)² −β

= (x−vt)(1−β²)

β· (x−vt) +^p(x−vt)²− (x×β)^2.

Our long and messy expression for the field is therefore

E=eγ³_tr 1

R²(1−β²)(R^ˆ −β)

=e β· (x−vt) +R^∗ R^∗(1−β²)

!3

(1−β²)²

(_β· (_x−_vt) +R^∗)²(1−β²)(x−vt)(1−β²) β· (_x−_vt) +R^∗

This gives us our final result

E=e 1

(R^∗)³(1−_β²)(_x−_vt) (65) As a small test we observe that we get the expected result

E=e x

|x|^{3 (66)}

for the β=0 case.

When v = Ve₁this also recovers equation (38.6) from the text as desired, and if we switch to primed coordinates

x⁰ =γ(x−vt) (67)

y⁰ =y (68)

z⁰ =z (69)

(1−β²)r⁰²= (x−vt)²+ (y²+z²)(1−β²), (70) we recover the field equation derived twice before in previous problem sets

E= ^e

(r⁰)³(x⁰, γy⁰, γz⁰) (71) 2.5. 2. Solution. EM fields from a uniformly moving source along x axis.

Initially I had errors in the vector treatment above, so tried with the simpler case using uniform velocity v along the x axis instead. Comparison of the two showed where my errors were in the vector algebra, and that’s now also fixed up.

Performing all the algebra to solve for t_rin

|x−vtre₁| =c(t−tr), (72) I get

ctr = ^ct−xβ−^p(x−vt)²+ (y²+z²)(₁−β²)

1−β² = −γ(βx⁰+r⁰) (73)

This matches the vector expression from 57 with the special case of v = ve₁ so we at least started off on the right foot.

For the retarded distance R =ct−ctrwe get

R= ^β(x−vt) +^p(x−vt)²+ (y²+z²)(1−β²)

1−β² = γ(βx⁰+r⁰) (74)

This also matches 58, so things still seem okay with the vector approach. What’s our vector retarded distance

R=x−βctre₁

= (x−βct_r, y, z)

= ^x−vt+βp

(x−vt)²+ (y²+z²)(1−β²)

1−β² , y, z

!

= γ(x⁰+βr⁰), y⁰, z⁰
So

Rˆ = ¹

γ(βx⁰+r⁰) ^γ(x⁰+βr⁰)_{, y}⁰_{, z}⁰

= ¹

βx⁰+r⁰



x⁰+βr⁰,y⁰ γ,z⁰

γ



Rˆ −_β= ¹

γ(βx⁰+r⁰) ^γ(x⁰+βr⁰)_{, y}⁰_{, z}⁰
− (_{β, 0, 0})

= ¹

βx⁰+r⁰



x⁰(1−β²),y⁰ γ,z⁰

γ



= ¹

γ(βx⁰+r⁰)(x−vt, y, z)

For ∂t_r/∂t, using ˆRcalculated above, or from73calculating directly I get

∂tr

∂t = ^r

0+βx⁰

r⁰(1−β²) = ^γ(r⁰+βx⁰)

R^∗ , (75)

where, as in §38 of the text, we write R^∗ =

q

(x−vt)²+ (y²+z²)(₁−β²)_{. (76)} Putting all the pieces together I get

E=e(1−β²)(x−vt, y, z)

 γ(βx⁰+r⁰)



 γ(r⁰+βx⁰)

R^∗

3

1

 γ²(βx⁰+r⁰)² so we have

E= e1−β²

(R^∗)³ (x−vt, y, z) (77)

This matches equation (38.6) in the text.

3. Problem 3.

FIXME: TODO.

4. Grading notes.

Only the first question was graded (I lost 1.5 marks). I got my units wrong when I integrated 11, and my ω/c factor in the result is circled. Should have also done a dimensional analysis check.

There was also a remark that the integral is zero if t <y/c, which introduces a θ function. I think that I incorrectly dropped my θ function, and should have retained.

FIXME: Revisit both of these and make sure to understand where exactly I went wrong.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. 2.2