617
21-1 COU LOM B’S L AW
Then from Eq. 21-7, we can write the net force
on particle 1 as
Because the forces
and
are not directed along the
same axis, we cannot sum simply by combining their magnitudes. Instead, we must add them as vectors, using one of
the following methods.
Method 1. Summing directly on a vector-capable calculator.
For
, we enter the magnitude
and the angle
1.15 ϫ 10
Ϫ24
F
:
12
F
:
14
F
:
12
F
:
1,net ϭ F
:
12 ϩ F
:
14 .
F
:
1,net
and 3 have charge of opposite signs, particle 1 is attracted
to particle 3.Thus, force
is directed toward particle 3, as indicated in the free-body diagram of Fig. 21-7d.
Three particles: To find the magnitude of
, we can
rewrite Eq. 21-4 as
We can also write
in unit-vector notation:
The net force
on particle 1 is the vector sum of F
:
12
F
:
1,net
F
:
13 ϭ (2.05 ϫ 10
Ϫ24
N)i ˆ .
F
:
13
ϭ 2.05 ϫ 10
Ϫ24
N.
ϫ
(1.60 ϫ 10
Ϫ19
C)(3.20 ϫ 10
Ϫ19
C)
(
3
4 )
2
(0.0200 m)
2
ϭ (8.99 ϫ 10
9
Nиm
2
/C
2
)
F 13 ϭ
1
4p´ 0
͉q 1 ͉͉q 3 ͉
(
3
4 R)
2
F
:
13
F
:
13
and
; that is, from Eq. 21-7, we can write the net force
on particle 1 in unit-vector notation as
.
( A n s w e r )
Thus,
has the following magnitude and direction (relative
to the positive direction of the x axis):
9.00 ϫ 10
Ϫ25
N and 0Њ.
( A n s w e r )
(c) Figure 21-7e is identical to Fig. 21-7a except that particle 4
is now included. It has charge q 4 ϭ Ϫ3.20 ϫ 10
Ϫ19
C, is at a
distance
from particle 1, and lies on a line that makes an
3
4 R
F
:
1,net
ϭ (9.00 ϫ 10
Ϫ25
N)i ˆ
ϭ Ϫ(1.15 ϫ 10
Ϫ24
N)i ˆ ϩ (2.05 ϫ 10
Ϫ24
N)i ˆ
F
:
1,net ϭ F
:
12 ϩ F
:
13
F
:
1,net
F
:
13
angle u ϭ 60Њ with the x axis. What is the net electrostatic
force
on particle 1 due to particles 2 and 4?
KEY IDEA
The net force
is the vector sum of
and a new force
F
:
12
F
:
1,net
F
:
1,net
acting on particle 1 due to particle 4. Because particles 1
and 4 have charge of opposite signs, particle 1 is attracted to
particle 4. Thus, force
on particle 1 is directed toward
F
:
14
F
:
14
particle 4, at angle
60Њ, as indicated in the free-body diagram of Fig. 21-7f.
Four particles: We can rewrite Eq. 21-4 as
ϭ 2.05 ϫ 10
Ϫ24
N.
ϫ
(1.60 ϫ 10
Ϫ19
C)(3.20 ϫ 10
Ϫ19
C)
(
3
4 )
2
(0.0200 m)
2
ϭ (8.99 ϫ 10
9
Nиm
2
/C
2
)
F 14 ϭ
1
4p´ 0
͉q 1 ͉͉q 4 ͉
(
3
4 R)
2
u ϭ
180Њ. For
, we enter the magnitude
and the
2.05 ϫ 10
Ϫ24
F
:
14
angle 60Њ.Then we add the vectors.
Method 2. Summing in unit-vector notation. First we
rewrite
as
Substituting
N for F 14 and 60Њ for u, this becomes
.
Then we sum:
(Answer)
Method 3. Summing components axis by axis. The sum of
the x components gives us
The sum of the y components gives us
The net force
has the magnitude
(Answer)
To find the direction of
, we take
u ϭ tan
Ϫ1
F 1,net,y
F 1,net,x
ϭ Ϫ86.0Њ.
F
:
1,net
F 1,net ϭ 2F
2
1,net,x ϩ F
2
1,net,y ϭ 1.78 ϫ 10
Ϫ24
N.
F
:
1,net
ϭ 1.78 ϫ 10
Ϫ24
N.
ϭ (2.05 ϫ 10
Ϫ24
N)(sin 60Њ)
F 1,net,y ϭ F 12,y ϩ F 14,y ϭ 0 ϩ F 14 sin 60Њ
ϭ Ϫ1.25 ϫ 10
Ϫ25
N.
ϭ Ϫ1.15 ϫ 10
Ϫ24
N ϩ (2.05 ϫ 10
Ϫ24
N)(cos 60Њ)
F 1,net,x ϭ F 12,x ϩ F 14,x ϭ F 12 ϩ F 14 cos 60Њ
Ϸ (Ϫ1.25 ϫ 10
Ϫ25
N)i ˆ ϩ (1.78 ϫ 10
Ϫ24
N)j ˆ .
ϩ (1.025 ϫ 10
Ϫ24
N)i ˆ ϩ (1.775 ϫ 10
Ϫ24
N)j ˆ
ϭ Ϫ(1.15 ϫ 10
Ϫ24
N)i ˆ
F
:
1,net ϭ F
:
12 ϩ F
:
14
F
:
14 ϭ (1.025 ϫ 10
Ϫ24
N)i ˆ ϩ (1.775 ϫ 10
Ϫ24
N)j ˆ
2.05 ϫ 10
Ϫ24
F
:
14 ϭ (F 14 cos u)i ˆ ϩ (F 14 sin u)j ˆ .
F
:
14
However, this is an unreasonable result because
must
F
:
1,net
Additional examples, video, and practice available at WileyPLUS
have a direction between the directions of
and
. To
correct u, we add 180Њ, obtaining
Ϫ86.0Њ ϩ 180Њ ϭ 94.0Њ.
( A n s w e r )
F
:
14
F
:
12