Download Lesson 3 Electric Potential You have no doubt noticed that TV sets

Lesson 3
Electric Potential
You have no doubt noticed that TV sets, light bulbs, and other electric appliances
operate on 115 volts, but electric ovens and clothes dryers usually need 220 volts.
Batteries may be rated at a harmless 1.5, 6, 9, or 12 volts, but a high-tension electric
transmission line may provide electric power at 400,000 volts. Now just what physical
quantity is measured by all these volts? How do volts relate force, energy, and power,
about which you have learned in earlier lessons? The answer is that volts measure
electric potential difference (sometimes called “voltage”), which is derived from the
potential energy acquired by electrically charged objects as a result of the electric forces
they experience. Even though your familiarity with volts probably stems from electric
power supplied to your household, your introduction to the concept of electric potential
in this lesson will be in the context of the interaction of stationary (static) electric
charges.
3-1: Definitions and Relations
OBJECTIVES: Relate electric potential to (a) work done on a displaced charge, (b) the
electric field, and (c) electric potential energy. Use the electron volt to
express energy and solve simple problems applying energy
conservation.
State and interpret the conservative nature of the electrostatic field.
PREREQUISITES: Evaluating work integrals and applying the work-energy theorem to
solve problems
Calculating potential energy and identifying conservative forces
Calculating electrostatic forces and fields
Reading Assignment
Study in your textbook Chapter 23.
Commentary
The reading in this chapter includes a fairly thorough review of the work done by a
conservative force. This is an essential concept for understanding electric potential,
and since it has been some time since we have applied it, the review is well worthwhile.
In particular, make sure that you understand the sign convention for negative and
positive work done on a particle in a force field: positive work must be done by an
external agent to move a positive test charge from a region of lower potential to a region
of higher potential. Another way of remembering this convention is to note that electric
flux line point in the direction of decreasing potential.
As you study the reading assignment notice also that the focus of the discussion
sometimes shifts from potential difference between two points (
) to the potential
at a point ( ). To accomplish this transition, a certain point at infinity, at the coordinate
origin, or at some other point having symmetry in relation to the charges and fields is
arbitrarily chosen as the reference point in order to give potential as a function of
position in relation to that point (“absolute potential”). Then the electric potential is
simply a (scalar) function of the coordinates of any given point with the reference point
at zero potential by definition. The convention of using infinite separation as a reference
point for electric potential is consistent with the definition of electric potential energy as
the work required to assemble a system of charges by bringing them together from an
infinite distance. You may find it useful to explore the analogy between electrostatic
forces and gravitation by asking yourself such questions as “What would be the unit of
gravitational potential?” and “How does the variation in this potential as one approaches
a mass differ from the variation of electric potential as one approaches a charge?”.
EXAMPLE 3-1: Consider a region of constant electric field in the
field having magnitude
the
. The four points
coordinates
, and
direction, with the
and
have
,
.
,
There is no gravitational
field.
(a) An external agent slowly moves a body with charge
from
to . How much work is done on the body by the electric field, and
how much is done on the body by the external agent?
sketch to illustrate.
Use a
(b) Answer the questions in part (a) for displacement of the same body
(i) from to , (ii) from to , and (iii) from to .
(c) A bead of mass
and charge
long a frictionless wire between points
is permitted to slide a
and . At which of the two
points should it be released from rest in order that it will get to the
other one, and with what speed will it arrive?
(d) Find the differences in electric potential between points
points
and
, and points
and
and
; compare them with your
answers to parts (a) and (b).
Solution: (a) The situation is depicted at
right. From the definition of
work,
.
For a constant field, as in this
case,
done by field.
The external agent exerts a force
(b) (i)
and hence does
;
.
(ii)
.
(iii)
,
of work.
(c) The negative charge in this part experiences a force to the left, hence it
must be released at
and will acquire speed sliding toward . The
kinetic energy gained is equal to the work done on it by the electric field,
which is
.
Hence
, and
.
(d)
.
.
