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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2007
Problem Set 2 Solutions
Problem 1: Closed Surfaces
Four closed surfaces,
S
1
through
S
4
, together with the charges
–2
Q
,
Q
, and –
Q
are sketched in the figure at right. The
colored lines are the intersections of the surfaces with the
page.
Find the electric flux through each surface.
By Gauss’s Law, the flux through the closed surfaces is equal
to the charge enclosed over
ε
0
.
So,
12
3
4
00
0
SS
S
S
2;
0
;
;
QQ
ε
Φ=
−
Φ= Φ=
−
Problem 2:
Field on Axis of a Line Charge
A wire of length
has a uniform positive linear charge density and a total charge
.
Calculate the electric field at a point
located along the axis of the wire and a distance
from one end:
l
Q
P
a
a.
Give an integral expression for the electric field at point
P
in terms of the
variables used in the above figure
()
2
ˆ
al
e
dx
k
a
x
λ
+
′
=−
′
∫
Ei
b.
Evaluate this integral.
11
1
ˆˆ
ˆ
ˆ
ee
e
e
lQ
kk
k
k
a
x
al a
aal
λλ
+
⎡⎤
−
==
−
=
=
−
⎢⎥
′
++
i
i
i
+
⎣⎦
c.
In the limit that the length of the rod goes to zero, does your answer reduce to the
right expression?
2
ˆ
In the limit that
goes to zero, our expression becomes
, as we desire.
e
Q
lk
⎛⎞
⎜⎟
a
⎝⎠
G
PS02 Solution  1
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View Full Document Problem 3: Charged Slab & Sheets
An infinite slab of charge carrying a charge per unit volume
ρ
has its boundaries located at
x =
−
2 meters and
x =
+2
meters. It is infinite in the
y
direction and in the
z
direction
(out of the page).
Two similarly infinite charge sheets (zero
thickness) are located at
x =
−
6 meters and
x =
+6 meters,
with area charge densities
σ
1
and
σ
2
respectively.
In the
accessible regions you’ve measured the electric field to be:
ˆ
06
ˆ
10V/m
6 m < x < 2 m
ˆ
10V/m
6 m > x > 2 m
x
⎧
<−
⎪
−
⎪
=
⎨
⎪
i
i
E
i
G
m
ˆ
m
x
⎪
>+
⎩
i
(a)
Use Gauss’s Law to find the charge density
of the slab.
You must indicate the
Gaussian surface you use on a figure.
You can use the symbol
ε
o
in your answer.
0
2
2
LL
RR
S
Ad
dE
EA
A
⋅=⋅+⋅=
=
⇒=
∫∫
EAEA EA
G
GG
G
w
A
()
23
0
5 V/m
C/m
⎡⎤
=
⎣⎦
00
2 10 V/m
4 m
d
εε
=
(b)
Use Gauss’s Law to find the two surface charge densities of the left and right charged
sheets.
You must indicate the Gaussian surface you use on a figure.
You can use the
symbol
o
in your answer.
Using Gauss’s law with the Gaussian pillboxes indicated in the figure, we have
2
10
(10
10 V/m
C/m
S
qA
enc
1
V/m)
dA
σ
σε
⎡
⎤
⋅=
−
=
−
⋅
=
=
⇒
⎣
⎦
∫∫
G
G
w
2
20
10 V/m
C/m
⎡
⎤
=−
⋅
In a similar manner,
⎦
⎣
PS02 Solution  2
Problem 4: Charge Slab
Consider a slab of insulating material which is infinitely large in two of its three
dimensions, and has a thickness
d
in the third dimension. An edge view of the slab is
shown in the figure. The slab has a uniform positive charge density
ρ
.
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This homework help was uploaded on 04/07/2008 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.
 Fall '07
 Hudson

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