(a) By applying charge conservation to a volume V, derive the integral form
of the equation of charge conservation. Deduce the equation of continuity,
**∇∙j** + (∂/∂t)ρ = 0.

(b) Show that the equation of continuity is a consequence of Maxwell's
equations.

(c) A region containing a charge Q_{0} is produced at time
t = 0 inside a conductor of conductivity
σ. By
considering the flow of current across the surface of the region, show that at
subsequent times the charge inside the region is Q(t) = Q_{0}exp(-σt/ε_{0}).

HINT: Use Gauss' Law and Ohm's Law in the form **j **= σ**E**.

(d) Explain why you expect Q --> 0 as t --> ∞
.

Solution:

- Concepts:

Gauss' theorem, Maxwell's equations, properties of a conductor - Reasoning:

(a) If we have a conserved quantity in a volume V, then the rate at which it flows out of the volume must equal the rate it decreases inside the volume.

For charge this is expressed as ∫_{closed_A}**j∙n**dA = -(∂/∂t)∫_{V}ρdV for any volume V.

Gauss' theorem then yields ∫_{V}**∇∙j**dV = -(∂/∂t)∫_{V}ρdV for any volume V.

Therefore**∇∙j**= -∂ρ/∂t.

(b) Maxwell's equations relate the sources and the fields. - Details of the calculation:

**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t (SI units),**∇∙**(**∇**×**B**) = 0 --> μ_{0}**∇∙j**+ (1/c^{2})(∂/∂t)**∇∙E**= 0.

**∇∙E**= ρ/ε_{0}, μ_{0}**∇∙j**+ (1/(ε_{0}c^{2}))(∂ρ/∂ t) = 0.

μ_{0}ε_{0}= 1/c^{2},**∇∙j**+ (∂ρ/∂t) = 0.

(c) ∫_{V}**∇∙j**dV = -(∂/∂t)∫_{V}ρdV.

For a conductor we have**j**= σ**E**.

∫_{V}σ**∇∙E**dV = ∫_{V}(σρ/ε_{0})dV = (σ/ε_{0})Q = -(∂Q/∂ t).

Q = Q_{0}exp(-σt/ε_{0}).

(d) In a conductor charges are free to move. They will move and distribute themselves over the surface until the electric field inside the conductor is zero.

Q_{inside}≠ 0 implies**E**_{inside}≠ 0, therefore we expect Q(t) = 0 as as t --> ∞ .

(a) Show that Maxwell's equations are consistent with the
conservation of electric charge.

(b) Show that the power P injected into a circuit by an electric field is
given by
∫**j∙E **dV. Verify that in steady state this reproduces
the Ohmic heat loss in a “thin wire” approximation.

Solution:

- Concepts:

Maxwell's equations - Reasoning:

(a) If we have a conserved quantity in a volume V, then the rate at which it flows out of the volume must equal the rate it decreases inside the volume.

For charge this is expressed as ∫_{closed_A}**j∙n**dA = -(∂/∂t)∫_{V}ρdV for any volume V.

Gauss' theorem then yields ∫_{V}**∇∙j**dV = -(∂/∂t)∫_{V}ρdV for any volume V.

Therefore**∇∙j**= -∂ρ/∂t. - Details of the calculation:

Maxwell's equations relate the sources and the fields.

**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t (SI units),**∇∙**(**∇**×**B**) = 0 --> μ_{0}**∇∙j**+ (1/c^{2})(∂/∂t)**∇∙E**= 0.

**∇∙E**= ρ/ε_{0}, μ_{0}**∇∙j**+ (1/(ε_{0}c^{2}))(∂ρ/∂ t) = 0.

μ_{0}ε_{0}= 1/c^{2},**∇∙j**+ (∂ρ/∂ t) = 0.

Maxwell's equations are consistent with the conservation of electric charge.

(b) For a single charge, the rate of doing work by external fields**B**and**E**is q**v∙E**, in which**v**is the velocity of the charge. Let ρdV = dq be the amount of charge in a volume element dV.

dW/dt = dq**v∙E**= ρ**v∙E**dV =**j∙E**dV = rate at which work is done by the field on the charges in dV. The power P injected into a circuit is obtained by integration dW/dt over the volume of the circuit, P = ∫**j∙E**dV. If we consider a steady current to a thin wire, then for an element of length d**l**and cross section d**A**of the wire, dV = d**l∙**d**A**. In this approximation**j**is parallel to d**l**and**E**does not vary appreciably over the cross section of the wire. Hence ∫**j∙E**dV = ∫**j∙**d**A**∫**E∙**d**l**= IV = I^{2}R which is the Ohmic heat loss.

