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6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
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6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2006
Quiz 1 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 1
z
q
ε0
z
σ=∞
z
-q
image
charge
Figure 1: Charge q of a mass m above a perfectly conducting ground plane and its image charge −q. (Image
by MIT OpenCourseWare.)
A
Question: What is the velocity of the charge as a function of position z?
Solution: Force = ma
dv
= F due to image charge
dt
dv
q(−q)
q 2
dvz
⇒m
=
=−
iz ⇒ m
2
dt
4πε0 (2z)
dt
4πε0 4z 2
dvz
dvz
q2
dvz dz
dvz
d vz2
⇒
=−
Use
of
chain
rule:
=
=
v
=
z
dt
16πε0 mz 2
dt
dz dt
dz
dz 2
⇒m
1
Spring 2006 Quiz 1
q2
=−
⇒
16πε0 mz 2
q2
+C
⇒ vz2 = +
8πε0 mz
Use I.C. vz (z = d) = 0
d
⇒
dz
vz2
2
6.641, Spring 2005
Z
d
vz2
2
=
Z
−
q2
dz
16πε0 mz 2
2
⇒ C = − 8πεq0 md
q2
1 1
vz2 =
−
8πε0 m z
d
q
2
⇒ vz (z) = − 8πεq 0 m
1
z
−
1
d
since particle is moving towards the conducting ground plane v = vz iz (vz has minus sign)
B
Question: How long does it take the charge to reach the z = 0 ground plane?
Hint:
Z
dz
1
z
−
p
3
= − zd(d − z) + d 2 tan−1
1
1 2
d
r
z
d−z
Solution:
dz
vz =
=−
dt
⇒
Z
s
dz
q
1
z
−
1
d
q2
8πε0 m
=
Z
−
s
1 1
−
z
d
q2
dt
8πε0 m
p
3
− zd(d − z) + d 2 tan−1
r
s
z
q2
= −t
+ |{z}
C
d−z
8πε0 m
constant
We use I.C. to find constant ⇒ at t = 0, z(t = 0) = d.
r
3
π 3
d
−1
⇒ d 2 tan
= C ⇒ C = d2
2
| {z 0}
π
2
plugging in z = 0 gives the time (T) it takes for the charge to reach the z = 0 ground plane.
s
r
3
0
q2
π 3
=−
T + d2
⇒ 0 + d 2 tan−1
d
8πε0 m
2
| {z }
0
2
Spring 2006 Quiz 1
π 32
2d
⇒T = q
q2
8πε0 m
6.641, Spring 2005
=
s
q
π 2 3 8πε0 m
2π 3 d3 ε0 m
T
=
d
⇒
q2
2
4
q
Problem 2
y
λ
ε0
+
+
+ + + + + + +
-L
-a
I
+
+
+
a
+
II
a
φ
+ + + + + + +
III
L
Figure 2: Uniformly distributed line charge λ. (Image by MIT OpenCourseWare.)
Φ(r) =
Z
l′
λ(r ′ )dl′
4πε0 |r − r ′ |
E(r) =
Z
l′
λ(r ′ )ir′ r dl′
2
4πε0 |r − r ′ |
A
Question: Find the potential at the point (x = 0, y = 0, z = 0)
Z
Z π
Z L
λdx
λadφ
λdx
+
+
4πε
(−x)
4πε
a
4πε
0
0
0 (x)
x=−L
φ=0
x=a
−a
π
L
λ
=
− ln x + φ + ln x
4πε0
−L
0
a
L
a
λ
λ
− ln + π + ln
= 4πε
(π + 2 ln La )
=
0
4πε0
L
a
Φ(x = 0, y = 0, z = 0) =
−a
B
Question: Find the electric field (magnitude and direction) at (x = 0, y = 0, z = 0).
3
x
Spring 2006 Quiz 1
6.641, Spring 2005
Solution:
Z
Z π
Z L
λix dx
λ(−ir )adφ
λ(−ix )dx
+
+
2
2
2
4πε0 a
x=−L 4πε0 x
φ=0
x=a
4πε0 x
Z
π
1 −a
1 L
λ
− cos φix − sin φiy
− ix +
dφ + ix
=
4πε0
x −L
a
x a
φ=0


 
 
!
π
π
�
�
�
�
1�
1�
1
λ 
1
1
− � + � ix +
− sin φ ix + cos φ iy  +  � − �  ix 
=
| {z } 0
4πε0
−L
a | {z } 0
L
�a
�
�−a �
0
−2
−2
λ
iy ⇒ E(x = 0, y = 0, z = 0) = − 2πελ0 a iy
=
4πε0
a
E(x = 0, y = 0, z = 0) =
−a
Problem 3
y
III
I
-∞
�0
y= a
a
φ
x
II
-∞
y = -a
I
I
Figure 3: Current. (Image by MIT OpenCourseWare.)
Question: What is the magnetic field H at the point (x = 0, y = 0, z = 0)?
Hint: One or more of the following indefinite integrals may be useful.
√
R dx
a.
x2 + a2
1
= ln x +
2
2
[x +a ] 2
b.
c.
d.
e.
R
1
[x2 +a2 ] 2
R
[x2 +a2 ] 2
R
R
xdx
dx
[x2 +a2 ]
1
= x2 + a2 2
=
dx
1
a
tan−1
x
3
=
3
=−
xdx
[x2 +a2 ] 2
x
a
1
a2 [x2 +a2 ] 2
1
1
[x2 +a2 ] 2
4
Spring 2006 Quiz 1
6.641, Spring 2005
Solution:
1
H(r) =
4π
Z
Idl′ × ir′ r
2
|r − r ′ |
H(x = 0, y = 0, z = 0)
1
=
4π
(Z
0
Z
π
2
Iiφ adφ × (−ir )
+
+
3
2
2
π
a2
(x + a ) 2
x=−∞
φ=− 2
(Z
)
Z 0
Z π2
0
aiz dx
aiz dx
I
iz dφ
+
=
+
2
2 3
2
2 3
4π
a
x=−∞ (x + a ) 2
x=−∞ (x + a ) 2
φ=− π
2
0
0
1
π2
1
1
x
x
I
=
iz + φ π iz +
iz
4π a (x2 + a2 ) 12
−∞
a − 2
a (x2 + a2 ) 32
−∞
n
o
I
π
π
=
iz 0 − (−1) + + + 0 − (−1)
4πa
2
2
=
I
4πa
Iix dx × (−xix + aiy )
(2 + π) iz
5
Z
0
x=−∞
I(−ix )dx × (−xix − aiy )
3
(x2 + a2 ) 2
)