.
.
The electron volt (
.
.
), introduced in Section 23.2, is a unit of energy commonly used in
atomic physics. It has the advantage that when an electron moves in an electric field
across a potential difference of, say,
, its kinetic energy changes by
. In
other words, the potential difference gives the energy directly in electron volts if you are
dealing with a proton, electron, or other singly charged atomic particle. For most other
purposes, the electron volt is impractically small. The system of International Units (SI)
asks that the electron volt only be used in atomic physics. Otherwise is should be
converted to joules (
). Remember to convert to SI units before
using masses in kilograms and velocities in meters per second with an energy originally
given in electron volts.
EXAMPLE 3-2: In the figure at right, electrons acquire an increased speed by moving in
an electric field between two plates with slits through which the
electrons can pass. The electric potential difference between the plates
is
.
(a) By how much does the
electric potential energy of
an electron change as it
passes through the two-plate
system? (Give answer in
electron volts.)
(b) By how much does the
kinetic energy of an electron
changes as it passes
through
the
two-plate
system? (Give answer in
electron volts.)
(c) With what speed does an electron leave the space between the
plates if it enters with a speed of
?
Solution: (a) It decreases by
(b) It increases by
.
.
(c)
.
.
.
.
One of your objectives in this section is to be able to verify that electrostatic forces are
conservative. You know that a force field is conservative (permits the definition of a
potential energy function) if and only if (a) the work done by the force on a particle
moved from point to point is independent of the path taken from to ; or (b) the
work done by the force on a particle moving through a closed path back to its starting
point is zero. Since the electrostatic force at any point is directly proportional to the
electric field at that point,
,
(1)
The two equivalent conditions on the work stated above lead to two equivalent
conditions on the field:
(a)
is independent of path from
to
(2)
is independent of path from
to .
(3)
implies that
(b)
by any closed path
(4)
by any closed path.
(5)
implies that
(Note: the line integral in Eqs. (4) and (5) along a closed path is sometimes indicated by
a small circle on the integral sign:
. Do not confuse this symbol with the
same symbol used in many texts to represent a surface integral over a closed surface,
as in Gauss’s law. You can distinguish between the two uses of this symbol by looking
carefully at the infinitesimal element under the integral sign to see if it refers to a surface
or to a displacement.)
We shall now show that Coulomb field caused by a single point charge is conservative
by evaluating the integral in Eq. (3) and observing that its value depends only on the
end points, not on the path. First, we note that the magnitude of the electric field
caused by the point charge depends only on the distance (not on the entire vector )
from the charge and is given by
,
where
and
(6)
is the distance from
Figure 3-1 illustrates the geometric relationships.
. The direction of the field is radial.
Figure 3-1. The value of the line integral
element
is found in the limit as the path
approaches zero.
In Figure 3-1(a), you can see points
and
arrows representing four electric field values
We must now calculate
, a particular path we have chosen, and
and four infinitesimal line elements
.
, which we shall do with the help of the enlarged
diagram in Figure 3-1(b). This figure shows the angle
between the path and the
radial direction and also the radial increment
. Using the definition of the
dot product, we find
.
(7)
In other words, because the field is radially directed, only the radial increments and not
the angular increments of the path contribute to the line integral.
With the integrand in Eq. (7), we can calculate the integral over the radius variable ,
which varies from
to
:
.
(8)
Evidently, this result depends only on the end points and not on the path. Our proof is
hereby completed.
Our conclusion of path independence can be extended easily to an electric field that is
caused by several point charges. Since such a field is obtained by adding the fields
from the individual charges, the line integral can be expressed as a sum of line integrals
like that in Eq. (8), for each of which path independence has been established. It is not
so easy to extend the proof to the fields caused by continuous charge distributions. As
a matter of fact, more advanced treatments of the theory of static electric fields always
begin with the postulate that the fields are conservative.
Practice Exercise 3-1
Write your solutions to the following problems in your notebook.
1. An upward-directed, uniform electric field of
exists in a certain
region.
