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Solutions
1. Plum Pudding Model
(a) Find the corresponding electrostatic potential inside and outside the atom.
For r ≤ R
∇2Vin = − ρε00 .
The solution can be found by integrating twice,
ρ0 2
r + ra2 + b,
Vin = − 6ε
0
where a and b are constants. The potential at the center of the atom has to be finite, so a = 0.
Finally,
ρ0 2
Vin = − 6ε
r + b.
0
For r > R
∇2Vout = 0.
The solution can be found by integrating twice,
Vout = − cr + d.
But for large r, the potential has to go to zero, therefore d = 0.
Using the boundary condition that the potential has to be continuous at r = R:
ρ0 2
c
6ε0 R + b = R .
1
(b) Find the electrostatic vector-field inside and outside the atom.
For r R:
!Eout = −!∇Vout =
c
r̂.
r2
For r ≤ R:
!Ein = −!∇Vin =
ρ0!r
3ε0 .
Using the continuity of !E at r = R:
c
R2
=
ρ0 R
3ε0
⇔c=
ρ0 R3
3ε0
In summary:
!Ein =
ρ0!r
3ε0 ,
!Eout =
ρ0 R3 r̂
3ε0 r2
=
1 Qtotal
4πε0 r2 r̂,
"
2
ρ0 ! 2
0R
b = − ρ6ε
, and Vin = − 6ε
r + R2
0
0
(c) Electrons inside the Plum Pudding Model atom will oscillate. Explain why!
The potential energy of an electron inside the atom is
! 2
"
2 = Ar 2 + const
0
U = eV (r) = − eρ
r
+
R
6ε0
and the force on the electron is:
!F = q!Ein = − eρ0!R
3ε0
(d) Find the frequency of this oscillation.
Use
2ε0
2
0r
m∂t2 r = − eρ
2ε0 ⇔ ∂t r = − eρ0 r
Using
r = r0 eiωt
we can find
ω=
#
2
ρ0 e
3mε0
2. Dielectric Sphere
(a) Determine the boundary conditions for the given setup.
The boundary conditions can be summarized as follows
1. limr→∞ V (!r) = −E0 r cos θ
2. limr→0 V (!r) remains finite.
3. V (R)in = V (R)out
4. εr,in ∂rVin − ∂rVout = 0
(b) Find the potential inside and outside the dielectric sphere. Explain your approach! Use the
method of separation of variables in spherical coordinates to determine the potential inside and
outside the sphere. In combination with the first two boundary conditions, you will find that
Vin = ∑ Al rl Pl (cos θ ) and Vout = −E0 r cos θ ∑ rDl+1l Pl (cos θ )
l
l
Using the remaining boundary conditions, Fourier’s Trick and the orthogonality of the Legendre
Polynomials, we can find:
cos θ
3
0
Vin = − ε3E
z and Vout = −E0 r cos θ + εεrr −1
+2 R E0 r2
r +2
(c) Find the electric field !E and polarization !P inside the dielectric sphere.
!Ein = −!∇Vin = −∂zVin ẑ = − 3E0 ẑ
εr +2
and for a linear dielectric:
!P = ε0 χe !E =
3
εr −1
εr +2 ε0 χe E0 ẑ
(d) Sketch the electric field lines for all regions of this setup.
(e) Find the bound volume charge density ρb , and all the bound surface charge densities σb .
Since there are no free charge inside the sphere
ρb = 0.
The bound surface charge is
σb = !P · n̂ =
E0 (εr −1)
εr +2 ε0 χe cos θ .
4
3. Inductance
(a) In the quasistatic approximation, find the induced electric field as a function of distance s from
the wire. In the quasistatic approximation, the magnetic field of a wire is
B=
µ0 I
2πs .
Using Faraday’s law, we can find for the electric field:
%
$
%s
!E d!l = E(s0 )l − E(s)l = − d !B · d!a = − µ0 I dI 1% ds%
dt
2π dt s
s0
&
'
(
)
µ
I
ω
⇔ !E(s) = − 02π0 ln s sin(ωt) + K ẑ = 6 × 10−6 sin(ωt) + K ẑ.
(b) Is your answer valid for the limit s → ∞? Explain your answer. No, since ln s diverges for
large s The quasistatic approximation only holds for s ' ct.
(c) Find the self-inductance L of the rectangular coil.
The magnetic field inside a toroid is given by
B=
µ0 NI
2πs .
So, the flux through a single turn is
%
Φ = !B · d!a =
b
µ0 NI % 1
µ0 NIh
b
h
2π
s ds = 2π ln a .
a
Which means that the total flux is N times this, and the self-inductance is
Φ = LI ⇔ L =
µ0 N 2 h
b
2π ln a .
(d) In the quasi-static approximation, what emf is induced in the coil?
In the quasistatic approximation:
!B =
µ0
2πs φ̂ .
So,
φ1 =
µ0 I
2π
%b 1
a
µ0 Ih
b
s h ds = 2π ln a .
5
This is the flux through only one turn, so the total flux is N times Φ1 :
Φ=
µ0 Nh
b
2π ln a I0 cos(ωt).
So,
E = − dφ
dt =
µ0 Nh
b
−4
2π ln a I0 ω sin(ωt) = 2.61 × 10 sin(ωt)
V,
using
ω = 377 1s .
(e) Find the current I(t) in the resistor R.
