We have seen that the difference in electric potential between two arbitrary points in space is a function of the electric field which permeates space, but is independent of the test charge used to measure this difference. Let us investigate the relationship between electric potential and the electric field.
Consider a charge
which is slowly moved an infinitesimal distance
along the
-axis. Suppose that the difference in electric potential between the final and initial positions of the charge is
. By definition, the change
in the charge's electric potential energy is given by
, electric field strength has dimensions of potential difference over length. It follows that the units of electric field are volts per meter (
. Of course, these new units are entirely equivalent to newtons per coulomb: i.e.,
Consider the special case of a uniform
-directed electric field
generated by two uniformly charged parallel planes normal to the
-axis. It is clear, that if
is to be constant between the plates then
must vary linearly with
in this region. In fact, it is easily shown that
where
is an arbitrary constant. , the electric potential
decreases continuously as we move along the direction of the electric field. Since a positive charge is accelerated in this direction, we conclude that positive charges are accelerated down gradients in the electric potential, in much the same manner as masses fall down gradients of gravitational potential (which is, of course, proportional to height). Likewise, negative charges are accelerated up gradients in the electric potential.
, the
-component of the electric field is equal to minus the gradient of the electric potential in the
-direction. Since there is nothing special about the
-direction, analogous rules must exist for the
- and
-components of the field. These three rules can be combined to give
Here, the
derivative is taken at constant
and
, etc. The above expression shows how the electric field
, which is a vector field, is related to the electric potential
, which is a scalar field.
We have seen that electric fields are superposable. That is, the electric field generated by a set of charges distributed in space is simply the vector sum of the electric fields generated by each charge taken separately. Well, if electric fields are superposable, it follows that electric potentials must also be superposable. Thus, the electric potential generated by a set of charges distributed in space is just the scalar sum of the potentials generated by each charge taken in isolation. Clearly, it is far easier to determine the potential generated by a set of charges than it is to determine the electric field, since we can sum the potentials generated by the individual charges algebraically, and do not have to worry about their directions (since they have no directions).
looks rather forbidding. Fortunately, however, it is possible to rewrite this equation in a more appealing form. Consider two neighboring points
and
. Suppose that
is the vector displacement of point
relative to point
. Let
be the difference in electric potential between these two points. Suppose that we travel from
to
by first moving a distance
along the
-axis, then moving
along the
-axis, and finally moving
along the
-axis. The net increase in the electric potential
as we move from
to
is simply the sum of the increases
as we move along the
-axis,
as we move along the
-axis, and
as we move along the
-axis:
But, ,
, etc. So, we obtain
which is equivalent to
where
is the angle subtended between the vector
and the local electric field
. Note that
attains its most negative value when
. In other words, the direction of the electric field at point
corresponds to the direction in which the electric potential
decreases most rapidly. A positive charge placed at point
is accelerated in this direction. Likewise, a negative charge placed at
is accelerated in the direction in which the potential increases most rapidly (i.e.,
). Suppose that we move from point
to a neighboring point
in a direction perpendicular to that of the local electric field (i.e.,
). In this case, it follows that the points
and
lie at the same electric potential (i.e.,
). The locus of all the points in the vicinity of point
which lie at the same potential as
is a plane perpendicular to the direction of the local electric field. More generally, the surfaces of constant electric potential, the so-called equipotential surfaces, exist as a set of non-interlocking surfaces which are everywhere perpendicular to the direction of the electric field. Figure 14 shows the equipotential surfaces (dashed lines) and electric field-lines (solid lines) generated by a positive point charge. In this case, the equipotential surfaces are spheres centred on the charge.
we found that the electric field immediately above the surface of a conductor is directed perpendicular to that surface. Thus, it is clear that the surface of a conductor must correspond to an equipotential surface. In fact, since there is no electric field inside a conductor (and, hence, no gradient in the electric potential), it follows that the whole conductor (i.e., both the surface and the interior) lies at the same electric potential.