(a) What is the potential difference between a certain point
origin and the point (i)
diagonally downward to the right
above?
(ii)
that we shall call the
to the left?
(iii)
?
(b) How much work does an external agent to when it moves a charge of
very slowly from the origin to the point (i) in (a-ii)? (ii) in (a-ii)? (iii) in
(a-iii)?
(Hint: Draw a diagram and remember the definitions.)
2. An electron is placed at a distance of
to move freely until it is only
from a proton. It is then allowed
from the proton.
(a) What is the potential difference between the two points in the proton’s electric
field?
(b) How much kinetic energy does the electron gain during its motion? Express the
result in electron volts.
(c) With what speed does the electron arrive at the second point, assuming that it
starts from rest?
(Hint: Draw a diagram and remember the definitions.)
3. Given the two conditions that establish the conservative nature of the electric field:
is independent of path from
to
(3)
by any closed path
(5)
Show that these two conditions are equivalent.
Check your answers to Practice Exercise 3-1 with the Module 1 Answer Key and review
Section 3-1 as necessary before you begin working on the next section.
3-2: Finding Potentials from Charges
OBJECTIVES: Use the definition of electric potential and/or the superposition principle
to find the potential caused by (a) one or more given point charges, and
(b) continuous charge distributions with planar, cylindrical, or spherical
symmetry.
PREREQUISITES: Finding the electric field using Gauss’s law and describing the
electric field near conductors
Finding the electric field using Coulomb’s law and using field lines
to describe the electric field
Commentary
After the derivation of the potential due to a point charge in Section 23.2, the discussion
in the reading assignment is all presented in terms of absolute potential by setting
. With this approach, we can apply the superposition principle and find the
potential at any point that results from an arrangement of charges by simply taking the
(scalar) sum of the potentials due to the individual charges.
Another technique that is sometimes useful for analyzing the potential due to an
arrangement of point charges is to graph the potential as a function of position along a
single axis. This method is applied in the following example.
EXAMPLE 3-3: A point charge of magnitude
magnitude
(
is located at (
).
is located at (
) and a point of
), as in the figure below.
(a) Find a mathematical expression for the electric potential on the axis, and sketch it on a graph. (Draw the sketch from about six to
eight well-placed points.)
(b) Find a mathematical expression for the electric potential along the
-axis, and sketch it on a graph. (Draw the sketch from about six to
eight well-placed points.) How would you expect the potential to
vary along the -axis? Explain.
(c) Find the locus of points where the electric potential caused by the
two charges is equal to zero.
Solution: The electric potential caused by the two charges is
,
Where
charges
and
to
the
and
point
are the distances from the locations of the
(
):
and
.
(a) Since
along the
-axis,
and
.
(The
absolute values appear because distances are always to be taken as
positive.)
Hence, the potential on the -axis is given by
.
Rather than plotting
though, we will plot the dimensionless
quantity
).
We first make a data table for values of this quantity at several points on
the -axis:
We further note that
, because the charges
as
and
and
as
are located at these points.
Using the eight values in the table and asymptotes through
and
, we can sketch the graph.
(b) Along
the
-axis,
and
,
so
that
. Our data table is thus
and the -axis potential is sketched below.
The potential along the -axis should be identical to that along the -axis
since the
and
axes are symmetric with respect to the -axis on which
the charges are located. In fact, the potential is the same at all points of
any circle lying in the -plane and centered at the origin.
We note further that the potential is positive along the -axis between
and
, but that it is negative elsewhere on the -axis and on the other
axes. This is to be expected, since the negative charge had the greater
magnitude; hence the positive potential is confined to a region near the
positive charge. At large distances from both charges, where the distance
between them is negligible, the potential caused by the two-charge system
resembles that caused by a charge
at the origin.
(c) We saw from the graphs and qualitative reasoning that the potential would
be positive near the positive charge and negative elsewhere. Therefore
there should be a finite surface on which the potential is zero. Setting
, we find that
or
.
