Physical Chemistry , 1st ed.

Equations 8.2 and 8.3 involve the force due to electrical charges in a vac-
uum. If the electrical charges are in some medium other than vacuum, then a
correction factor called the dielectric constant,r, of that medium appears in
the denominator of the equation for the force. Equation 8.3 becomes

F

4 

q

0

1







q

r

2

r^2

 (8.4)

Dielectric constants are unitless. The higher the dielectric constant, the smaller
the force between the charged particles. Water, for example, has a dielectric
constant of about 78.
The electric field Eof a charge q 1 interacting with another charge q 2 is de-
fined as the force between the charges divided by the magnitude of the charge
itself. Therefore, we have

E

q

F

1



4 

q

0

2

r^2

 (8.5)

in a vacuum. (Again, for a nonvacuum medium, we would add the dielectric
constant of the medium in the denominator.) The magnitude of the electric
field E(the electric field is technically a vector) is the derivative with respect
to position of some quantity called the electric potential :

E 

r



Electric potential represents how much energy an electric particle can acquire
as it moves through the electric field. We can rewrite this equation and inte-
grate with respect to position r:

Edrd

( Edr)d


Edr

Since we have an expression for Ein terms ofr(equation 8.5), we can substitute:





4 

q

0

2

r^2

dr

This integral is solvable, since it is a function ofr(that is,rto the second power
in the denominator; all other variables are constant). We get





4 

q 2

 0



r

1

2 dr

Evaluating:



4 

q



2

0 r

 (8.6)

The units for electric potential, based on this expression, are J/C. Since we will
be working with electric potentials quite a bit, we define a new unit, volt (V),
such that

1 V 1 J/C (8.7)

The unit volt is named in honor of the Italian physicist Alessandro Volta, who
enunciated many fundamental ideas about electrochemical systems.

8.2 Charges 209