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.
15
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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
−
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 Fall '08
 Joshi
 Electrical Engineering, Charge, Electric charge, ax, charge density, Eφ

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