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Chapter 21
The Electric Field I: Discrete Charge Distributions
22
•
A charge equal to the charge of Avogadro’s number of protons
(
N
A
= 6.02
×
10
23
) is called a
faraday
. Calculate the number of coulombs in a faraday.
Picture the Problem
One faraday =
N
A
e
.
We can use this definition to find the
number of coulombs in a faraday.
Use the definition of a faraday to calculate the number of coulombs in a faraday:
(
29 (
29
C
10
63
.
9
C/electron
10
602
.
1
electrons
10
02
.
6
faraday
1
4
19
23
A
×
=
×
×
=
=

e
N
26
•
A point charge
q
1
= 4.0
μ
C is at the origin and a point charge
q
2
= 6.0
C is on the
x
axis at
x
= 3.0
m. (
a
) Find the electric force on charge
q
2
. (
b
) Find the electric force on
q
1
. (
c
) How would your answers for
Parts (
a
) and (
b
) differ if
q
2
were –6.0
C?
Picture the Problem
We can find the electric forces the two charges exert on
each by applying Coulomb’s law and Newton’s 3
rd
law. Note that
i
r
ˆ
ˆ
2
,
1
=
because
the vector pointing from
q
1
to
q
2
is in the positive
x
direction. The diagram shows
the situation for Parts (
a
) and (
b
).
(
a
) Use Coulomb’s law to express
the force that
q
1
exerts on
q
2
:
2
,
1
2
2
,
1
2
1
2
,
1
ˆ
r
F
r
q
kq
=
Substitute numerical values and evaluate
2
,
1
F
:
(
29 (
29 (
29
(
29
(
29
i
i
F
ˆ
mN
24
ˆ
m
3.0
μC
6.0
μC
4.0
/C
m
N
10
8.988
2
2
2
9
2
,
1
=
⋅
×
=
(
b
) Because these are actionand
reaction forces, we can apply
Newton’s 3
rd
law to obtain:
(
29
i
F
F
ˆ
mN
24
2
,
1
1
,
2

=

=
(
c
) If
q
2
is

6.0
C, the force between
q
1
and
q
2
is attractive and both force vectors
are reversed:
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29 (
29 (
29
(
29
(
29
i
i
F
ˆ
mN
24
ˆ
m
3.0
μC
6.0
μC
4.0
/C
m
N
10
8.988
2
2
2
9
2
,
1

=

⋅
×
=
and
(
29
i
F
F
ˆ
mN
24
2
,
1
1
,
2
=

=
38
•
Two point charges, each +4.0
μ
C, are on the
x
axis; one point charge is at the origin and the other is
at
x
= 8.0 m.
Find the electric field on the
x
axis at (
a
)
x
= –2.0 m, (
b
)
x
= 2.0 m, (
c
)
x
= 6.0 m, and (
d
)
x
= 10
m. (
e
) At what point on the
x
axis is the electric field zero? (
f
) Sketch
E
x
versus
x
for –3.0 m <
x
< 11 m.
Picture the Problem
Let
q
represent the point charges of +4.0
C and use
Coulomb’s law for
E
due to a point charge and the principle of superposition for
fields to find the electric field at the locations specified.
Noting that
q
1
=
q
2
, use Coulomb’s law and the principle of superposition to
express the electric field due to the given charges at point P a distance
x
from the
origin:
(
29
(
29
(
29
(
29
(
29
(
29
(
29

+
⋅
=

+
=

+
=
+
=
P
,
2
P
,
2
2
P
,
2
P
,
2
1
P
,
2
2
P
,
2
1
2
1
2
1
2
1
2
1
ˆ
m
0
.
8
1
ˆ
1
/C
m
kN
36
ˆ
m
0
.
8
1
ˆ
1
ˆ
m
0
.
8
ˆ
q
q
q
q
q
q
q
q
r
x
r
x
r
x
r
x
kq
r
x
kq
r
x
kq
x
E
x
E
x
E
(
a
) Apply this equation to the point at
x
=

2.0 m:
(
29
(
29
(
29
(
29
(
29
(
29
(
29
i
i
i
E
ˆ
kN/C
4
.
9
ˆ
m
10
1
ˆ
m
0
.
2
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 Spring '08
 Turner
 Physics, Charge

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