MIT OpenCourseWare
http://ocw.mit.edu
6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2009
Additional Problems - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 1.1
I
A
Figure 1: A diagram of two identical conducting balls suspended by essentially weightless strings of length l.
Figure 2: A diagram showing the forces on one of the balls (Image by MIT OpenCourseWare).
QQ
T =
Fgravity
Fcoulomb
Mg
2 2
⇒
=
=
cos θ
sin θ
cos θ
4π�0 s2 sin θ
M g4π�0 s2 sin θ
Q2
4
cos θ
= 1 ⇒ tan θ =
Q2
16π�0 s2 M
1
Additional Problems - Solutions
sin θ =
s
2
l
=
6.641, Spring 2009
� s �2
s
Q2
Q2
⇒ tan θ sin2 θ =
=
2
2l
16π�0 s M g 2l
64π�0 l2 M g
II
A
Figure 3: A diagram showing two charges, one at the center of another orbiting charge.
Figure 4: A diagram showing the forces on the orbiting charge (Image by MIT OpenCourseWare).
Fcentrif ugal = Fcoulomb
mω 2 R =
Q1 Q2
⇒ω=
4π�0 R2
Q1 = e
Q2 = ze
�
Q1 Q2
4π�0 R3 m
B
nh
L = mvr = mωR =
⇒ m2 ω 2 R4 =
2π
2
2
⇒R=
�
nh
2π
�2
=
2
4π�0 n4πh2
�0 h2 2
=
n
mQ1 Q2
me2 zπ
C
Hydrogen atom ⇒ z = 1
e = 1.6022 × 10−19 C
m = 9.1094 × 10−31 kg
2
m2 R4 Q1 Q2
mQ1 Q2 R
=
4π�0 R3 m
4π�0
Additional Problems - Solutions
6.641, Spring 2009
0 = 8.8542 × 10−12 Fm−1
h = 6.6261 × 10−34 Js
0 h2 2
1 = 5.2917 × 10−11 m ≈ 0.0529 nm
me2 1π
nh
1h
∼ 2.19 × 106 m/s
mvR =
⇒v=
=
2π
2πmRmin
Rmin =
III
Figure 5: A diagram showing a charge q moving at velocity v0 ix at a distance L from a screen in an electric
field E0 iz moving in a curved path until finally contacting the screen at y = h.
m
dx
d2 x
x
=0⇒
= v0 ⇒ x = v0 t ⇒ t =
v0
dt2
dt
d2 z
dz
q
0
*
= qE0 ⇒
= E0 t + vz
0
2
dt
dt
m
0
0
qE0 2
qE0 2
9E0 x2
*
t +
vz0
t+
z
⇒z=
t =
>
0 ⇒z =
2m
2m
2m v02
m
⇒ h = z(x = L) ⇒ h =
9E0 L2
2mv02
IV
A
1
1
2e
mvx2 − eV1 = mV02 ⇒ vx2 = v02 + V1
2
2
m
B
r
Ey
2e
From Lorentz force eqn: Ey − vx B0 = 0 ⇒ vx =
= v02 + V1
B0
m
s
V2
V2
2e
V22
2eV1
Since Ey =
⇒ 2 2 2 = v02 + V1 ⇒ v0 =
−
s
s B0
m
s2 B02
m
3
Additional Problems - Solutions
6.641, Spring 2009
Figure 6: A diagram of a cathode-ray tube.
C
mvx2
mvx
m
evx B0 =
⇒R=
=
B0 e
R
B0 e
�
v02 +
2e
V1
m
D
�
� m �2
�m�
1 m
2e
R=
v02 + V1 ⇒ R2 B02 =
v02 + 2V1
B0 e
m
e
e
� m �2 2V � m � R2 B 2
e
v02
1
0
�
−
+ 2
=
0
⇒
=
⇒
e
V0
e
V02
m
V12 + B02 R2 v02 − V1
V
d�v
�
� where H
= q�v × (µ0 H)
dt
�
� ix iy iz
�
d�v
⇒m
= qµ0 �� vx vy vz
dt
� 0 0 H0
m
= H0 iz ,
�
�
�
�
�
�
�v = vx ix + vy iy + vz iz
d�v
qµ0
=
(vy H0 ix − vx H0 iy )
dt
m
dvx
qµ0
dvy
qµ0
dvz
⇒
=
vy H0 (1),
=−
vx H0 (2),
= 0 (3)
dt
m
dt
m
dt
Taking the time derivative of (1) and using (2), get differential equation for vx and vy
⇒
d2 vx
qµ0 H0 dvy
−q 2 µ20 2
=
=
H0 v x
dt2
m
dt
m2
d 2 vy
qµ0 H0 dvx
d2 vy
q 2 µ20 H02
=−
⇒
=−
vy
2
2
dt
m
dt
dt
m2
�
�
�
�
qµ0 H0
qµ0 H0
⇒vx = A cos
t + B sin
t
m
m
�
�
�
�
qµ0 H0
qµ0 H0
vy = B cos
t − A sin
t
m
m
vz = vz0
4
Additional Problems - Solutions
6.641, Spring 2009
Initial conditions:
vx (t = 0) = vx0 = A ,
vy (t = 0) = vy0 = B
�
�
qµ0 H0
t
� m �
0 H0
vy0 cos qµm
t
vx (t) = vx0 cos
⇒
vy (t) =
vz (t) = vz0
−
�
�
qµ0 H0
t
� m �
0 H0
vx0 sin qµm
t
+ vy0 sin
NOTE:
|�v | =
�
vx2 + vy2 + vz2 ,
ω0 =
qµ0 H0
m
�
����
�0�
�
= vx20 cos2 (ω0 t) + vy20 sin2 (ω0 t) + 2v�
sin(ω
t)
cos(ω0 t)
x0 v
y0
�
� �
�1/2
����
2
2
2
�0�
�
+vy0
cos2 (ω0 t) + vx0
sin2 (ω0 t) − 2v�
sin(ω
t)
cos(ω
t)
+
v
x0 v
y0
0
�
z0
� �
=
�
�
�
�
�
2
2
2
vx0
cos2 (ω0 t) + sin2 (ω0 t) + vy0
cos2 (ω0 t) + sin2 (ω0 t) + vz0
=
�
vx20 + vy20 + vz20 ⇒ constant in time
A
Velocity in xy-plane is vxy =
�
Centripetal acceleration is
�
�
� v2 �
� xy �
�m
�
� r �
� �� �
2
vxy
where r is the radius of the circle
r
�
�
v2
�
� �� ⇒ m xy = qvxy µ0 H0 ⇒ r = mvxy
= �q�v × µ0 H
r
qµ0 H0
�
��
�
force due to
centripetal
acceleration
⇒ r=
vx20 + vy20
m
qµ0 H0
force caused
by magnetic
field
�
2 + v2
vx0
y0
5
Additional Problems - Solutions
6.641, Spring 2009
B
Figure 7: A diagram of a mass spectrograph.
� + q�v × µ0 H
�
f� = 0 = qE
� + q(v0 iy ) × (µ0 H0 iz )
= qE
� = −v0 µ0 Hix
E
V = −v0 µ0 H0 s
⇒
C
d = 2r = 2
⇒ d=
mv0
V
; and from part (b): v0 =
qB0
µ0 H0 s
for V = 100Volts
s = 1cm = 0.01m
µ0 H0 = B0 = 1Tesla
q = e = 1.6022 × 10−19 C
m = 1.67 × 10−27 kg (mass of proton and neutron)
2mV
qB02 s
2 × 24 × 1.67 × 10−27 × 100
= 5 × 10−3 ≈ 5.003mm
1.6022 × 10−19 × 1 × 0.01
⇒ for Mg24
d=
⇒ for Mg25
d ≈ 5.2116mm
⇒ for Mg26
d ≈ 5.42mm
6
Additional Problems - Solutions
6.641, Spring 2009
Problem 1.2
Figure 8: A diagram of two parallel plates connected to a power source creating a magnetron.
A
�
�
dvx
V0
m
= −e − + vy B0
dt
s
m
dvy
m dvy
dvx
m d2 vy
⇒
= evx B0 ⇒ vx =
=
dt
eB0 dt
dt
eB0 dt2
⇒
d 2 vy
ω 2 V0
eB0
+ ω02 vy = 0
where ω0 =
2
dt
B0 s
m
⇒ vy = A1 sin(ω0 t) + A2 cos(ω0 t) +
vy (t = 0) = 0 ⇒ A2 = −
vx (t = 0) = 0 ⇒
⇒ vy =
x=
�
V0
B0 s
V0
B0 s
�
dvy ��
= 0 ⇒
A1 = 0
dt �t=0
(1 − cos(ω0 t)) , vx =
vx dt =
V0
B0 s
1 dvy
=
ω0 dt
V0
B0 s
sin(ω0 t) = vx
�
V0
V0
(1 − cos(ω0 t)) , y = vy dt =
B sω
B0 s
� 0 0 ��
�
�
use B.C. x|t=0 = 0
�
�
sin(ω0 t)
t−
ω0
��
�
use B.C. y|t=0 = 0
B
xmax =
2V0
2V0 m
2V0 m
<s⇒
< s ⇒ B02 >
for cut-off
B0 sω0
es2
B0 s eB0
7
Additional Problems - Solutions
6.641, Spring 2009
C
Figure 9: A diagram of two concentric conducting cylinders connected to a voltage source creating a mag­
netron.
Electrons injected from r = a, φ = 0 with zero initial velocity
ir = cos φix + sin φiy , iφ = − sin φix + cos φiy
�
� dφ
dir
dφ
vφ
= − sin φix + cos φiy
= iφ
=
iφ
dt
dt
dt
r
�
� dφ
diφ
vφ
= − cos φix − sin φiy
= −ir
dt
dt
r
v = vr ir + vφ iφ
Acceleration
�v
�
� v �
dv
dvr
dir
dvφ
diφ
dvr
dvφ
φ
φ
= ir
+ vr
+ iφ
+ vφ
= ir
+ vr
iφ + iφ
+ vφ − ir
dt
dt
dt
dt
dt
dt
r
dt
r
�
�
�
2 �
vφ
dv
v v
r
a = ir dv
+ iφ dtφ + φr r
dt − r
a=
D
m
�
�
dv
−V0
= −e E + v × B ; E =
ir
dt
r ln ab
�
vφ2
dvr
−
ir component ⇒
dt
r
�
=
−eEr
evφ B0
