468 Chapter 16Vectors
EXAMPLE 16.20Force and potential energy
By a generalization to three dimensions of the discussion of conservative forces
in Section 5.7 (see also Example 16.13), the components of a conservative force are
derivatives of a potential-energy function V,
The force is therefore (minus) the gradient of V,
(16.69)
0 Exercises 48–50
EXAMPLE 16.21Coulomb forces
The potential energy of interaction of charges q
1
and q
2
separated
by distance risV 1 = 1 q
1
q
2
24 πε
0
r(see Example 5.19). If ris the
position of q
2
relative to q
1
(Figure 16.31) then the force acting
onq
2
due to the presence of q
1
is
The unit vector fromq
1
toq
2
is Therefore
(16.70)
The force has strength , and acts along the line fromq
1
toq
2
for like charges,
fromq
2
toq
1
for unlike charges (see also Example 5.17). In addition, the force per
unit charge acting at point rdue to the presence of chargeq
1
isE 1 = 1 F 2 q
2
. Then
r
12
0
2
4 πε
F
r
==r
r
r
12
0
3
12
0
2
44 ππεε
ˆ
ˆr.
r
=
r
=
r
12
0
3
4
r
πε
=++
=
qq
x
r
y
r
z
r
r
x
12
0
33 3
12
0
3
4
4
π
π
ε
ε
ijk ( i+++yzjk)
Fi=−∇ =− j
∂
∂
+
∂
∂
+
∂
∂
V
xr yr
12
0
4
11
πε zzr
1
k
Fijk=−∇ =−
∂
∂
∂
∂
∂
∂
V
V
x
V
y
V
z
F
V
x
F
V
y
F
V
z
xyz
=−
∂
∂
,=−
∂
∂
,=−
∂
∂
q
1
q
2
r
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Figure 16.31