DIFFERENTIAL OPERATIONS
25
The
RHS
of
this expression can be recognized
as
the total differential change of
Q,
due to a differential change of position
dF.
The result can be written entirely in vector
notation as
dQ,
=
VQ,
*
dF.
(2.33)
From Equation 2.33, it
is
clear that the largest
value
of
dQ,
occurs when
dT
points in
the same direction as
VQ,.
On the other hand, a displacement perpendicular to
VQ,
produces no change in
Q,
because
d@
=
0.
This means that the gradient will always
point perpendicular to the surfaces
of
constant
Q,.
At the beginning of this chapter, we discussed the electric potential function
generated by two line charges.
The
electric field was generated by taking the gradient
of
this scalar potential, and was plotted in Figure
2.1.
While we could use this example
as a model for developing a physical picture for the gradient operation, it is relatively
complicated. Instead, we will look at the simpler two-dimensional function
Q,
=
-xy.
(2.34)
A
plot
of
the equipotential lines of
@
in the xy-plane is shown in Figure 2.4. The
gradient operating on
this
function generates the vector field
-
VQ,
=
-ye,
-xi?,. (2.35)
Now
imagine we are at the point (1,2) and move to the right, by an infinitesi-
mal amount
dr
along the positive x-axis. The corresponding change in
Q,
can be
determined by calculating:
dQ,
=vQ,
.dF
=
(-26,
-
16,)
.
(dr
C,)
=
-2dr.
(2.36)
Y
Figure
2.4
Surfaces
of
Constant
@
=
-xy