−
m
m
=
eV0
− ω0 vφ
mr ln ab
�
vφ vr
evr B0
dvφ
+
=
iφ ⇒
= ω0 vr
dt
r
m
�
8
where ω0 =
eB0
m
Additional Problems - Solutions
6.641, Spring 2009
Use of hint:
dvr
dvr dr
dvr
d
=
= vr
=
dt
dr dt
dr
dr
�
1 2
v
2 r
�
dvφ
dvφ dr
dvφ
=
= vr
dt
dr dt
dr
iφ component
�
�
� v
vφ vr �
dvφ
vφ
φ
⇒ vr
+
= vr
+
= ω0 vr
dt
r
dr
r
�
��
�
1 d
r dr (vφ r)
1 d
(vφ r) = ω0
r dr
ω0 r2
⇒ rvφ =
+ constant
2
�
vφ (r = a) = 0 ⇒ vφ = ω20 r −
⇒
a2
r
�
�
a2
r−
r
ir component
�
vφ2
dvr
−
dt
r
�
d
=
dr
�
1 2
v
2 r
�
ω2
− 0
4r
�2
eV0
ω02
=
−
2
mr ln ab
�
a2
r−
r
�
�
��
�
�
�
eV0
ω02
a2
a2
eV0
ω02
a4
−
r
−
1
+
=
−
r
1
−
r2
r4
4
r
4
mr ln ab
rm ln ab
�
�
�
1 2
eV0
r
ω02 r2
a4
2
⇒ vr =
ln −
+ 2 − a ← using B.C. vr |(r=a) = 0
2
a
4 2
2r
m ln ab
�
d
dr
=
�2
eV0
r
ω2 �
ln − 02 r2 − a2
b
a 8r
m ln a
�
vr =
2eV0
b
m ln a
1 2
v
2 r
�
�
⇒
=
ln ar −
ω02
4r 2
[r2 − a2 ]
2
E
For cut-off ⇒ vr (r = b) < 0
⇒
�2
2eV0 �
b
ω02 � 2
�
ln
<
b − a2
2
b � a
�
4b
m�
ln a
�2
2eV0
e2� B02 � 2
⇒ � <
b − a2
2
2
m
�
4b m�
8b2 mV0
e(b2 −a2 )2
< B02 ⇐ cut-off condition
9
Additional Problems - Solutions
6.641, Spring 2009
Check: cylindrical geometry approaches planar geometry of (a) if b = a + s where s << a
�
�2
� 2
�2
2
b − a2 → (a + s) − a2
� �
�2
s �2
2
2
→ a 1+
−a
a
� �
�
�2
2s
→ a2 1 +
− a2
a
�
�2
2
2
→ �
a�
+ 2as − �
a�
2
→ (2as)
b2 ∼ a2
B02 >
8b2 mV0
e (b2 − a2 )
2
for s << a
B02 >
B02 >
8a2 mV0
2
8a�
mV0
= �
2 2
e (2as)
e4�
a�
s
2
2mV0
← agrees with (b)
es2
Problem 1.3
By problem:
�
ρ=
ρb r
b ;
ρa ;
r<b
b<r<a
� = 0 for r > a
Also, no σs at r = b, but non zero σs such that E
A
By Gauss’ Law:
�
Sr
� · d�a =
�0 E
�
ρdV ; SR = sphere with radius r
Vr
� has only radial component: E
� = Er lˆr
As shown in class, symmetry ensures E
LHS of Gauss’ Law:
�
Sr
� · d�a =
�0 E
2π
�
0
�
0
π
�
�
�0 Er lˆr ·
�
�
r2 sin θdθdφîr
�
��
�
d�
a in spherical coordinates
2
= �4πr
�� � Er �0
where A is the surface area of a sphere of radius r
A
10
Additional Problems - Solutions
6.641, Spring 2009
RHS of Gauss’ Law:
For r<b:
�
� r � 2π � π
ρb r
ρdV =
b
VR
0
0
0
����
���
� ����
r
φ
r2 sin θdθdφdr
�
��
�
dV - diff. vol. element
θ
4 πr4
πr4 ρb
ρb =
4 b
b
For r>b & r<a: (b<r<a):
=
�
� �
2π
�
π
ρb r 2
r sin θdθdφdr +
b
VR
0 0
0
�
�
3
4πρa r3 − b3
4πρ
bb
�
=
+
3
4
�
�
�
4
= πρb b3 + πρa r3 − b3
b<r<a
3
r
�
�
2π
�
ρdV =
b
0
B
Equating LHS and RHS
⎧
4
�πr ρ ;
⎪
⎨ 4
r<b
b
4b
� 3
4πr2 Er �0 =
3
3
4πρb b
4πρa (a − b )
⎪
⎩ �
+
; b<r<a
3
4
�
⎧ 2
⎪ r ρb ;
⎨
r<b
4�0 b
Er =
3
3
3
⎪
⎩ b ρb + ρa (r − b ) ; b < r < a
4�0 r2
3�0 r2
C
�
�
Again: n̂ · �0 E a − �0 E b = σs
� (r = a+ ) = 0
E
By part (a):
�
�
ρa a3 − b3
ρb b3
+
Er (r = a− ) =
4�0 a2
3�0 a2
�
�
� (r = a− ) , So:
σs = îr · −�0 E
�
�
��
ρa a3 − b3
ρ b b3
σs = −
+
4�0 a2
3�0 a2
11
0
π
ρa r2 sin θdθdφdr
Additional Problems - Solutions
6.641, Spring 2009
D
�
3
r<b
2
2
Qσ (r = a) = σs 4πa = −4πa
Qb = πb ρb
�
�
b < r < a Qa = 43 π a3 − b3 ρa
�
��
ρa a3 − b3
ρ b b3
+
4�0 a2
3�0 a2
Qσ = Qb + Qa + Qσ = 0
Problem 1.4
A
Figure 10: A diagram of a wire with z directed volume current with −z directed surface current at r = a.
(Image by MIT OpenCourseWare).
We are told current in +z direction inside cylinder r < b
Current going through cylinder:
�
= Itotal =
S
J� · d�a =
�
0
b
2π
�
0
� �
� J 2πb2
J0 r
0
îz · rdφdriˆz =
b
3
� �� � � �� �
�
J�
�|=
|K
d�
a
Total current in sheet
length of sheet (i.e., circumference of circle of radius a)
12
Additional Problems - Solutions
� ’s units are
Thus, K
�|=
|K
2
2
3 J0 πb
=
2πa
Amps
m
6.641, Spring 2009
, whereas J�’s units are
Amps
m2
J 0 b2
3a
� = − J0 b2 iˆz
K
3a
B
Figure 11: A diagram of the current carrying wire with a contour circle C centered on the z-axis with r < b
(Image by MIT OpenCourseWare).
Ampere’s Law
�
C
� · d�s =
H
�
S
J� · d�a +
d
dt
�
�
S
� · d�a
�0 E
�
��
� field, term is 0
no E
Choose contour C as a circle and S as the minimum surface that the contour bounds (as shown in
Figure 11)
Now solve LHS of Ampere’s Law
�
C
� · d�s =
H
2π
�
0
(Hφ iˆφ ) · (rdφ)iˆφ = 2πrHφ
� �� � � �� �
�
H
d�
s
13
Additional Problems - Solutions
6.641, Spring 2009
We assumed Hz = Hr = 0. This follows from the symmetry of the problem. Hr = 0 because
0. In particular choose S as shown in Figure 12a.
�
S
� · d�a =
µ0 H
Figure 12: A diagram of the wire with the choice for S as well as a diagram of the wire with the choice of
contour C (Image by MIT OpenCourseWare).
Hz is more difficult to see. It is discussed in Haus & Melcher, Chapter 1. The basic idea is to use the
contour, C (depicted in Figure 12b), to show that if Hz �= 0 it would have to be nonzero even at ∞, which
is not possible without sources at ∞.
Now for RHS of Ampere’s Law:
r<b
�
S
J� · d�a =
2π
�
0
�
r
0
�
J0 r � ˆ � � � ˆ �
iz · r dr dφiz
b
��
�
� �� � �
�
d�
a
J�
2J0 r3 π
3b
a > r > b
=
�
S
J� · d�a =
2π
�
0
�
b
�
0
J0 r � ˆ
iz
b
� �
� � 2π � r �
� �
�
� �
ˆ
· r dr dφiz +
0 · iˆz · r� dr� dφiˆz
b
�0
��
�
0
2
J0 b2 π
3
Equating LHS & RHS:
=
2πrHφ =
⎧
⎨
2
3
3b J0 r π ;
2
2
3 J0 b π ;
⎩
0;
r < b
a > r > b
r>a
14
Additional Problems - Solutions
⎧
J0 r2 ˆ
⎪
⎪
⎪
⎨ 3b iφ ;
�
J0 b2 ˆ
H=
iφ ;
⎪
⎪
⎪
⎩ 3r
0;
6.641, Spring 2009
r<b
a>r>b
r>a
Problem 1.5
Demos 1.3.1, 1.5.1 Coulombs’s Force Law and Measurements of Charge
• Rubbing of inflated balloons with a dry cloth
– Accumulation of charge on balloon surfaces
– Balloons repel each other because they have been charged to same polarity
– Charges on balloons induce image charges of opposite polarity on conducting surfaces
∗ Balloons are then attracted to these surfaces
• If we insert balloons in a Faraday cage
– We can measure the charges on the balloons
– It makes no difference to the measurement if balloons make contact with the inner surface
