The Chemistry Maths Book, Second Edition

16.5 The scalar (dot) product 461

EXAMPLE 16.14Charges in an electric field

The force experienced by a charge qin the presence of an electric fieldEisF 1 = 1 qE.

If the field is an electrostatic field (constant in time) then this force is conservative

so that, by equation (16.40), the components of the field,E 1 = 1 F 2 q, are (minus) the

derivatives of a functionφ 1 = 1 V 2 q:

(16.43)

The function φis the electrostatic potential function(potential energy per unit

charge) of the field.

If the field Eis a uniformfield, constant in space, then integration of these

equations gives (see (5.58) for a constant force)

φ(r) 1 = 1 −(xE

x

1 + 1 yE

y

1 + 1 zE

z

) 1 + 1 C 1 = 1 − 1 r 1

·

1 E 1 + 1 C (16.44)

whereφ(r) is the electrostatic potential at positionr 1 = 1 (x,y, z)and Cis an arbitrary

constant. The potential energy of charge qat rin the field is then

V 1 = 1 qφ(r) 1 = 1 −qr 1

·

1 E 1 + 1 qC (16.45)

The potential energy of a system of charges, q

1

at r

1

, q

2

at

r

2

=,q

N

atr

N

, is the sum of the energies of the individual charges,

For the potential (16.44), we therefore have

(16.46)

= 1 −

μ

1

·

1 E 1 + 1 QC

where μis the dipole moment of the system of charges and Qis the total charge. The

term QCis zero for an electrically neutral system or if the potential φis chosen to be

zero at the origin (the usual choice). Then

V 1 = 1 −

μ

1

·

1 E 1 = 1 −μE 1 cos 1 θ

0 Exercise 30

VqqC q

NN

=− + =−

==























∑∑

i

ii i

i

ii

rE r

11

()

·

1



=

























• 
  

∑

11

·

E

i

i

1

N

qC

Vq q q q

NN

i

N

i

=+ ++ =

=

∑

11 22

1

φφ() ()rr φ( )r rφ()

i



E

x

E

y

E

z

xyz

=−

∂

∂

,=−

∂

∂

,=−

∂

∂

φφφ

E

μ

θ

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Figure 16.23