The Chemistry Maths Book, Second Edition

16.3 Components of vectors 451

EXAMPLE 16.5The centre of mass of a system of Nmasses, m

1

with position

vectorr

1

(‘at positionr

1

’),m

2

atr

2

,=,m

N

atr

N

(Figure 16.13) is

(16.12)

where is the total mass.

The cartesian components of a position vector r

of a point are the cartesian coordinates of the point;

r 1 = 1 (x, 1 y, 1 z). The vector equation (16.12) therefore

corresponds to three ordinary (scalar) equations, one

for each coordinate ofR 1 = 1 (X, Y, Z):

(16.13)

(see equation (5.40) for the one-dimensional case).

0 Exercise 11

EXAMPLE 16.6Dipole moments

The system of two charges shown in Figure 16.14,

with−qatr

1

andqatr

2

, defines an electric dipole

with vector dipole moment

μ 1 = 1 −qr

1

1 + 1 qr

2

1 = 1 q(r

2

1 − 1 r

1

) 1 = 1 qr (16.14)

More generally, a system of Ncharges,q

1

atr

1

,q

2

at

r

2

,=,q

N

atr

N

has dipole moment with respect to the

originO as point of reference,

(16.15)

This quantity depends on the position of the reference point if the total charge

is not zero. Thus, the position of chargeq

i

with respect to some point Ris

r

i

1 − 1 R, and the dipole moment of the system of charges with respect to Ris

(16.16)

andμ(R) 1 = 1 μ( 0 )only ifQ 1 = 10.

μμ()RrRr( )

i

ii

i

ii

i

=−=













−

===

∑∑∑

111

NNN

qqq

ii

RR













=−μμ() 0 Q

Qq=

∑ i

μμ=+++ =

=

∑

()qq q q

NN

N

11 2 2

1

rr r r

i

ii

X

M

mx Y

M

my Z

M

== =mz

== =

∑∑∑

111

i 111

ii

i

ii

i

ii

NNN

Mm

N

=

=

∑

i

1

Rrr r r

i

ii

=+++=

=

∑

11

11 2 2

1

M

mm m

M

m

NN

N

()

. .... ... ... .... .... ... .... ... .... ... .... ... .... ... .... .... ... ... .... .... .

r

1

r

2

r

N

m

1

m

2

m

N

o

.

.....

......

.......

.....

......

.......

.....

......

.......

.....

.......

......

.....

.......

......

.....

.......

......

....

.....

......

.

..

.........

........

.......

.........

............

...........

............

...........

..........

............

...........

............

...........

............

..........

...........

............

...........

............

...........

..........

............

...........

............

...........

............

..........

...........

............

...........

.....

.......

......

.......

......

.......

..

...............

.............................

....

.................

...................

.................

...................

.................

....................

.................

...................

.................

...................

.................

............................

................................

.......

........

.........

........

....

Figure 16.13

........

............................................

.......................................

......................................

............................................

.......................................

..............................................

..........

...........

...

....

...................

..

........

.......................................

...........................................

.......................................

.....

..

..........

.........

........

..........

.........

........

..........

.........

..........

........

.........

..........

........

.........

..........

........

.........

..........

........

.........

..

......

.....

.......

......

.

........................

...................

..........

........

.........

..........

........

.........

..........

........

.........

..........

........

.........

..........

........

.........

...

......

.......

.....

......

.......

.....

......

.......

.....

......

.......

......

.....

.......

......

.....

.......

......

.....

.........

........

..........

........

......

..

.....

......

....

.....

......

.....

.

.....

......

−q

q

r

2

r

1

r

o

Figure 16.14