– If balloon is broken in the Faraday cage the charge is not removed when the broken balloon pieces
are removed
Demo 11.7.1: Steady state magnetic levitation
• Demonstrates magnetic forces due to conduction currents
• A pancake coil is excited by 60 Hz current and placed on an aluminum ground plane
• Typical currents of 20-30 Amps
• As current is raised, Lorentz force can overcome the coils weight and the pancake coil rises
• For ground plate thickness ∼ skin depth, the coil rises
• For ground plate thickness less than skin depth, the coil lifts up at higher current
• For ground plate thickness much less than skin depth, the coil does not lift up, because most of the
magnetic field penetrates through the ground plane.
15
Additional Problems - Solutions
6.641, Spring 2009
Problem 2.1
A
The idea here is similar to applying the chain rule in a 1D problem
d
dx
�
1
f (x)
�
�
d
=
df
�
1
f (x)
�� �
�
�
df
−f (x)
= 2
dx
f (x)
f (x) corresponds to |r − r� |. So, by diff. f (x) we get part of the answer to the derivative of
can just do it directly too.
�
|r − r� | = (x − x� )2 + (y − y � )2 + (z − z � )2
�
�
�
�
�
�
�
�
1
∂
1
∂
1
∂
1
= îx
+ îy
+ îz
�
|r − r� |
∂x |r − r� |
∂y |r − r� |
∂z |r − r� |
1
f (x) .
But we
So we can apply the trick above by just considering x, y, and z components separately.
�
∂
∂ ��
|r − r� | =
(x − x� )2 + (y − y � )2 + (z − z � )2
∂x
∂x
x − x�
= �
(x − x� )2 + (y − y � )2 + (z − z � )2
=
x − x�
|r − r� |
Similarly,
∂
∂x
∂
y − y�
∂
z − z�
�
|
|
|r − r� | =
and
r
−
r
=
.
∂y
|r − r� |
∂z
|r − r� |
�
1
|r − r�
�
=
∂
− ∂x
|r − r� |
|r − r� |2
and so on for y and z.
|r − r� |2 = (x − x� )2 + (y − y � )2 + (z − z � )2
so:
�
�
� − (x − x� )î + (y − y � )iˆ + (z − z � )iˆ
x
y
z
1
�
=
3/2
|r − r� |
[(x − x� )2 + (y − y � )2 + (z − z � )2 ]
�
Denominators = |r − r� |3/2 . Thus,
�
�
1
−(r − r� )
−1 (r − r� )
�
=
=
|r − r� |
|r − r� |3
|r − r� |2 |r − r� |
=
−îr� r
|r − r� |2
B
Follows from (A) immediately by substitution. Remember � is derived in terms of unprimed x, y, z. � does
not affect x� , y � , z � .
16
Additional Problems - Solutions
C
Φ(r) =
V
6.641, Spring 2009
ρ(r )dV 4πε0 |r − r |
C
ρ(r ) = charge density in mC3 . We have λ in units of m
. In this sense, ρ → ∞ at the ring. We can represent
this in cylindrical coordinates by ρ(r ) = λ0 δ(z)δ(r − a). Then we can evaluate the triple integral
λ0 δ(z)δ(r − a)rdrdφdz
=
4π0 |r − r |
2π
0
λ0 adφ
4π0 |r − r |
But, we can skip that unnecessary work by simply considering infinitesimal charges (adφ)λ0 around the ring.
y
x
dφ a
adφ
Figure 13: A ring of line charge with infinitesimal charge elements qd = λ0 adφ. (Image by MIT OpenCourseWare).
We only care about z axis in this problem as well, so, by symmetry, there is no field in the x and y
directions.
Φ(r) =
Φ(r) =
2π
0
λ0 (adφ)
4πε0
(a2 + z 2 )1/2
distance from
the charge
element λ0 adφ
to the point z
on the z-axis
λ0 a
2ε0 (a2 + z 2 )1/2
on the z-axis.
⎛
0
0
⎞
∂ ∂ ⎟
⎜ ∂ E = −∇Φ(r) = − ⎝ˆix Φ + îy Φ + îz Φ⎠
∂y
∂z
∂x
E = −îz
∂
∂z
λ0 a
2
2ε0 (a + z 2 )1/2
17
Additional Problems - Solutions
6.641, Spring 2009
aλ0 z
2ε0 (a2 + z 2 )3/2
Using the equation from the Problem 2.2 Statement with z component only (symmetry) and with ρ(r� )dV � →
λ0 adφ
� 2π
λ0 adφ cos θ
z
Ez (z) =
, cos θ = 2
2
2
4πε0 (z + a )
(a + z 2 )1/2
0
� 2π
λ0 az
dφ
=
2
2
3/2
(a + z ) 4πε0
0
λ0 az
=
2ε0 (a2 + z 2 )3/2
Limit |z | → ∞
�
a2 + z 2 → |z |
E = îz
λ0 a
2πλ0 a
Q
≈
≈
4πε0 |z|
4πε0 |z|
2ε0 (a2 + z 2 )1/2
Q = 2πλ0 a (total charge on loop). Φ(z) looks like potential from point charge in far field.
� Q
z>0
λ0 az
λ0 az
2
Ez =
≈
=
4πε−0Q|z|
3
2
2
3/2
2ε0 |z|
2ε0 (a + z )
z < 0
4πε0 |z|2
Φ(z) ≈
D
0r
From (C), Φ = 2ε0 (r2λ+z
2 )1/2 for a ring of radius r. But now we have σ0 , not λ0 . How do we express λ0 in
terms of σ0 ? Take a ring of width dr in the disk (see Figure 14). Total charge in the ring = (r)(2π) (dr)σ0 .
� �� �
circumference
a
r
dr
Figure 14: A line charge ring of width dr in the disk (Image by MIT OpenCourseWare).
charge
Line charge density = λ0 = total
= σ0 dr
length
So: λ0 = σ0 dr
σ0 rdr
dΦ =
2ε0 (r2 + z 2 )1/2
� a
� a
σ0 rdr
σ0
rdr
=
Φtotal =
2 + z 2 )1/2
2 + z 2 )1/2
2ε
2ε
(r
(r
0 0
0
0
��r=a
�
σ0 �� 2
σ0 �� 2
�
2
=
r +z �
=
a + z 2 − |z
|
2ε0
2ε0
r=0
�
�
σ0
1
1
E = −�Φtotal =
z √ −√
iz
2ε0
a2 + z 2
z2
18
Additional Problems - Solutions
6.641, Spring 2009
E
As z → ∞,
2
2 1/2
(a + z )
Φtotal
→
a2
→ |z| +
;
2|z |
2
2 −1/2
(a + z )
1
→
|z |
�
�
a2
1− 2
2z
πa2 σ0
4�0 π |z |
2
¯ → πa σ0 iz
E
4πε0 z 2
just like a point charge of σ0 πa2 .
F
√
As a → ∞, z in the a2 + z 2 can be neglected, so
σ0
Φtotal →
[a − |z|]
2ε0
�
� � σ0
z>0
σ0 z 1
ε0
Ez →
− 0 = 2−σ
0
2ε0 |z|
z<0
2ε
0
just like a sheet charge.
G
For λ(φ) = λ0 sin φ
� 2π
λ(φ)a
a
√
√
dφ =
λ0 sin φdφ
4π�0 a2 + z 2
4π�0 a2 + z 2 0
0
�2π
�
aλ0
√
=
(− cos φ)�� = 0 along z axis
2
2
4π�0 a + z
0
The electric potential along the z axis is zero. It is not possible to find the electric field along the z-axis
using the above result for the scalar electric potential value along the z-axis.
�
2π
Φ=
One cannot use Equation (2) from the problem statement to find E in this case.
H
�
ρ(r� )ir� r
λ(r� )ir� r dl�
�
dV
=
� 2
� 2
V � 4π�0 |r − r |
L� 4π�0 |r − r |
The charge density along the hoop will have a sinusoidal variation as a function of φ as shown in Figure 16
with zero net charge on the hoop.
Due to odd symmetry of the charge distribution the z component of the E field will cancel out as shown
in Figure 15.
�
E(r) =
i.e., dEz |φ = −dEz |φ+π
Similarly due to symmetry with respect to the yz plane, the x component of the E field also cancels as
shown in Figure 17.
Thus the resulting E field is in −y direction only!
19
Additional Problems - Solutions
6.641, Spring 2009
z
d E1

