The Chemistry Maths Book, Second Edition

(Grace) #1

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


qq


r


12

0

2

4 πε


F


r


==r


qq


r


qq


r


12

0

3

12

0

2

44 ππεε


ˆ


ˆr.


r


=


r


=


qq


r


12

0

3

4


r


πε


=++








=


qq
x

r


y


r


z


r


qq


r


x


12

0

33 3

12

0

3

4
4

π
π

ε
ε

ijk ( i+++yzjk)


Fi=−∇ =− j









 +









 +




V


qq


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

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