Do the fields **E** = **i** E_{0}cos(ωt − kx ),
**B** = 0 satisfy Maxwell's equations?

If a special condition for ρ and **j** is needed, what is it?

Solution:

- Concepts:

Maxwell's equations

**∇∙E**= ρ/ε_{0},**∇**×**E**= -∂**B**/∂t,**∇∙B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t - Reasoning

We are supposed to check if**E**=**i**E_{0}cos (ωt − kx) is consistent with Maxwell's equations. - Details of the calculation:
**∇**•**E**= ∂[E_{0}cos (ωt − kx)]/∂x = kE_{0}sin(ωt − kx) ≠ 0, ρ ≠ 0 .

ρ = kε_{0}E_{0}sin(ωt − kx).

Since**B**= 0,**∇∙B**= 0.**∇**×**E**= 0, ∂**B**/∂t = 0,**B**= 0 is a possible solution.

The equation of continuity follows from Maxwell's equations.

**∇∙j**= -(∂ρ/∂t) = -ωkε_{0}E_{0}cos(ωt − kx).

Since**B**= 0,**∇**×**B**= 0. This requires**j**= (-1/μ_{0}c^{2})∂**E**/∂t = (ω/μ_{0}c^{2})E_{0}sin(ωt − kx)**i.**= ∂[(ω/μ

∇∙j_{0}c^{2})E_{0}sin(ωt − kx))]/∂x = (-kω/μ_{0}c^{2})E_{0}cos(ωt − kx) = -ωkε_{0}E_{0}cos(ωt − kx).

The fields**E**=**i**E_{0}cos(ωt − kx ),**B**= 0 satisfy Maxwell's equations

with ρ = kε_{0}E_{0}sin(ωt − kx) and**j**= (ω/μ_{0}c^{2})E_{0}sin(ωt − kx)**i.**

(a) Write down Maxwell's equations in vacuum for a charge
density and current density free medium (ρ = 0 and** j** = 0).

(b) Show that the electric field and the magnetic field satisfy a wave equation.

(c) Write down plane-wave solutions of the wave equations for the electric
field and magnetic field. How are they related to each other? Find the group
velocity and phase velocity of the electromagnetic waves.

Solution:

- Concepts:

Maxwell's equations - Reasoning:

Maxwell's equations in free space yield the wave equation for both**E**and**B**. - Details of the calculation:

(a)**∇**∙**E**= 0,**∇**×**E**= -∂**B**/∂t,**∇**∙**B**= 0,**∇**×**B**= (1/c^{2})∂**E**/∂t.

(b)**∇**×(**∇**×**E**) =**∇**(**∇∙E**) -**∇**^{2}**E**= -(∂/∂t)(**∇**×**B**) = -μ_{0}ε_{0}∂^{2}**E**/∂t^{2}.

**∇**^{2}**E**= μ_{0}ε_{0}∂^{2}**E**/∂t^{2}.

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= μ_{0}ε_{0}(∂/∂t)(**∇**×**E**) = -μ_{0}ε_{0}∂^{2}**B**/∂t^{2}.

**∇**^{2}**B**= μ_{0}ε_{0}∂^{2}**B**/∂t^{2}.

(c) Plane wave solutions**E**=**E**((**k**/k)∙**r**- vt),**B**=**B**((**k**/k)∙**r**- vt) exist.

(v = (μ_{0}ε_{0})^{-½}= c) All plane wave solutions are linear superpositions of harmonic waves of the form sin(**k**∙**r**- ωt) and cos(**k**∙**r**- ωt), with ω/k = c.

The phase velocity is ω/k = c, the group velocity is dω/dk = c. There is no dispersion for EM waves in free space.

Consider an electromagnetic traveling wave with electric
and magnetic fields given by

E_{x} = E_{0}cos(kz
– ωt) +
φ),

and

B_{y} = B_{0}cos(kz
– ωt) +
φ).

Using Maxwell's equations show that B_{0} can be
written in terms of E_{0}.

Solution:

- Concepts:

Maxwell's equations - Reasoning:

Maxwell's equations in free space yield the wave equation for both**E**and**B**. They can also be used to show that**E**⊥**B**,**E**⊥**k**,**B**⊥**k**,**B**= (μ_{0}ε_{0})^{½}(**k**/k)×**E**. - Details of the calculation:

From Maxwell's equations:×

∇**E**= -∂**B**/∂t.×

∇**E**= ∂E_{x}/∂z**j**= -kE_{0}sin(kz – ωt + φ)**j**, -∂**B**/∂t = -ωB_{0}sin(kz – ωt + φ)**j**.