d E

d E2
y
a

x
Figure 15: A diagram showing that the E field generated from the hoop of line charge with odd symmetry
in angle φ is transverse to the z axis in the −iy direction (Image by MIT OpenCourseWare).
|dE sin θ| gives the magnitude of the E-field on the xy plane due to small line charge component dq = λadφ
λ(r� ) = λ0 sin φ
r = ziz , r� = air
r − r� = ziz − air , |r − r� | =
ir� r =
r − r�
,
|r − r� |
�
z 2 + a2
dl� = adφ
Figure 16: A graph showing the line charge density λ0 sin φ along the hoop as a function of φ (Image by
MIT OpenCourseWare).
20
Additional Problems - Solutions
6.641, Spring 2009
y
Contribution due
to charge at
3

= and =
etc.
2
2

x
Figure 17: A diagram of the electric field components from hoop charge elements looking down along the
z-axis showing that the x components cancel and the y components add in the −y direction (Image by MIT
OpenCourseWare).
2π
λ0 sin φ(ziz − air )adφ
, ir = cos φix + sin φiy
4π�0 (z 2 + a2 )3/2
φ=0
�� 2π
�
� 2π
−λ0 a2 sin φ cos φdφ
Ex =
=
0
sin
φ
cos
φdφ
=
0
2
2 3/2
φ=0 4π�0 (z + a )
0
�� 2π
�
� 2π
−λ0 a2 sin2 φdφ
−λ0 a2
2
Ey =
=
sin
φdφ
=
π
2
2 3/2
4�0 (z 2 + a2 )3/2
φ=0 4π�0 (z + a )
0
�� 2π
�
� 2π
λ0 z sin φadφ
Ez =
=0
sin φdφ = 0
2
2 3/2
φ=0 4π�0 (z + a )
0
�
E=
Contribution due to
d E∣ and d E∣
y
Contribution due to
d E∣− and d E∣2 −
E

