Download I. Charge Densities

We began our study of electromagnetism with the consideration of electric point charges
under static conditions. We have considered the electric field in the vicinity of these
charges as well as the electric potential surfaces that are generated. What if instead of a
single point charge you have a continuous charge distribution spread along a line, a
surface or throughout a volume? How do we determine the electric field or electric
potential due to continuous charge distributions? This is the goal of our second unit of
electromagnetism. Before taking the second test of the third quarter (test 302) each
student should be familiar with the material outlined below. Students should also review
the corresponding equations from the equation map of the previous unit.
I.
Charge Densities
A. Linear Charge Distributions and 
B. Surface Charge Distributions and 
C. Volume Charge Distributions and 
II.
Electric Field of Continuous Distributions by Direct Integration
A. Linear Distributions
1. Long straight wire
2. Semi-circle of wire
B. Surface Distributions
1. Axis of ring
2. Axis of disk
3. Above the mid-point of a square plane of charge
III.
Potential of Continuous Distributions by Direct Integration
(Same situations as in II)
IV.
Relationship Between E and V
1. Differentiation of V to get E
2. Integration of E to get V
Lesson 3-09
Read Section 23-6
Charge Densities
Charge can be distributed along a straight line or curve, over a two-dimensional surface
or throughout a solid. In this lesson we consider each of these possible charge
distributions. Table 23-2 shows the symbol and each of the dimensions. You should be
able to recognize each of these and associate the correct symmetry with the correct form
of charge density.
Linear Charge Density ()
Charge can be distributed along a line or curve. We use the symbol lamda that is
the Greek equivalent of the letter L. Remember “L” for
 = q/L or  = dq/dx
line charge, “L” for lambda. Lamda has units of C/m.
uniform
or variable density
Consider the example problems using a line charge:
Example #1
Suppose the rod in figure 23-11b has a length of 60 cm and = 300nC/m. Determine the
total charge on the rod if it is centered on the origin.
Example #2
Suppose that the rod of the previous problem has a charge density that varies according
the distance from the origin. Find the total charge on the rod if it has a density according
to  = +400nC/m2 |x|. What fraction of the charge is between  L/2 and L /4?
Example #3
Refer to figure 23-42. Assume that x = 0 is at point P. Let the charge density on the rod
be  = C/|x| where C is a constant. What is the total charge on the rod in terms of C, L
and a? What does your answer reduce to if L = a?
Surface Charge Density (}
The Greek letter sigma is used for the charge distributed over a surface.
Remember “s” for surface and “s” for sigma. Expect
 = q/A or  = dq/dArea
to see this type of charge density when working with
uniform
or variable density
a sheet of charge or a solid conductor such as a
conducting sphere or conducting cylinder. This
charge density is also very common when dealing with capacitors. In most cases the
differential area can be written in terms of a single variable but not always.
Example #4
Refer to figure 23-12 that shows a charged disk of radius R. Find the net charge on the
disk if the uniform charge has a value of 250 C/m2. The disk has a diameter of 12 cm.
What is the differential charge shown in the yellow differential area in terms of r and dr?
Example #5
Suppose that the charge density of the disk is not uniform. Instead the concentration of
charge is greater near the outer edge of the disk. Find the total charge on the disk if the
charge density on the surface is  = (200mC/m3)r.
Volume Charge Density ()
Sometimes charge is distributed throughout a volume. We use these three dimensional
models of charge for the nucleus of an atom. We could also get volume charge
distributions by exposing non-conducting solids to
 = q/Vol or  = dq/dVol
ionizing radiation. The symbol for volume charge
uniform or variable density
density or any three dimensional density is the
Greek letter “rho”. I have no idea why this was
chosen except that there is probably no Greek letter for “V”? Use of this charge density
is most often seen in problems that deal with three-dimensional, non-conducting solids.
Example #6
A sphere has a uniform charge density of 42 C/m3. The radius of the sphere is 9 cm.
What is the total charge of the sphere? What percent of the charge is within 5 cm from
the center?
Example #7
A hollow sphere has an inner radius of 6 cm and an outer radius of 10 cm. The total
charge of 900 pC is uniformly distributed throughout the hollow sphere. What is the
charge density?
Example #8
A sphere with a diameter of 8 cm has a non-uniform charge distribution of  = Cr where
C has a value of 0.002. What are the dimensions of C? What is the total charge of the
sphere? Hint: The differential volume is a spherical shell of thickness dr.
Lesson 3-10
Read Section 23-6&7
Electric Fields from Continuous Charge
Now that we have considered spreading charge along 1, 2 or 3 dimensions the next
logical step is to consider the electric field due to this charge. For a uniform, linear
charge distribution the electric field will be radial with no component along the axis.
Picture those spinning, horizontal brushes at an automatic carwash and you have a good
idea. For spherical distributions either uniform or that vary only according to “r” the
field lines will remain like that of the spiny sea urchin. When outside of spheres you can
generally treat them as point charges and when outside of cylinders you can treat them as
line charges but there are other symmetries. What about a semi-circle of charge, a ring of
charge or even a disk of charge? The electric fields of these charge distributions are the
goals of today’s lesson.
Finding E
In order to find the electric field at a point in space for a continuous charge
distribution one must draw a picture and consider the
dE = kdq/r2
small amount of electric field, dE, contributed by a
small amount of charge, dq. In order to get the total
electric field one must sum all of the dE’s due to the dq’s. One must be careful of two
things when constructing the integral. First draw two differential charges that will create
two symmetric dE’s. This will enable you to determine if you are adding the x part only,
y-part only or the total dE from each dq. Second, draw a third dq that produces a third
dE that is slightly off-set from the first dq. By comparison of the first and third you
should be able to determine what is a variable in the integrand and what is a constant.