Therefore kE = ωB_{0}, B_{0}= (k/ω)E_{0}.

In free space k/ω = 1/c.

Use Maxwell's equations to
find the magnetic field of an EM wave in vacuum for which the electric field is
given by **
E** = (E

Solution:

- Concepts:

Maxwell's equations - Reasoning:

Maxwell's equation in vacuum are

**∇∙E**= 0,**∇**×**E**= -∂**B**/∂t,**∇∙B**= 0,**∇**×**B**= (1/c^{2})∂E/∂t.

Using these equations we can derive the homogeneous wave equation for**E**and**B**and show that⊥

E**B**,**E**⊥**k**,**B**⊥**k**,**B**= (1/c^{2})∂(**k**/k)×**E**. - Details of the calculation:

For the given plane wave:

(**∇**×**E**)_{x}= ∂E_{z}/∂y - ∂E_{y}/∂z = k E_{0y}cos(ωt – kz + φ) = -∂B_{x}/∂t

Therefore B_{x}= (k/ω)E_{0y}sin(ωt – kz + φ) = -(E_{0y}/c)sin(ωt – kz + φ)

(**∇**×**E**)_{y}= ∂E_{x}/∂z - ∂E_{z}/∂x = -k E_{0x}cos(ωt – kz + φ) = -∂B_{y}/∂t

Therefore B_{y}= (k/ω)E_{0x}sin(ωt – kz + φ) = (E_{0x}/c)sin(ωt – kz + φ)

(**∇**×**E**)_{z}= ∂E_{y}/∂x - ∂E_{x}/∂y = -∂B_{z}/∂t = 0, therefore B_{z}= 0.

(We are not interested in constant fields.)

**B**= (1/c)(E_{0x}**j**- E_{0y}**i**)sin(ωt – kz + φ).

Starting with Maxwell's equations:

(a) Derive the wave equations for a light wave in vacuum. Write out solutions
for these equations for **E** and **B**.

(b) Show that the electric and magnetic fields are in phase, perpendicular to
each other and perpendicular to the direction of motion.

(c) Determine the relative magnitude of the **E** and **B** fields.

Solution:

- Concepts:

Maxwell's equations - Reasoning:

In regions where ρ and**j**are zero Maxwell's equations lead to the homogeneous wave equation for**E**and**B**. All solutions can be viewed as linear superpositions of sinusoidal plane wave solutions. Inserting these solutions into Maxwell's equations we derive (b) and (c). - Details of the calculation:

(a) Maxwells equations in SI units are

**∇∙E**= ρ/ε_{0},**∇**×**E**= -∂**B**/∂t,**∇∙B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t

Assume ρ and**j**are zero in the medium.

**∇**×(**∇**×**E**) =**∇**(**∇∙E**) -**∇**^{2}**E**= -(∂/∂ t)(**∇**×**B**) = -μ_{0}ε_{0}∂^{2}**E**/∂t^{2}.

**∇**^{2}**E**= μ_{0}ε_{0}∂^{2}**E**/∂t^{2}.

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= μ_{0}ε_{0}(∂/∂ t)(**∇**×**E**) = -μ_{0}ε_{0}∂^{2}**B**/∂t^{2}.

**∇**^{2}**B**= μ_{0}ε_{0}∂^{2}**B**/∂t^{2}.

μ_{0}ε_{0 }= 1/c^{2}.

(b) Each Cartesian component of**E**and**B**satisfies the 3-dimensional, homogeneous wave equation.

Sinusoidal plane wave solutions**E**(**r**,t) =**E**_{0 }exp(i(**k**_{i}**∙r**- ωt)),**B**(**r**,t) =**B**_{0 }exp(i(**k**_{i}**∙r**- ωt)) exist.**∇∙E**= ∂E_{x}/∂x + ∂E_{y}/∂y +∂E_{z}/∂z = i**k∙E**= 0**∇∙E**= 0 requires that**E∙k**= 0 for radiation fields, i.e. that**E**is perpendicular to**k**.

Similarly,**∇∙B**= 0 requires that**B∙k**= 0 for radiation fields.**∇**×**E**= -∂**B**/∂t requires that i**k**×**E**= iω**B**, i.e.**B**is perpendicular to**E**and**k**.

(c) B = (k/ω)E = E/c.