x
Figure 18: A diagram of the electric field components looking down along the z-axis (Image by MIT OpenCourseWare).
21
Additional Problems - Solutions
E=−
6.641, Spring 2009
λ0 a2
1
λ0 a2
πiy =
−
iy
2
2
3/2
4π�0 (a + z )
4�0 (a2 + z 2 )3/2
λ0 a2
−λ0 a2
−py
lim E(r = 0, z) = lim −
i
≈
iz =
y
�
�
3
� a �2 3/2
z →∞
z→∞
4�0 |x|
4π�0 |z |3
� �� �
4�0 z 3 1 +
z
z>>a
p = λ0 πa2 iy → acts like field due to a dipole charge with dipole moment py .
I
Surface charge density σ(φ) = σ0 sin φ
We can use the method of superposition using the result obtained in part (a)
Figure 19: A diagram showing the method of superposition using discretized rings of line charge to approx­
imate a surface charge distribution (Image by MIT OpenCourseWare).
a
�
2π
�
Φ=
0
a
�
0
2π
�
=
0
0
�
σ(φ)
�
r� dφdr�
4π�0 (r� )2 + z 2
σ sin φ
�0
dφr� dr�
�
2
2
4π�0 (r ) + z
��
�
0
= 0 along z axis
So it is not possible to use Equation (2) in the problem statement to find the electric field along the
z-axis.
J
Again using method of superposition with the result of part (b) for a hoop of radius r�
dE = −
dλ0 (r� )2
4�0 ((r� )2 + z 2 )
i =
3/2 y
−σ(r� )2 dr�
iy
4�0 ((r� )2 + z 2 )3/2
22
Additional Problems - Solutions
⇒E=
�
a
r � =0
−
6.641, Spring 2009
σ0 (r� )2
4�0
((r� )2
+
i dr
3/2 y
z2)
�
��
�
� �a
�
�
�
−�
+ ln r + (r� )2 + z 2 �
�
2
2
��
(r ) + z
r =0
�
�
�
�√ � �
�
σ0
a
=−
iy − √
+ ln a + a2 + z 2 − ln z 2
4�0
a2 + z 2
�
�
��
√
σ0
a
a + a2 + z 2
E=−
iy − √
+ ln
4�0
|z|
a2 + z 2
σ0
=−
iy
4�0
�
r�
⎛
⎞
�
�
�
σ0 ⎝ −a|z|
a
a 2
�
=−
+ ln
+ ( ) + 1 ⎠ iy
4�0
|z|
|z|
( a )2 + 1
|z|
=
−σ0 a3
−σ0 a3
i
=
iy
y
4�0 3|z |3
12�0 |z |3
−py
−σ0 a3
=
4π�0 |z |3
12�0 |z |3
py =
σ0 πa3
3
Problem 2.2
A
By the divergence theorem:
i
�
V
�
� · (� × A)dV
=
�
S
� · d�a
(� × A)
where S encloses V . By Stokes’ Theorem:
ii
�
S�
� ) · d�a =
(� × A
�
C
� · d�l
A
Suppose S is as in Figure 20
�
and S is as in Figure 21
�
�
i.e. S is the same as S, except for the curve C, which makes S slightly unclosed. Now consider limit as
C → 0 (Figure 22)
�
�
�
� · d�l = 0. By equation (ii), (� × A
� ) · d�a = 0. By equation
In limit C → 0, S → S. If C is 0, then C A
S
�
� )dV = 0. Since V can be any volume, argument of integral must be identically 0.
(i), V � · (� × A
� =0
� · (� × A)
23
Additional Problems - Solutions
6.641, Spring 2009
S
Figure 20: Closed surface S (Image by MIT OpenCourseWare).
S
!
C
�
Figure 21: Open surface S (Image by MIT OpenCourseWare).
Figure 22: Limit as C → 0 (Image by MIT OpenCourseWare).
B
� = Ax i�x + Ay i�y + Az i�z
A
�
� i�x
� ∂
�
� × A = �� ∂x
� A
x
∂
∂
∂
� = i�x
+ i�y
+ i�z
∂x
∂y
∂z
�
�
�
�
�
�
�
i�z ��
∂
∂ �
�x ∂ Az − ∂ Ay + i�y ∂ Ax − ∂ Az + i�z ∂ Ay − ∂ Ax
=
i
∂y
∂z �
∂y
∂z
∂x
∂y
∂z
∂x
Ay Az �
�
�
�
�
�
�
∂ ∂Az
∂Ay
∂ ∂Ax
∂Az
∂ ∂Ay
∂Ax
�
−
−
−
� · (� × A) =
+
+
∂x ∂y
∂z
∂y
∂z
∂x
∂z
∂x
∂y
=
i�y
∂ 2 A�
∂ 2 A�
∂ 2 A�
∂ 2 A�
∂ 2 A�
∂ 2 A�
z
y
x
z
y
x
� − � + � − � + � − � = 0 using interchangability of partial derivatives
∂x∂y
∂x∂z
∂y∂z
∂y∂x
∂z∂x
∂z∂y
�
�
�
�
�
�
� = 0 in spherical coordinates
� · (� × A)
24
Additional Problems - Solutions
6.641, Spring 2009
� = Ar i�r + Aθ i�θ + Aφ i�φ
A
∂
1 ∂Aφ
� = 1 ∂ (r2 Ar ) + 1
�·A
(sin θAθ ) +
2
r ∂r
r sin θ ∂θ
r sin θ ∂φ
�
�
�
�
1
∂Ar
1
∂Aθ
1 ∂Aφ
= 2 2rAr + r2
+
cos θAθ + sin θ
+
r
∂r
r sin θ
∂θ
r sin θ ∂φ
=
2
∂Ar
1
1 ∂Aθ
1 ∂Aφ
Ar +
+ cot θAθ +
+
r
∂r
r
r ∂θ
r sin θ ∂φ
�
�
�
�
�
1
∂
1 ∂Aθ
1 ∂Ar
1 ∂
1 ∂
1 ∂Ar
�
�
−
(sin θAφ ) −
+iθ
(rAφ ) +iφ
(rAθ ) −
r sin θ ∂θ
r sin θ ∂φ
r sin θ ∂φ
r ∂r
r ∂r
r ∂θ
�
�
�
�
�
�
��
1
∂Aφ
1 ∂Aθ
1 ∂Ar
1
∂Aφ
−
−
= i�r
cos θAφ + sin θ
+ i�θ
Aφ + r
+
r sin θ
∂θ
r sin θ ∂φ
r sin θ ∂φ
r
∂r
� �
�
�
1
∂Aθ
1 ∂Ar
−
+i�φ
Aθ + r
r
∂r
r ∂θ
� = i�r
�× A
�
�
�
�
�
�
�
1
1 ∂Aφ
1 ∂Aθ
1 ∂Ar
1
∂Aφ
1
∂Aθ
1 ∂Ar
�
�
�
−
− Aφ −
−
= ir
cot θAφ +
+ iθ
+ iφ Aθ +
r
r ∂θ
r sin θ ∂φ
r sin θ ∂φ
r
r
∂r
r ∂θ
∂r
�
�
�
�
2 1
1 ∂Aφ
1 ∂Aθ
∂ 1
1 ∂Aφ
1 ∂Aθ
�
−
−
⇒ � · (� × A) =
cot θAφ +
+
cot θAφ +
r r
r ∂θ
r sin θ ∂φ
∂r r
r ∂θ
r sin θ ∂φ
�
�
�
�
1
1 ∂Ar
1
∂Aφ
1 ∂
1 ∂Ar
1
∂Aφ
− Aφ −
+ cot θ
+
− Aφ −
r
r sin θ ∂φ
r
∂r
r ∂θ r sin θ ∂φ
r
∂r
�
�
1
∂ 1
∂Aθ
1 ∂Ar
−
+
Aθ +
r sin θ ∂φ r
∂r
r ∂θ
� 1 ∂A�
� 1
�
2
2 �
∂A
∂A�
�� 2 ∂A�
�� 1
θ
�θAφ + ��φ −
�θAφ + cot�θ� φ − ��φ
= 2�
cot
− 2�
cot
�
2
2
2
�
r�
sin θ ∂φ �
r
�
�r �� � �
�r ��∂θ � �
�r �� ∂r � �
�r ��∂θ �
�
��
� � �� � �
VII
+
IV
II
IV
�
�
�
�
1 ∂2A
1 1 �∂A
1 �
∂ 2�
A�
1 cos �
θ ∂A
1
∂A�
�� 1
�
�
φ
θ
θ
r
φ
��
�
�
+
−
+
−
cot
θA
−
cot
θ
�
�
�
φ
�
2�
2� 2
2�
�
�
�
∂φ
r
r
∂r∂θ
r
sin
θ
∂φ
r
sin
θ
∂r∂φ
r
r
∂r
sin
θ
�
� �� � �
� �� � �
�
��
�
�
��
� �
�
��
� �
��
� �
VI
−
VII
III
III
V
VII
I
II
�
2 �
�
�
�
1 cos �
θ ∂A
1 �
∂�
Ar
1 ∂A
1 ∂2A
1 �
∂A
1 �
∂ 2�
A�
�
�
r
φ
φ
θ
θ
��
+
−
−
+
+
�
�
�
�
�
2�∂θ
2�
�
�
r2�
r2�
sin θ ∂θ∂φ �
r
r
∂θ∂r
r
sin
θ
∂φ
r
sin
θ
∂φ∂r
sin2 θ ∂φ �
� �� � �
�
��
� �
��
� � �� � �
�
��
� �
�
��
�
I
IX
IV
VI
2 �
1
∂�
Ar
− 2 ��
�
r
sin
θ
∂φ∂θ
�
�
��
�
IX
=0
25
III
V
Additional Problems - Solutions
6.641, Spring 2009
Problem 2.3
A
Cartesian
hx = 1
hy = 1
hz = 1
Cylindrical
hr = 1
hφ = r
hz = 1
Spherical
hr = 1
hθ = r
hφ = r sin θ
B
∂f
∂f
∂f
du +
dv +
dw
∂u
∂v
∂w
= �f · dl
�
�
= �f · hu duiu + hv dviv + hw dwiw
df =
(�f )u =
1 ∂f
1 ∂f
1 ∂f
; (�f )v =
; (�f )w =
hu ∂u
hv ∂v
hw ∂w
1 ∂f
1 ∂f
1 ∂f
iu +
iv +
iw
hu ∂u
hv ∂v
hw ∂w
�f =
C
dSu = hv hw dvdw; dSv = hu hw dudw; dSw = hu hv dudv
dV = hu hv hw dudvdw
D
�
Φ=
S
A · dS =
�
�u
�
Au hv hw dvdw −
Au hv hw dvdw
��
� � u−Δu ��
�
1
�
�
� 1
+
Av hu hw dudw − Av hu hw dudw
v+Δv
v
��
�
�
��
� �
2�
2
�
�
+
Aw hu hv dudv −
Aw hu hv dudv
w+Δw
w
��
�
�
��
� �
3�
3
�
�
Au hv hw |u − Au hv hw |u−Δu
Av hu hw |v+Δv − Av hu hw |v
Aw hu hv |w+Δw − Aw hu hv |w
=
+
+
ΔuΔvΔw
Δu
Δv
Δw
�
�
A· dS
A · dS
S
S
�·A=
lim
=
ΔV
h
h
h
u v w ΔuΔvΔw
Δu → 0
Δv → 0
Δw → 0
26
Additional Problems - Solutions
=
1
hu hv hw
�
6.641, Spring 2009
∂(hv hw Au ) ∂(hu hw Av ) ∂(hu hv Aw )
+
+
∂u
∂v
∂w
�
Curl
hv  v
hw  w
v , w
L
Figure 23: A diagram depicting how to calculate Curl for generalized right-handed orthogonal curvilinear
coordinates (Image by MIT OpenCourseWare).
�
(� × A)u =
�
L
A · dl
L
lim
Δv → 0
hv hw ΔvΔw
Δw → 0
�
� �
�
A · dl = Av hv Δv|w − Av hv Δv|w+Δw + Aw hw Δw|v+Δv − Aw hw Δw|v
1
hv hw
(� × A)u =
��
� �
��
Av hv |w − Av hv |w+Δw
Aw hw |v+Δv − Aw hw |v
+
Δw
Δv
lim
Δv → 0
Δw → 0
�
�
1
∂(hw Aw ) ∂(hv Av )
−
=
hv hw
∂v
∂w
Similarly
�
∂(hu Au ) ∂(hw Aw )
−
∂w
∂u
�
�
1
∂(hv Av ) ∂(hu Au )
−
(� × A)w =
hu hv
∂u
∂v
1
(� × A)v =
hu hw
�
E
1
� f = � · (�f ) =
hu hv hw
2
�
∂
∂u
�
hv hw ∂f
hu ∂u
�
∂
+
∂v
27
�
hu hw ∂f
hv ∂v
�
∂
+
∂w
�
��
hu hv ∂f
hw ∂w
Additional Problems - Solutions
6.641, Spring 2009
Problem 2.4
Demo 4.7.1: Charge Induced in Ground Plane by Overhead Conductor
• This problem is analogous to high voltage power line over earth problem.
• We can use image charge assumption to determine the induced charge on ground plane.
• We make use of a plane probe, insulated from the ground plane, to measure the charge induced on it
by the cylindrical conductor.
• The surface charge density distribution is proportional to the voltage applied to the cylinder.
• Because the probe voltage is time derivative of the applied voltage, the probe signal is 90◦ out-of-phase.
• As the probe is moved out from below the conductor cylinder, charge induced on its surface decreases.
• Electric field at low frequencies do not penetrate conducting body.
Demo 10.2.1: Edgerton’s Boomer
• Illustrates induction of a current in a conductor, subjected to a time varying magnetic field.
• Illustrates the interplay of laws of Faraday, Ampere, and Ohm.
• MQS conditions apply.
• 4kV capacitor voltage.
• Use of a coil probe to determine voltage induced by time varying magnetic field.
• Magnetic field produced by the coil is non-uniform, similar to that of a dipole.
• Prediction of an arc due to high electric field intensity between almost touching wires.
• Can be used to chape materials.
• Or shoot up metal plates.
Problem 3.1
Demo 8.4.1: Surface used to define flux linkage
• Copper wires were wound on a circular wooden rod, and coil inductances are measured.
• If number of turns is doubled, the measured inductance value is quadrupled
• When number of coils is doubled as well as the length, the inductance value is doubled
• Results agree well with the theoretical equation: L =
µ0 AN 2
d
Demo 8.2.1: Field of circular cylindrical solenoid
• Magnetic field of a large cylindrical solenoid is measured using magnetometer probe and transverse
probe
• Field insude is uniform and in the axial direction, the field outside the solenoid is close to 0.
28
Additional Problems - Solutions
6.641, Spring 2009
• Circular cylindrical solenoid is analogous to plane parallel capacitors for having uniform B-field inside
and zero B-field outside
• A transverse probe (which measures B-field intensity transverse to its surface) is used to determine
that the B-field inside is axial.
• Through a slit cut on the cylindrical solenoid, the transverse probe is used to observe the discontinuity
of the magnetic field intensity between inside and outside.
Demo 8.2.2: Field of square pair of coils
• 2 square coils 45cm apart, each has 50 turns and size of 45cm length.
• Axial magnetometer probe is used to measure the intensity of axial magnetic field
• Theoretical curve is well matched (see Figure 24)
Same direction
currents
Opposite direction
currents
1
d
−d
x
0
Figure 24: Two graphs showing current along the x-axis of the coil with same direction current and opposite
direction current (Image by MIT OpenCourseWare).
Problem 4.1
A
Linear media ⇒ J = σE , by symmetry E is only in x direction, thus current density J is also in x direction
E = Ex ix , J = Jx ix
σ = σ(x) − a function of x
Conservation of charge � · Jf +
∂ρf
In DC steady state
=0
∂�
⇒ � · Jf = 0 ⇒
For σ(x) =
∂ρf
∂t
= 0
d
Jx = 0 ⇒ Jx = J0 = constant
dx
⇒ J = J0 ix
σ0
1 + xs
29
Additional Problems - Solutions
6.641, Spring 2009
x
l
s
−
V0

