CHAPTER
4
ENERGY
AND
POTENTIAL
In the previous two chapters we became acquainted with Coulomb's law and its
use in finding the electric field about several simple distributions of charge, and
also with Gauss's law and its application in determining the field about some
symmetrical charge arrangements. The use of Gauss's law was invariably easier
for these highly symmetrical distributions, because the problem of integration
always disappeared when the proper closed surface was chosen.
However, if we had attempted to find a slightly more complicated field,
such as that of two unlike point charges separated by a small distance, we would
have found it impossible to choose a suitable gaussian surface and obtain an
answer. Coulomb's law, however, is more powerful and enables us to solve
problems for which Gauss's law is not applicable. The application of
Coulomb's law is laborious, detailed, and often quite complex, the reason for
this being precisely the fact that the electric field intensity, a vector field, must be
found directly from the charge distribution. Three different integrations are
needed in general, one for each component, and the resolution of the vector
into components usually adds to the complexity of the integrals.
Certainly it would be desirable if we could find some as yet undefined scalar
function with a single integration and then determine the electric field from this
scalar by some simple straightforward procedure, such as differentiation.
83
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View Full DocumentThis scalar function does exist and is known as the
potential
or
potential
field
. We shall find that it has a very real physical interpretation and is more
familiar to most of us than is the electric field which it will be used to find.
We should expect, then, to be equipped soon with a third method of finding
electric fieldsÐa single scalar integration, although not always as simple as we
might wish, followed by a pleasant differentiation.
The remaining difficult portion of the task, the integration, we intend to
remove in Chap. 7.
4.1 ENERGY EXPENDED IN MOVING A
POINT CHARGE IN AN ELECTRIC FIELD
The electric field intensity was defined as the force on a unit test charge at that
point at which we wish to find the value of this vector field. If we attempt to
move the test charge against the electric field, we have to exert a force equal and
opposite to that exerted by the field, and this requires us to expend energy, or do
work. If we wish to move the charge in the direction of the field, our energy
expenditure turns out to be negative; we do not do the work, the field does.
Suppose we wish to move a charge
Q
a distance
d
L
in an electric field
E
.
The force on
Q
due to the electric field is
F
E
±
Q
E
²
1
³
where the subscript reminds us that this force is due to the field. The component
of this force in the direction
d
L
which we must overcome is
F
EL
±
F
Á
a
L
±
Q
E
Á
a
L
where
a
L
±
a unit vector in the direction of
d
L
:
The force which we must apply is equal and opposite to the force due to the
field,
F
appl
±´
Q
E
Á
a
L
and our expenditure of energy is the product of the force and distance. That is,
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 Spring '10
 Wilton
 Electrostatics, Electromagnet, Electric charge, Fundamental physics concepts, eText Main Menu

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