CHAPTER 2
2.1. Four 10nC positive charges are located in the
z
=
0 plane at the corners of a square 8cm on a side.
A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the
magnitude of the total force on this fifth charge for
=
0
:
Arrange the charges in the
xy
plane at locations (4,4), (4,4), (4,4), and (4,4). Then the fifth charge
will be on the
z
axis at location
z
=
4
√
2, which puts it at 8cm distance from the other four.
By
symmetry, the force on the fifth charge will be
z
directed, and will be four times the
z
component of
force produced by each of the four other charges.
F
=
4
√
2
×
q
2
4
π
0
d
2
=
4
√
2
×
(
10
−
8
)
2
4
π(
8
.
85
×
10
−
12
)(
0
.
08
)
2
=
4
.
0
×
10
−
4
N
2.2. A charge
Q
1
=
0
.
1
µ
C is located at the origin, while
Q
2
=
0
.
2
µ
C is at
A(
0
.
8
,
−
0
.
6
,
0
)
. Find the
locus of points in the
z
=
0 plane at which the
x
component of the force on a third positive charge is
zero.
To solve this problem, the
z
coordinate of the third charge is immaterial, so we can place it in the
xy
plane at coordinates
(x, y,
0
)
. We take its magnitude to be
Q
3
. The vector directed from the first
charge to the third is
R
13
=
x
a
x
+
y
a
y
; the vector directed from the second charge to the third is
R
23
=
(x
−
0
.
8
)
a
x
+
(y
+
0
.
6
)
a
y
. The force on the third charge is now
F
3
=
Q
3
4
π
0
Q
1
R
13

R
13

3
+
Q
2
R
23

R
23

3
=
Q
3
×
10
−
6
4
π
0
0
.
1
(x
a
x
+
y
a
y
)
(x
2
+
y
2
)
1
.
5
+
0
.
2[
(x
−
0
.
8
)
a
x
+
(y
+
0
.
6
)
a
y
]
[
(x
−
0
.
8
)
2
+
(y
+
0
.
6
)
2
]
1
.
5
We desire the
x
component to be zero. Thus,
0
=
0
.
1
x
a
x
(x
2
+
y
2
)
1
.
5
+
0
.
2
(x
−
0
.
8
)
a
x
[
(x
−
0
.
8
)
2
+
(y
+
0
.
6
)
2
]
1
.
5
or
x
[
(x
−
0
.
8
)
2
+
(y
+
0
.
6
)
2
]
1
.
5
=
2
(
0
.
8
−
x)(x
2
+
y
2
)
1
.
5
2.3. Point charges of 50nC each are located at
A(
1
,
0
,
0
)
,
B(
−
1
,
0
,
0
)
,
C(
0
,
1
,
0
)
, and
D(
0
,
−
1
,
0
)
in free
space. Find the total force on the charge at
A
.
The force will be:
F
=
(
50
×
10
−
9
)
2
4
π
0
R
CA

R
CA

3
+
R
DA

R
DA

3
+
R
BA

R
BA

3
where
R
CA
=
a
x
−
a
y
,
R
DA
=
a
x
+
a
y
, and
R
BA
=
2
a
x
. The magnitudes are

R
CA
 = 
R
DA
 =
√
2,
and

R
BA
 =
2. Substituting these leads to
F
=
(
50
×
10
−
9
)
2
4
π
0
1
2
√
2
+
1
2
√
2
+
2
8
a
x
=
21
.
5
a
x
µ
N
where distances are in meters.
14
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2.4. Let
Q
1
=
8
µ
C be located at
P
1
(
2
,
5
,
8
)
while
Q
2
= −
5
µ
C is at
P
2
(
6
,
15
,
8
)
. Let
=
0
.
a) Find
F
2
, the force on
Q
2
: This force will be
F
2
=
Q
1
Q
2
4
π
0
R
12

R
12

3
=
(
8
×
10
−
6
)(
−
5
×
10
−
6
)
4
π
0
(
4
a
x
+
10
a
y
)
(
116
)
1
.
5
=
(
−
1
.
15
a
x
−
2
.
88
a
y
)
mN
b) Find the coordinates of
P
3
if a charge
Q
3
experiences a total force
F
3
=
0 at
P
3
: This force in
general will be:
F
3
=
Q
3
4
π
0
Q
1
R
13

R
13

3
+
Q
2
R
23

R
23

3
where
R
13
=
(x
−
2
)
a
x
+
(y
−
5
)
a
y
and
R
23
=
(x
−
6
)
a
x
+
(y
−
15
)
a
y
. Note, however, that
all three charges must lie in a straight line, and the location of
Q
3
will be along the vector
R
12
extended past
Q
2
. The slope of this vector is
(
15
−
5
)/(
6
−
2
)
=
2
.
5. Therefore, we look for
P
3
at coordinates
(x,
2
.
5
x,
8
)
. With this restriction, the force becomes:
F
3
=
Q
3
4
π
0
8[
(x
−
2
)
a
x
+
2
.
5
(x
−
2
)
a
y
]
[
(x
−
2
)
2
+
(
2
.
5
)
2
(x
−
2
)
2
]
1
.
5
−
5[
(x
−
6
)
a
x
+
2
.
5
(x
−
6
)
a
y
]
[
(x
−
6
)
2
+
(
2
.
5
)
2
(x
−
6
)
2
]
1
.
5
where we require the term in large brackets to be zero. This leads to
8
(x
−
2
)
[
((
2
.
5
)
2
+
1
)(x
−
6
)
2
]
1
.
5
−
5
(x
−
6
)
[
((
2
.
5
)
2
+
1
)(x
−
2
)
2
]
1
.
5
=
0
which reduces to
8
(x
−
6
)
2
−
5
(x
−
2
)
2
=
0
or
x
=
6
√
8
−
2
√
5
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 Spring '08
 Brown
 Charge, Electromagnet, Electric charge, ax, charge density, ρs

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