 ,  x =
0
1
x
s
0
Depth d
Figure 25: A diagram of parallel plate electrodes enclosing a lossy dielectric with x dependent conductivity
(Image by MIT OpenCourseWare).
J = σE ⇒ Jx = σ(x)Ex ⇒ Ex =
�
s
s
�
Ex dx = V0 =
0
0
⇒ J0 =
Jx
J0
J0 �
x�
= σ0 =
1+
σ(x)
σ0
s
1+ x
s
J0 �
x�
J0
1+
dx =
σ0
s
σ0
�
��s
x2 ��
J0 �
s � 3 J0 s
x+
=
s
+
=
2s �0
σ0
2
2 σ0
2σ0 V0
3s
⇒ Ex (x) =
�
2�
σ�
x�
0 V0
1+
=
3s�
σ�
s
0
2V0
3s
�
�
1 + xs
Total current ⇒ I = J0 (Area) = J0 ld =
⇒ Resistance =
2σ0 V0 f d
3s
V0
3s
=
I
2σ0 ld
B
� · D = ρf ⇒ ρf = �
ρf =
dEx
d
=�
dx
dx
�
�
2V0 �
x�
1+
3s
s
�2V0
3s2
Boundary conditions at x = 0 and x = s will provide the surface charges densities (see Figure 26).
30
Additional Problems - Solutions
6.641, Spring 2009
D2
ix
− − −− − −− − −− − − − − −− − −− − −− − −
−
V0
D1