In other words,
.
By combining the -dependent terms, we find
.
The other terms also combine and permit cancellation of a factor . The
final equation is
,
which is a sphere of radius
about the point
passes through the origin and the point
.
The sphere
, where we had already
found the potential to be zero.
Example 3-3 illustrates the kind of quantitative results that can be obtained by applying
a bit of combined algebraic and geometric reasoning to a simple charge distribution. It
also yields a generalizable result: the electric potential of two point charges of opposite
sign is always zero on a spherical surface of certain radius and center that surrounds
the smaller of the two charges. You can prove this result without much difficulty by
using charges at the origin and
at the point (
), and proceeding as we did
above. What do you expect for
?
Finding the potential due to a continuous charge distribution (e.g., a charged conductor)
is fundamentally an application of Gauss’s law, and thus has similar limitations in
usefulness- to cases involving uniform sheets of charge and distributions with spherical
or cylindrical symmetry. In these cases, the sum that gives the net potential by means
of the superposition principle is replaced by an easily evaluated integral.
EXAMPLE 3-4: A point charge of magnitude
is located at the center of a hollow
conducting spherical shell that is electrically neutral. The interior and
exterior radii of the shell are
and , respectively. Find the electric
potential as a function of radius from the shell’s center, and draw a
sketch of the dimensionless combination
.
Solution: Since the system of charges is spherically symmetric, the electric field outside
the shell can be found from Gauss’s law. It is directed radially and has the
magnitude
,
.
In the shell itself, the electric field is zero because it is a conductor,
(9)
,
.
(10)
Inside the hollow cavity of the shell, the field is again found from Gauss’s law
(or memory):
,
.
(11)
To find the potential as a function of radius, relative to infinity as the reference
point, we integrate the electric field:
.
(12)
The important and new step (compared to evaluating this integral for a point
charge) is to break up the integral into several parts since the electric field is
given in different mathematical expressions for different intervals of the
radius. We also use the fact that is radial so that
.
Then in the first interval (
) we have one expression for
,
For the second interval (
for
.
for
For the third interval (
(13)
) we break the integral into two parts, one
using Eq. (9) and a second for
for
:
using Eq. (10):
;
.
(14)
) we apply the result for
integrate for the potential difference to
from Eq. (14) and
using Eq. (11):
.
In order to graph
, we arbitrarily set the ratio
(15)
.
Practice Exercise 3-2
Write your solutions to the following problems in your notebook.
1. Two point charges are placed on the -axis, one of magnitude
one of magnitude
at the point
at the origin, and
.
(a) Find the mathematical expression for the electric potential on the -axis. Draw
an approximate graph of your result.
(b) Find the point(s) on the -axis where the potential is zero.
(c) Find an approximate expression for the potential at large distances from the
origin along the -axis and explain its physical significance.
2. A home electrostatic air cleaner has a wire of radius
metal cylinder of radius
. The wire is at a potential of
at the center of a
relative
to the cylinder.
(a) Find the electric charge per unit length needed on the wire to establish this
potential difference.
(b) Show that the electric field at the surface of the wire is greater than
, a field at which air becomes ionized. (By the ionization of air,
charged particles are produced that can attach themselves to the dust, thereby
making the dusk subject to removal by the action of the electric forces.)
Check your answers to Practice Exercise 3-2 with the Module 1 Answer Key and review
section 3-2 as necessary.
3-3: Finding Fields from Potentials
OBJECTIVES: Determine the electric field when given an electric potential that is a
function of one position variable only.
Use equipotential surfaces and field lines for describing the potential
and field semiquantitatively near several given point charges and/or
simply shaped metallic surfaces.
Commentary
Study Figure 23.24 in the text carefully now in light of your new understanding of the
relationship between and . To find from graphically, we simply draw the flux
lines everywhere perpendicular to the equipotential surfaces. The method for using
partial derivatives of to find the components of in three dimensions is outlined in the
text. We will focus our problem-solving efforts on one-dimensional cases- those that
can be solved for by differentiating with respect to a single variable- but note that
the analysis is essentially the same when
depends on more than one variable. Study
the following examples before doing the practice exercise.
