Chapter
2
Electrostatics
Problem
2.1
(a) I Zero.!
(b)
IF
=
_
4
1
q~,
I
where
r
is the
distance
from center
to each numeral.
F points
toward
the
missing
q.
11"1=0
r
Explanation:
by superposition,
this is equivalent
to (a), with an extra
q
at 6 o'clocksince
the force of all
twelve is zero, the net force is that
of
q
only.
(c)
I
Zero.
I
(d)
1
4
1
q~,
I
pointing
toward the missing
q.
Same reason as (b). Note, however, that
if you explained
(b) as
1r€0r
a cancellation
in pairs of opposite
charges (1 o'clock against
7 o'clock; 2 against
8, etc.), with one unpaired
q
doing the job, then you'll need a
different
explanation
for (d).
Problem
2.2
(a) "Horizontal"
components
cancel. Net vertical field is:
Ez
= 4;<02~ cos
O.
q
I
q
From far away,
(z
»
d),
the field goes like E ~
4;<0
~
z, which, as we shall see, is the field of a .dipole. (If we
set
d
+ 0, weget E = 0, as is appropriate; to the extent that this configurationlookslike a singlepoint charge
from far away, the net charge is zero, so E + 0.)
1
2qz
A
Here
12
=
Z2
+ (~)2 ;
cosO
=
i,
so
I
E
=
41r€0(Z2
+ (~)2)3/2 z.
When
z»
d
you're so far away it just
looks like a single charge
2q;
the field
should reduce to E
=
_
4
1
~ z. And it
aoes
(just setd + 0 in the formula).
?r<o
z
(b) This time the
"vertical"
components
cancel, leaving
E =
_
4
1
2~ sinOx, or
?r<0
'V
E
1
qd
A
=
41r€0 (z2
+ (~)2)3/2 x.
E
x
E
x
22
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Problem
2.3
Ez
1
rL),
dx
()
(
2
2
2
411"fO
JO
~
cos
;
1
=
Z
+
x
; COS()
= ~)
/Ldq
=
)"dx
Ex
1).
.z
rL
1
dx
411"fO
JO
(z2+x2)3/2
[
]
I
L
1).
.Z
1
x

1~~
411"fO
Z2~
0 
411"fO
z~'
1
rL
)'dx
sin()

1).
.
I
xdx
411"fOJo
~

411"fO
(X2+z2)3/2
[
]
I
L
[
]
 1
)..

L
  1
)..
1 
L
411"fO
~
0

411"fO
Z
~.
x

x
E
1)"
[
(
1
Z
)
A
(
L
)
A
]
 

+
x+
z

4m:0
Z
..,j
Z2
+
L2
..,j
z2
+
L2
. ,
For
z
»
L
you expect it to look like a point char ge
q
=
)"L:
E + _
4 1
¥z.
It checks, for with
z
»
L
the x
1I"fO
Z
termt 0,andthe z term + _
4
1
~kz.
1I"fOz z
Problem
2.4
From Ex. 2.1, with
L
+
~
and
z
+
vz2
+ (~)2 (distance from center of edge to
F),
field of
one
edge is:
El
=
~
)..a
.
41f1:0
vz2
+
a2
vz2
+
a2
+
a2
4
4
4
There are 4 sides, and we want vertical components
only, so multiply by 4 cos () = 4
!i+?
:
z2+
a4
E
=
~
4).
.az
z.
47r£0
(Z2
+
a42)
V
Z2
+ ~2
Problem
2.5
"Horizontal"
components
cancel, leaving: E = 4;fO
{I
~
cos ()}z.
Here, 12= r2 +
z2,
cos() =
~
(both
constants),
while
I
dl
=
27rr.
So
E
=
~
)"(27rr)z
A
47r£0
(r2
+
Z2)3/2
z.
Problem
2.6
Break it into rings of radius r, and thickness
dr,
and use Prob. 2.5 to express the field of each ring.
Total
charge
of a ring
is 0"
.
27rr .
dr
=
)..
. 27rr, so
)..
=
O"dr
is the
"line
charge"
of each
ring.
1
(O"dr)27rrz
1
l
R
r
Ering=
_
4
3/2;
Edisk
=
_
4
27rO"z
3/2
dr.
7r£0
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 Electrostatics, Force, Electric charge, fo

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