D2
ix


D1
Figure 26: A diagram showing the parallel plate electrodes with displacement field vectors showing the
sources of the surface charge densities (Image by MIT OpenCourseWare).
n · (D2 − D1 ) = σsf
⇒ at x = 0 σsf = �Ex |x=0 =
at x = 0 σsf = −�Ex |x=s
2�V0
3s
4�V0
=−
3s
C
�
qtotal volume = ld
s
ρf dx = ld
0
2�V0
2�V0 ld
s=
3s2
3s
at x = 0 qsurface = ldσs f |x=0 = 2�V3s0 ρd
at x = s qsurface = ldσsf |x=s = − 4�V3s0 ld
�
⇒ qtotal surface = −
2�V0 ld
= −qtotal volume
3s
qsurface (x = 0) + qsurface (x = s) + qtotal volume = 0
Problem 4.2
A
ρf = 0 inside the dielectric media ⇒ use Gauss’ Laws for (a < r < b)
⇒ � · D = 0 (due to symmetry only radial D component exists)
⇒�·D =0=
1 ∂ 2
c
(r Dr ) = 0 ⇒ r2 Dr = c ⇒ constant ⇒ Dr = 2
2
r ∂r
r
31
Additional Problems - Solutions
6.641, Spring 2009
1
r =
b
1
r
a

v0
−
a
Figure 27: Concentric spherical electrodes enclose a dielectric with permittivity that varies with r with no
volume charge in the dielectric (Image by MIT OpenCourseWare).
�
�
Dr
c �
r�
c
1
1
= 2
1+
⇒ Er =
+
�(r)
r �1
a
�1 r 2
ra
�
�
�
��b
� b
� b
�
c
1
c
1 1
V0 =
Er dr =
r−2 + r−1 dr =
− + ln r ��
a
�1
r
a
r=a
r=a �1
a
�
�
c
1 1
1 1
=
− + ln b + − ln a
�1
a a
b a
�
�
c 1 1 1 b
V0 =
− + ln
�1 a b a a
⇒ �(r)Er = Dr ⇒ Er =
⇒ c=
⇒ E=
1
a
V 0 �1
− + a1 ln ab
1
a
1
b
V0
− 1b + a1 ln ab
�
1
1
+
r2
ra
�
ir for a < r < b
�
�
∂Φ
V0
1
11
E = −�Φ ⇒ Er = −
= 1 1 1 b
+
r2
ar
∂r
a − b + a ln a
�
� �
V0
1
11
⇒Φ=−1 1 1 b
+
dr
2
r
a
r
−
+
ln
a
b
a
a
⎛
⎞
V0
1 1
= − 1 1 1 b ⎝− + ln r + ����
d ⎠
r
a
−
+
ln
a
b
a
a
constant
�
�
V0
1 1
Φ|r=a = V0 = − 1 1 1 b − + ln a + d
a a
a − b + a ln a
1�� 1 1
1 � 1�� 1 �
− + ln b − �
ln a = − �
ln a − d
a
a
�a b a
�a �
�
1 1
d = − ln b or use Φ|r=b = 0 to get d
b a
�
�
V0
1 1 1 r
⇒ Φ(r) = − 1 1 1 b − + + ln
for a < r < b
r
b a b
a − b + a ln a
32
Additional Problems - Solutions
6.641, Spring 2009
B
n · (D2 − D1 ) = σSR
Using B.C. at r = a and at r = b :
ir
ir
D1
D2
a
b
Figure 28: The surface charge at r = a is D2r (r = a) and at r = b is −D1r (r = b) (Image by MIT
OpenCourseWare).
at r = a σsf = Dr |r=a =
1
a2
at r = b σsf = −Dr |r=b = −
1
a
−
1
b2
1
a
V0 �1
V0 �1
=
2
+ a1 ln ab
a − ab + a ln ba
1
b
−
V 0 �1
=
+ a1 ln ab
1
b
b2
a
−V0 �1
2
− b + ba ln ab
C
Total charge on inner electrode
q = σs |r=a 4πa2 =
4πa2 V0 �1
q
⇒ capacitance =
=
2
V
a − ab + a ln ab
1
a
−
4π�1
+ a1 ln ab
1
b
Problem 4.3
⎧
⎨ ρ0 r 2
ρf (t = 0) =
a20
⎩
0
0 < r < a0
r > a0
Charge relaxation
∂ρf
σ
∂ρf
Conservation of charge: � · J f +
= 0 ⇒ ρf +
=0
∂t
�
∂t
⎫
ρf ⎪
Gauss’ Law � · E =
⎬
ρf
�
� · J f = σ� · E = σ
⎪
�
⎭
Linear Media Jf = σE
33
⎫
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎭
⇒ ρf (t) = ρf (t = 0)e−t/τ , τ =
�
σ
Additional Problems - Solutions
6.641, Spring 2009
a0
a1
 f t
1 
0
Figure 29: An infinitely long cylinder of radius a0 with initial charge density ρf (t = 0) =
ρ0 r2
for r < a0
a20
and zero for r > a0 (Image by MIT OpenCourseWare).
⎧
⎨ ρ0 r2 −t/τ
e
⇒ ρf (r, t) =
a20
⎩
0
; 0 < r < a0
;r > a0
Using Gauss’ Law
Gaussian contour
a0
r
a1
length L
Figure 30: An infinitely long cylinder showing a Gaussian Contour for a0 < r < a1 (Image by MIT OpenCourseWare).
�
S
�E · da =
�
ρf dV
V
2πrLEr � = charge enclosed
34
Additional Problems - Solutions
�
��
2πr
LEr � = �
2π�
�
6.641, Spring 2009
r
0
ρf (r� , t)r� dr� ⇒ Er =
1
r�
r
�
0
�
ρf (r� , t)r� dr�
��
�
charge enclosed
⎧
�
�
�
1 ρ0 −t/τ r4
ρ0 r3 −t/τ
1 r ρ0 (r� )2 −t/τ � �
⎪
⎪
e
r
dr
=
e
=
e
⎪
⎪
⎪
r� 0
a20
r� a20
4
4a20 �
⎪
⎪
⎪
⎪
⎪
�
�
⎨ 1 � a0 ρ (r� )2
1 ρ0 −t/τ a40
ρ0 a20 −t/τ
0
−t/τ � �
E r = ir
e
r
dr
=
e
=
e
⎪
r� 0
a20
r� a20
4
4�r
⎪
⎪
⎪
⎪
⎪
�
�
� a0
⎪
⎪
1
ρ0 (r� )2 � �
1
ρ0 a40
ρ0 a20
⎪
⎪
⎩
dr
=
=
r
r�0 0
a20
r�0 a20 4
4�0 r
; 0 < r < a0
; a0 < r < a1
; r > a1
Using B.C., to find surface charge at r = a,
i r =n
D2
D1

a1
0
Figure 31: An infinitely long cylinder showing vectors D1 and D2 that determine the surface charge density
σsf (Image by MIT OpenCourseWare).
n · (D2 − D1 ) = σsf
σsf |r=a1 = �0 Er |r=a+ − �Er |r=a− = �0
1
⇒ σsf =
1
ρ0 a20
ρ0 a20 −t/τ
−�
e
4a1 �0
4a1 �
�
ρ0 a20 �
1 − et/τ
4a1
Problem 4.4
Goal: Place an image charge (or image charges) inside the cylinder such that cylinder remains an equipotential
surface. This simplifies the problem to that of infinitely long line charges.
Figure 33 shows the E field due to a line charge λ
�
�0
S
E · da =
�
ρdv
V
35
Additional Problems - Solutions
6.641, Spring 2009
c
R

