211
H
OMEWORK
2:
E
LECTRIC
C
HARGE AND
E
LECTRIC
F
IELD
Q21.14. ANSWER: The initial force on each charge is the same and depends only on the product of charges and
initial distance. Therefore, the acceleration will still be a
0
.
21.26.
IDENTIFY:
For a point charge,
2
q
Ek
r
.
SET UP:
E
is toward a negative charge and away from a positive charge.
EXECUTE:
(a)
The field is toward the negative charge so is downward.
9
9
2
2
2
3.00 10 C
(8.99 10 N m /C )
432 N/C
(0.250 m)
E
.
(b)
9
2
2
9
(8.99 10 N m /C )(3.00 10 C)
1.50 m
12.0 N/C
kq
r
E
EVALUATE:
At different points the electric field has different directions, but it is always directed toward the
negative point charge.
21.31.
IDENTIFY:
For a point charge,
2
.
q
r
The net field is the vector sum of the fields produced by each charge. A
charge
q
in an electric field
E
experiences a force
.
q
FE
SET UP:
The electric field of a negative charge is directed toward the charge. Point
A
is 0.100 m from
q
2
and
0.150 m from
q
1
. Point
B
is 0.100 m from
q
1
and 0.350 m from
q
2
.
EXECUTE:
(a)
The electric fields due to the charges at point
A
are shown in Figure 21.31a.
9
1
9
2
2
3
1
22
1
6.25 10 C
(8.99 10 N m /C )
2.50 10 N/C
(0.150 m)
A
q
r
9
2
9
2
2
4
2
2
12.5 10 C
(8.99 10 N m /C )
1.124 10 N/C
(0.100 m)
A
q
r
Since the two fields are in opposite directions, we subtract their magnitudes to find the net field.
3
21
8.74 10 N/C,
E
E
E
to the right.
(b)
The electric fields at points
B
are shown in Figure 21.31b.
9
1
9
2
2
3
1
1
(8.99 10 N m /C )
5.619 10 N/C
(0.100 m)
B
q
r
9
2
9
2
2
2
2
2
(8.99 10 N m /C )
9.17 10 N/C
(0.350 m)
B
q
r
Since the fields are in the same direction, we add their magnitudes to find the net field.
3
12
6.54 10 N/C,
E
E
E
to the right.
(c)
At
A
,
3
8.74 10 N/C
E
, to the right. The force on a proton placed at this point would be
19
3
15
(1.60 10
C)(8.74 10 N/C) 1.40 10
N,
F
qE
to the right.
EVALUATE:
A proton has positive charge so the force that an electric field exerts on it is in the same direction as
the field.
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Chapter 21
Figure 21.31
21.33.
IDENTIFY:
Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the
plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion; use
constant acceleration equations for the horizontal and vertical components of the motion.
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 Fall '05
 Ogilvie
 Charge, Acceleration, Force, Work

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