EXAMPLE 3-5: A certain spherically symmetric charge distribution generates the
potential
, where is the radius from the center. Find the
electric field.
Solution: Since
depends only on the radius, the electric field is radially directed and
has magnitude
.
If
is positive,
radially inward.
is directed radially outward. If
is negative,
is directed
EXAMPLE 3-6: They symmetry of the problem of the two point charges in Example 3-3
implies that on the -axis the electric field will be directed along the axis. Find the electric field on the -axis,
, using the result for
given in Example 3-3.
Solution:
We
must
calculate
from
the
result
.
To obtain the correct sign for
, we must eliminate the absolute-value signs
in the expression for . This involves three separate cases corresponding to
three intervals on the -axis.
Case (i):
,
,
.
Case (ii):
,
,
.
Case (iii):
,
,
.
Note that since the squares in the denominators of the electric-field terms
are always positive, you can identify the sign of a term form the sign in front
of it. Thus we see that
due to the charge at
is positive in the
interval
and negative for
, as it must be since this charge is
positive and thus repels another positive charge. The field along the -axis
due to the charge at
positively for
.
is directed negatively for
and
EXAMPLE 3-7: Two equal positive charges are placed on the -axis at equal distances
above and below the
-axis. Draw the equipotential lines and field
lines caused by this two-charge system in the
intersections of the equipotential surfaces with the
-plane (i.e., trace the
-plane).
Solution: For a problem like this, it is wise to exploit your knowledge of the potential of a
single point charge. Thus
(a) Near each charge, where the potential increases without limit, the
equipotential surfaces will be virtually unaffected by the presence of the
other charge.
The surfaces will approximate spheres, and their
intersection with the -plane will approximate circles centered around the
charge.
(b) Very far from all the charges, the whole system will act like a single point
charge whose magnitude is the algebraic sum of all the charges in the
system. Equipotential surfaces will again approximate spheres centered
around the “center” of the charges.
(c) Field lines are perpendicular to the equipotential surfaces and can
terminate only at charged bodies. After constructing the near- and far-field
lines with the help of (a) and (b), you must connect the two parts of the
diagram.
(d) At intermediate points, you may be able to use symmetry of the charges
and clues from the need to connect field lines at small and large distances
as described in part (c). You may also be able to locate special points
where the electric field is zero; at such points the field lines have no welldefined direction, and it is possible for two equipotential surfaces to
intersect. (Since these surfaces are perpendicular to the field lines, and a
field line can have only one perpendicular surface at a point, two
equipotential surfaces cannot ordinarily intersect.)
These ideas have been used to construct Figure 23.24(c) in the text. Note the small,
circular equipotentials near each charge, the large oval shapes that will become more
nearly circular at larger distances, the total number of field lines for the system (twice
that of each charge separately), and the field-free symmetry point half-way between the
charges where equipotentials intersect.
EXAMPLE 3-8: A point charge of magnitude
charge of magnitude
is located at
is located at
and a point
, with both
and
positive. Draw approximate equipotential lines and electric-field lines
in the -plane for this system of charges.
Solution: We shall begin with an overall analysis, following the procedure outlined in
Example 3-7.
(a) Near each charge there will be equipotential circles, representing positive
potentials near
and negative potentials near
.
(b) Far from the origin (that is, for
) there will be equipotential circles
representing the negative potentials of a charge
located approximately
at the origin.
(c) Field lines radiate outward from charge
at
many field lines radiate inward to charge
lines emanating from the charge at
the negative charge at
hemisphere at
, and two times as
at
. All of the field
go into the right hemisphere of
, whereas the lines coming into the left
come from infinity.
The latter are uniformly
distributed in angle at the large equipotential circles mentioned in (b).