A time-dependent, vacuum electromagnetic field in three dimensions (x, y, z) at time, t = 0, is shown in the figure.

It has the following form:

**E**(**r**, t = 0) = **i ** E_{0}exp(-(z/a)^{2}), **
B**(**r**, t = 0) = 0.

(a) Evaluate ∂**E**/∂t at t = 0.

(b) Evaluate ∂**B**/∂t at t = 0.

(c) Evaluate ∂^{2}**E**/∂t^{2} at t = 0 and
show that **E** satisfies the wave equation at t = 0.

(d) What are the values of the fields **E**(**r**, t ) and **B**(**r**,
t ) for a general time t, satisfying the inequality ct/a >> 1.

(e) Sketch in a single diagram the fields found in (d).

Solution:

- Concepts:

Maxwell's equations - Reasoning:

Maxwell's equations relate partial derivatives of**E**and**B**with respect to space and time and yield the wave equation - Details of the calculation:

**∇∙E**= 0,**∇∙B**= 0,_{}**∇**×**E**= -∂**B**/∂t,**∇**×**B**= (1/c^{2})∂**E**/∂t

(a) Here**E**(**r**, t = 0) =**i**E_{0}exp(-(z/a)^{2}),**B**(**r**, t = 0) = 0.

At t = 0,**∇**×**B**= (1/c^{2})∂**E**/∂t = 0, ∂**E**/∂t = 0.

(b)**∇**×**E**= -∂**B**/∂t =**j**(∂E_{x}/∂z), ∂**B**/∂t =**j**(2z/a^{2}) E_{0}exp(-(z/a)^{2}).

(c) ∂^{2}**E**/∂t^{2}= c^{2}(∂/∂t)(**∇**×**B**) = c^{2}**∇**×∂**B**/∂t.

At t = 0, ∂^{2}**E**/∂t^{2}= c^{2}**∇**×(**j**(2z/a^{2}) E_{0}exp(-(z/a)^{2})).

∂^{2}**E**/∂t^{2}= -c^{2}**i**(∂/∂z)[(2z/a^{2}) E_{0}exp(-(z/a)^{2}))] = c^{2}**i**(∂^{2}/∂z^{2})[E_{0}exp(-(z/a)^{2}))] = c^{2}**∇**^{2}**E**.

**∇**^{2}**E**-( 1/c2)∂^{2}**E**/∂t^{2}= 0.satisfies the wave equation at t = 0.

E

(d) The given**E**can only satisfy the wave equation if it is the sum of a function of z - ct and another function of c + zt. The same holds for**B**.**E**(**r**, t) =**i**[(E_{0}/2)(exp(-((z- ct)/a)^{2}) + exp(-((z + ct)/a)^{2})],**B**(**r**, t) =**j**[(E_{0}/(2c))(exp(-((z - ct)/a)^{2}) - exp(-((z + ct)/a)^{2})].

(e) We have two Gaussian-shaped electromagnetic pulses one traveling into the positive and one in the negative z-direction.

In unbounded free space the electric and magnetic fields satisfy

**∇**∙**E** = **∇**∙**B** = 0, **∇**×**E** = -∂**B**/∂t,
**∇**×**B** = (1/c^{2})∂**E**/∂t,

and therefore the homogeneous wave equation.

Assume that at t = 0 **E**(**r**, t = 0) = **j** f(x) and **B**(**r**,
t = 0) = 0.

Find **E**(**r**,t) and **B**(**r**, t) for t > 0.

Solution:

- Concepts:

Maxwell's equations - Reasoning:

Maxwell's equations relate partial derivatives of**E**and**B**with respect to space and time and yield the wave equation - Details of the calculation:

**E**and**B**must satisfy the homogeneous wave equation,**∇**^{2}**E**= (1/c^{2})∂^{2}**E**/∂t^{2}.

We have**E**⊥**B**,**E**,**B**⊥ direction of propagation,**E**×**B**pointing in the direction of propagation.

**E**is perpendicular to the xz-plane.**∇**×**E**=**k**∂E_{y}/∂x =**k**∂f(x)/∂x.

**B**points in the ±z-direction. The EM wave therefore propagates along the ±x-direction.

The given**E**can only satisfy the wave equation if it is the sum of a function of x - ct and another function of x + ct. The same holds for**B**.**E**(**r**, t) =**j**½(f(x - ct) + f(x + ct)),**B**(**r**, t) =**k**½(f(x + ct) - f(x + ct)).

We have two electromagnetic pulses, one traveling into the positive and one in the negative x-direction.