D
Figure 32: A diagram of an infinitely long line charge λ a distance D from the center of an infinitely long
cylinder of radius R with charge per unit length λc (Image by MIT OpenCourseWare).
C
GaussianContour
r

Figure 33: A diagram of the Gaussian Contour at radius r around an infinitely long line charge (Image by
MIT OpenCourseWare).
�0 Er 2πr�
L = λ�
L
Er =
λ
2π�0 r
Potential due to a line charge:
E = −�Φ ⇒ Er = −
⇒ Φ=−
d
λ
Φ=
dr
2π�0 r
λ
ln r + constant
2π�0
36
Additional Problems - Solutions
6.641, Spring 2009
Equipotential surfaces due to two line charges of magnitude λ and −λ are cylinders. See Figure 2.24 in
Zahn’s book.

−
Figure 34: A pair of opposite polarity line charges have circular cylindrical equipotential surfaces (Image by
MIT OpenCourseWare).
2
Thus, if an image charge −λ is placed inside the cylinder at RD (see Zahn p. 98) distance from the center,
the condition for an equipotential surface at the cylinder will be satisfied.
R
−

2
R
D
D
Figure 35: A diagram showing an image line charge −λ a distance R2 /D away from the center of the cylinder
and another line charge λ a distance D away from the center. (Image by MIT OpenCourseWare).
In order to satisify the condition that the surface of the cylinder holds a charge per unit length λc , and
maintaining that the surface is equipotential, another image charge must be placed at the center of the
cylinder. This image charge has value λ + λc to keep the total cylinder charge at λc .
Problem reduces to (see figure 36):
⇒ Force on cylinder is due to the force on the two image line charges −λ and λ + λc
λ
⇒ fx =
2π�0
�
−λ
(λ + λc )
+
R2
D
D− D
�
37
Additional Problems - Solutions
6.641, Spring 2009
−

c 
2
R
D
R
D
Figure 36: A diagram showing the locations of all the line charges and cylinder giving the total line charge
on the cylinder as λc (Image by MIT OpenCourseWare).
Problem 4.5
1.4.1 Magnetic field of line current
• Ampere’s law predicts B-field intensity is inversely proportional to radial distance r.
• Hall effect probe is used to measure B-field intensity.
• Hall effect proble measures B-field intensity perpendicular to its flat surface.
• Magnetic field due to a wire is non-zero only in φ direction.
H
1
r
r
Figure 37: A graph showing the B-field radial dependence of 1/r for a long line current as predicted by
Ampere’s Law (Image by MIT OpenCourseWare).
1.6.1 Voltmeter Reading Induced by Magnetic Induction
• Contour C is given by the coil shown in Figure 38
• Use the coil shown in Figure 38 to find the magnetic field associated for a current carrying wire.
38
Additional Problems - Solutions
6.641, Spring 2009
21 cm
70 turns
8 cm
Figure 38: A diagram showing Contour C (Image by MIT OpenCourseWare).
• Faraday integral law shows how a voltmeter reading induced by magnetic induction provides a mea­
surement of magnetic flux density.
• Observe a change in phase as coil is moved from below the wire to above the wire.
• Coil voltage is 90◦ out of phase with wire current.
6.6.1 An Artificial Dielectric
• Artificial dielectric is constructed of an array of conducting spheres (ping-pong balls with conductive
coating). See Figure 39
−
2R
S
v

Figure 39: A diagram showing an artificial dielectric constructed of an array of conducting spheres (Image
by MIT OpenCourseWare).
• Application of voltage v to the electrodes results in the spheres acquiring negative and positive charges
on their poles.
• Insertion of dielectric array between the plane parallel conductors increases the capacitance.
9.4.1 Measurement of B-H Characteristic
• Magnetizable material
• Polycrystalline and ferromagnetic materials at domain level have randomly oriented magnetic moments
that tend to cancel in the absence of applied field.
• Those domains align with the applied magnetic field.
• However phase delay develops between magnetization and applied field resulting in power dissipation.
• Hysteresis loop
39
Additional Problems - Solutions
6.641, Spring 2009
B
H
Figure 40: A magnetization hysteresis loop (Image by MIT OpenCourseWare).
Problem 5.1
Figure 41: A diagram showing the magnetic field lines from a line current I of infinite extent in free space
above a plane of material of infinite magnetic permeability.
Line current I of infinite extent above a plane of material of infinite permeability, µ → ∞.
A
� = µH
� ⇒ for µ → ∞, in order to have B finite, we need H zero ⇒ continuity of normal B
� and tangential
B
�
H at the surface.
B
Using method of images to satisify boundary conditions at y = 0 for medium where µ → ∞ requires
⇒ Hx = Hz = 0 at y = 0
For a line current at origin (see Figure 43)
40
Additional Problems - Solutions
6.641, Spring 2009
I
0
d
y
x
∞
image current
d
I
Figure 42: A diagram showing how to apply the method of images for a line current I in free space above
an infinite magnetic permeability material (µ → ∞)(Image by MIT OpenCourseWare).
Figure 43: A diagram depicting a Gaussian Contour to determine the magnetic field from an infinitely long
line current I (Image by MIT OpenCourseWare).
�
� · d�l = I ⇒ Hφ = I
H
2πr
C
� = µ0 I i�φ , since B
� =�×A
� ⇒ − ∂Az = Iµ0
⇒B
2πr
∂r
2πr
Iµ0
⇒ Az = −
ln r + constant
2π
⇒ for line currents I at z = d and I at z = −d
� ��
�
��
��
Iµ0
2
2
2
2
Az = −
ln
x + (y − d) + ln
x + (y + d)
2π
⇒ Az = −
��
��
Iµ0 �� 2
2
2
ln x + (y − d)
x2 + (y + d)
4π
C
� =�×A
� = − ∂ Az i�y + ∂ Az i�x
B
∂x
∂y
�
�
�
�
�
�
�
�
2
2
2
2
2
2
2
2
Iµ0 2x x + (y + d) + 2x x + (y − d) � Iµ0 2(y − d) x + (y + d) + 2(y + d) x + (y − d) �
�
��
�
�
��
�
iy −
ix
=
2
2
2
2
4π
4π
x2 + (y − d)
x2 + (y + d)
x2 + (y − d)
x2 + (y + d)
41
Additional Problems - Solutions
� = − Iµ0
B
2π
�
(y − d)i�x − xi�y
(y + d)i�x − xi�y
+
x2 + (y + d)2
x2 + (y − d)2
6.641, Spring 2009
�
D
Force is applied on the line current due to the image line current
Force per unit length:
� ← field due to image charge at (x = 0, y = −d)
F� = I� × B
�
�
µ0 I 1 �
= I�i�z ×
ix
2π 2d
= −
µ0 I 2 �
iy
4πd
so line current is attracted to the surface
42