(d) Since the charges are unequal, there is no simple right-left symmetry, but
there is up-down symmetry. Since the charges have opposite sign, the
field between them will be very strong. However, to the right of the
positive charge, which is weaker than the negative charge, there must be
a reversal of the field because near
the field lines point to the right,
whereas farther away, where the negative charge dominates, the field
lines point to the left. To find the zero-field position we need the root of
(see Example 3-6). This equation is solved most
easily by taking square roots of both sides and then collecting terms of the
linear equation, with the result that
.
At this point equipotential lines may cross to make a transition from the
behavior near the charges to the far behavior.
We can now draw the figure below as the solution to this problem.
Practice Exercise 3-3
Write your solutions to the following problems in your notebook.
1. A point charge of magnitude
is placed at the point
axis is given by
. Find the
. Its potential on the component of the electric field
associated with this potential by differentiating and justify your procedure. (Hint:
treat the absolute value carefully!)
2. A charge distribution at the coordinate origin gives rise to the potential
, where
is distance from the origin
and
is
given constant vector called the electric dipole moment.
(a) Find the
(b) Find the
component of the electric field for points on the -axis.
component of the electric field for points on the -axis. (Hint: First
state the potential on this axis.)
3. Draw approximate equipotentials and field lines in the
system in problem 1 of Practice Exercise 3-2.
-plane for the two-charge
4. The figure below shows a cross-section of equipotential surfaces in a region of
electric field with cylindrical (not circular) symmetry. Five points have been marked
by letters.
(a) At which point is the field strongest? Explain.
(b) At which point is the field weakest? Explain.
Check your answers to Practice Exercise 3-3 with those given in the Module 1 Answer
Key. Correct any errors in your solutions and review Section 3-3 as necessary before
going on to the self-check test. To see if you have achieved the objectives in Lesson 3,
try to solve the problems in Self-Check Test 3 without using any reference materials.
Self-Check Test 3
Write your solutions to the following problems in your notebook.
1. A long, hollow, cylindrical conductor of radius
has a charge of
coulombs per meter of length. On one side it has a tiny hole through which
electrons can emerge from the interior, but the hole has a negligible effect on the
electric field due to the cylinder. Use the hole as the reference (zero) point for the
electric potential.
(a) Find the electric potential
radius from the cylinder axis).
both outside and inside the cylinder (
is the
(b) How much kinetic energy is acquired from the field by an electron that emerges
from the hole and hits a screen
from the axis? Express the answer in
electron volts.
(c) With what speed does the electron hit the screen if it starts with negligible speed
in the hole?
(d) Suppose you had a frictionless guiding mechanism for the electron that made it
spiral around the cylinder rather than traveling straight from the hole to the
screen. How would that affect the answers to parts (b) and (c)? Explain briefly.
2. A
spherically
symmetric charge distribution gives rise to the potential
, where is the radius from the center of the charge, and and
are constants with dimensions of distance
potential and distance, respectively.
Find the electric field (magnitude and direction).
3. Two point charges of magnitudes
and
are placed with the separation
.
Make an approximate drawing of several equipotentials and field lines in a plane
including the two charges.
Check your answers with the key and review Lesson 3 as necessary.
Assignment 3
When you have demonstrated mastery of the content of this lesson, log into the
Mastering Physics website and work Homework Set 3.
Module 1 Review
Much of the rest of your work in this course will build upon what you have learned in this
unit, so it is essential that you have the important concepts of electrostatics firmly in
mind. You have considered these concepts from a number of different aspects; as you
review the unit try to “tie things together” in order to strengthen your understanding. For
example, lines of electric force (flux lines) can help you to visualize both equipotential
surfaces and Gaussian surfaces. The analogy between the gravitational field and the
electric field can also be useful; the only memory supplement you need for this analogy
is that the field is directed inward toward a negative charge. Use the objectives in
Lessons 1, 2, and 3 in this syllabus and the review-and-summary sections at the end of
Chapters 21-23 in the text to guide your review. When you have demonstrated mastery
of the content of this unit begin working on the next lesson.