Ir =
E
R
= 5.22 × 10−7 sin ωt A.
(f) Calculate the back emf in the coil, due to the current I(t). The back emf Eb is given by:
r
Eb = −L dI
dt ;
Now use the self-inductance of the square coils calculated in part (a):
L=
µ0 N 2 h
b
2π ln a
= 1.39 × 10−3 H.
Therefore,
Eb = −2.74 × 10−7 cos(ωt) V.
6
4. Momentum of Electromagnetic Fields
(a) What is the Poynting vector !S?
The Poynting vector is given by
!S =
1!
!
µ0 E0 × B0
The electric field of a parallel plate capacitor is given by
!E = − σ ẑ
ε0
So
!S =
σ B0
x̂
c2 µ0 ε0
(b) What is the momentum density of the electromagnetic field? The momentum density is given
by
!p =
1!
!0
E ×H
c2 0
⇒ !p = σ B0 x̂
(c) What is the total momentum of the electromagnetic field? The total momentum is given by
%
!Ptotal = !p dV = QB0 hx̂
(d) What is the impulse ∆!p in time ∆t experienced by each plate, as derived from the induced
electric field? How does this compare to the field momentum derived in part c)?
Now, we consider a closed loop in the x − z-plane of width a and height h. The magnetic flux
through this loop is:
%
!B0 d!a = Bb ahŷ
and for the induced electric field:
$
%
!Eind d!l = −∂t !B0 d!a = − B0 ah .
∆t
Taking the closed loop integral:
7
$
!Eind d!l = Eind 2a = − B0 ah ⇒ |Eind | = − B0 h
∆t
2∆t
The forces on the plates are now:
!F = q!E with !Eind = Eind x̂ at the top plate and !Eind = −Eind x̂ at the bottom plate.
So, the force on both plates with Q on the top plate and −Q on the bottom plate, is given by:
!F = Q B0 h x̂
2∆t
To find the momentum !p:
%
!p = !F dt =
QB0 h
2 x̂
So, each plate receives half of the momentum stored in the fields.
8
5. Dipole Radiation
(a) Find the electric field !E(r,t) and magnetic field !B(r,t) to leading order in powers of
terms of p̈0 (tr ) [i.e. the second time derivative of p0 (t) evaluated at the retarded time, tr ].
!1"
r
in
To find the electric field, use
!E = !∇V − ∂t !A
Since we only consider terms of powers
V (!r,t).
!∇V =
!1"
r
in terms of p̈0 (tr ), we only deal with the last term of
( !
")
r̂!p˙
1 1
1
˙ r ) r̂
∂
r̂
=
∂
r̂
!
p(t
r
r
4πε0
cr
4πε0 cr
!
"
1 r̂ ∂
˙ r ) ∂tr
r̂
!
p(t
= 4πε
cr
∂t
∂r
0
r
1 r̂!p¨
= − 4πε0 c2 r r̂
Now,
∂t !A =
µ0 ∂
4π ∂tr
*˙
!p(tr )
r
+
=
¨ r)
µ0 !p(t
4π r
Finally,
!E =
¨
¨
1 (r̂·!p)r̂−!p(tr )
2
r
4πε0 c
The magnetic field!B can be found, as
!B = !∇ × !A =
˙ r)
!p(t
µ0 !
4π ∇ × r
use the vector identity
*
+ *
+
!∇ × f!v = f !∇ ×!v = !v × !∇ f
*
+
! "
!∇ × 1 !p˙ = 1 !∇ × !p˙ −!p × !∇ 1
r
r
r
!1" 1
!
and ∇
= 2 r̂,which can be neglected.
r
Now,
r
!∇ × !p˙ = !∇ × p̂ ṗ = − p̂ × !∇ ṗ
! "
with !∇ ṗ = p̈!∇ (tr ) = p̈ − cr̂
9
So, finally:
!B =
"
µ0 !
µ0 1
¨
(
p̂
×
p̈r̂)
=
−
r̂
×
!
p
4π rc
4πrc
(b) Assume that !p(tr ) = p0 (t)ẑ. Show that !E(!r,t) = E(!r,t)θ̂ and !B(!r,t) = B(!r,t)φ̂ . Find expres-
sions for E(!r,t) and B(!r,t).
!E =
p̈((r̂·ẑ)r̂−ẑ)
1
r
4πε0 c2
=
p̈ cos θ r̂−ẑ
r
4πε0 c2
using
ẑ = cos θ r̂ − sin θ θ̂ ,
we finally find for the electric field:
!E =
p̈ sin θ θ̂
4πε0 c2 r
Similarly for the magnetic field:
!B = − µ0 p̈ (r̂ × ẑ) =
4πcr
µ0
4πcr p̈ sin θ
φ̂ .
(c) Find the power radiated to infinity by this time dependent charge distribution.
The total power is given by:
%
P = !S d!a
(d) Calculate the Poynting vector !S.
So, we need to find the Poynting vector:
!S =
=
1
µ0
*
+
!E × !B
µ0 p̈20
16π 2 c r2
sin2 θ r̂
So, the total power is then:
%
P = !S d!a =
=
µ0 p̈20 % sin2 θ 2
r sin θ
16π 2 c
r2
2
%
µ0 p̈0
3
8πc sin θ dθ
µ0 p̈20
= 6πc
10
dθ dφ