You should be able to determine the total electric field for each of the following
situations:
1. Find E at a point that is at a distance along the axis of a line of charge, fig. 23-11a.
2. Find E at a point along the perpendicular bisector of a line of charge, fig. 23-11b.
3. Find E at a point along the axis of a ring of charge, figure 23-9.
4. Find E at the focus of an arc of charge such as in figure 23-10.
5. Find E along the axis of a disk of charge such as in figure 23-12.
6. Find E above the mid-point of a square sheet of uniform charge. Hint: consider the
square to be composed of a series of rods, stacked side-by-side.
As a final note recognize that when you are very near a sheet of charge the electric field
expression should approach a limiting value of E = /(2o) for a non-conducting surface
and a limiting value of E = E = /o for a conducting surface.
Homework Problems Ch 23: 27, 29, 30, 31, 33, 34, 36, 38
Lesson 3-11
Read Section 25-8
Electric Potential of Continuous Charge
In the previous lesson we learned how to determine the electric field from a continuous
charge distribution. How would you determine the electric potential for the same
conditions? This is the object of today’s lesson.
The trick is to break the charge distribution into elements of charge, dq, with each
differential charge contributing a small portion of potential, dV.
In contrast to yesterday’s lesson this is a scalar sum so that you
dV = kdq/r
do not need to draw the symmetric differentials. If there is a high
degree of symmetry you can take advantage of halving the limits and doubling the
integral value. Once the expression for the total potential at a point is determined don’t
forget about using that to get work done or final kinetic energy as was done in the
previous unit for a point charge moved from infinity. Review the Sample problems from
section 25-8. You are responsible for finding V for each of the following:
1. Find E at a point that is at a distance along the axis of a line of charge, fig. 23-11a.
2. Find E at a point along the perpendicular bisector of a line of charge, fig. 23-11b.
3. Find E at a point along the axis of a ring of charge, figure 23-9.
4. Find E at the focus of an arc of charge such as in figure 23-10.
5. Find E along the axis of a disk of charge such as in figure 23-12.
6. Find E above the mid-point of a square sheet of uniform charge. Hint: consider
the square to be composed of a series of rods, stacked side-by-side.
Homework Problems Ch 25: 35, 36, 37, 38, 39, 40, 41, 42
Lesson 3-12
Read Section 25-9
Getting E from V
In the previous two lessons we have learned how to get expressions for the electric field
vector E and expressions for the scalar field, electric potential V. Both lessons involved
direct integration. The direct integration to get E is complicated by the fact that it is a
vector summation. The direct integration to get V was much easier due to the scalar
nature of the electric potential.
In today’s lesson we find an easier way to get the expression for the electric field
by differentiation of the electric potential equation.
E = - dV/dx for an equation
One must be careful that your potential expression
is a function of some variable instead of the equation E = - slope for graph of V vs. r
at a specific point. Consider equation 25-37 that gives
the electric potential at a point of z above the origin. In this case you can get the electric
field along the z-axis by using E = -dV/dz. Consider equation 25-35 that gives the
electric potential at a specific point, P. In this case the summation was constructed
specifically for a single point. You cannot use the above boxed method to get E at that
point. In order to use the above-boxed equation the problem should be analyzed for a
general distance of “x” instead of a specific value of “d”. Replacing the d with an x will
enable us to get the electric field in the x-direction by using E = -dV/dx. We still cannot
get the y component of the electric field at point P however. If the potential is shown
graphically instead of analytically then you can look at the negative slope of the graph in
order to get the electric field. This idea is often useful on the AP exam. Study Sample
Problem 25-7 and work the following exercises.
Homework Problems Ch 25: 45, 46, 47, 48(a), 49, 50(a), 51
Lesson 3-13
Getting V from E
Read Section 25-4&11
In the previous lesson we saw how to get an expression for E from an expression for V.
Is the reverse possible? You bet it is. In today’s
lesson we do the exact opposite. Yesterday’s
f
equation must be turned inside out. This is
V f  Vo   E  dS
shown in the box to the right as well as in
o
equation 25-19 of the text. You can use the
integral expression to determine the change in potential between two points, V. You
can also simplify the expression by making either the initial or final potential to be zero
volts. In most cases, but not necessarily always, the potential is defined to be zero at an
infinite distance. The text uses this equation to find the potential expression for a single
point charge. You should also review Sample Problem 25-2. We will use this equation
over and over again in the next unit. For this unit we will apply it to some of the
problems with spherical symmetry as an introduction. Recall the following facts for
spherical symmetry. 1) If you are outside of the total charge and the distribution does not
depend on any angular variables then you may treat the electric field as if it is from a
point charge. 2) When inside of conducting spheres the electric field is zero for static
conditions. 3) When inside of a uniform sphere of charge the electric field increases
linearly from zero at the center to that of a point charge at the surface.
Example #1
Let V = 0 at infinity and find the electric potential at the center of a conducting sphere
that has a total charge of +Q and a radius of “a”.
Example #2
Consider a uniform spherical charge distribution with a total charge of +Q and radius of
“a”. Find the potential at the center of the sphere is V = 0 at infinity.
Example #3
Find the potential at the origin of the following system if V = 0 at infinity. The charged
sphere of the previous problem is centered on the origin and placed at the center of a thin
conducting shell of charge +2Q. The thin conducting shell has a radius of 2a.
Example #4
Sketch rough graphs of both E(r) and V(r) for the previous three example problems.
Homework Problems Ch 25: 12, 14, 74, 75, 82(b)
Test Soon!