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The Electric Field
Concepts and Principles
Electric Charge, Electric Field and a Goofy Analogy
We all know that electrons and protons have electric charge. But what is electric charge and
what does it mean for a particle, like a neutron, to not have electric charge1? On one level, the
answer is that electric charge is the ability to create and interact with an electric field. Of
course, this begs the question, what is an electric field? To try to answer this question, let’s
look at mass and the gravitational field.
In Newton’s theory of gravitation, every object that has mass creates a gravitational field. The
object with mass is termed the source of the gravitational field. The source’s gravitational
field, which fills all of space, encodes information as to the location and mass of the source
into space itself. It’s almost as if an infinite number of little business cards have been printed
and distributed throughout space with detailed information concerning the source’s
characteristics. The gravitational field can be thought of as a huge number of business cards
embedded into the fabric of space. (No, I am not joking, it looks like this.)
1 kg source
down there
1 kg source
right here
1 kg source
over there
1 kg source
up there
1 kg source way
over there
1
Actually, neutrons are constructed from smaller particles called quarks that do have electric charge.
Neutrons are neutral because the sum of the quark charges inside of them is zero. The distinction
between an object having a net charge of zero and an object having no charge whatsoever is important
when considering polarization.
1
Typically when we draw the gravitational field we translate the message on the business cards
into a mathematically equivalent message. Here’s a picture of the gravitational field near the
surface of the earth.
9.8 N/kg
9.8 N/kg
9.8 N/kg
9.8 N/kg
9.8 N/kg
9.8 N/kg
9.8 N/kg
9.8 N/kg
9.8 N/kg
The relationship that allows you to “translate the business cards” is Newton’s expression for
the magnitude of the gravitational field surrounding a mass,
g=
Gm
r2
where
G is the gravitational constant, equal to 6.67 x 10-11 N m2/kg2,
m is the source mass, the mass that creates the gravitational field,
r is the distance between the source mass and the location of the business card,
and the direction of the gravitational field, g, is toward the source mass.
Every mass in the universe has a gravitational field described by this formula surrounding it.
Additionally, every mass in the universe has the ability to “read the business cards” of every
other mass. This transfer of information from one mass to another is the gravitational force
that attracts masses together. (Once one mass reads another masses’ business card, the first
mass feels a strange urge to go visit the second mass. The more enticing the business card (the
larger the value of g) the stronger the urge.)
Let’s drop the business card analogy and return to electric charges and electric fields.
Basically, every object that has electric charge surrounds itself with an electric field given by
a formula incredibly similar to the one for the gravitational field. How incredibly similar?
2
Below is the formula for the magnitude of the electric field at a particular point in space, from
a single source charge:
E=
kq
r2
where
k is the electrostatic constant, equal to 9.00 x 109 N m2/C2,
q is the magnitude of the source charge, the electric charge that creates the electric
field,
r is the distance between the source charge and the point of interest,
and the direction of the electric field is away from positive charges and toward
negative charges.
Every charged object is surrounded by a field given by this relationship. Every other charged
object in the universe can “read” this field and will respond to its information by feeling an
electric force. Objects without electric charge neither create nor interact with electric fields
(they can’t read the business cards).
In this chapter you’ll learn how to calculate the electric field produced by charged objects. In
the next chapter, you’ll learn how the electric field can be “sensed” by other electric charges
resulting in the electric force.
Charge and Charge Density
Macroscopic objects are normally neutral (or very close to neutral) because they contain equal
numbers of protons and electrons. All charged objects are charged because of either an excess
or lack of electrons. (It’s much easier to add or remove electrons from an object than trying to
add or remove the protons tightly bound inside the nuclei of its atoms.) Thus, the electric
charge of any object is always an integer multiple of the electric charge on an electron.
Because of its fundamental importance, the magnitude of the charge on an electron is termed
the elementary charge and denoted by the symbol e. In a purely logical world, the charge on
any object would be reported as a multiple of e. However, since the charge on a macroscopic
system can be many multiples of e, a more user-friendly unit, the coulomb (C), is typically
used to quantify electric charge. In this system,
e = 1.6 x10 −19 C
Thus, you can consider the charge on an electron as an incredibly small fraction of a coulomb,
or a coulomb of charge as an incredibly large number of electrons.
3
In many applications, in addition to knowing the total charge on an object you will need to
know how the charge is distributed. The distribution of charge on an object can be defined in
several different ways. For objects such as wires or other thin cylinders, a linear charge
density, λ, will often be defined. This is the amount of charge per unit length of the object. If
the charge is uniformly distributed, this is simply
λ=
Q
L
where Q is the total charge on the object2 and L its total length.
For objects such as flat plates or the surfaces of cylinders and spheres, a surface charge
density, σ, can be defined. This is the amount of charge per unit area of the object. If the
charge is uniformly distributed, this is
σ=
Q
Area
Lastly, for objects that have charge distributed throughout their volume, a volume charge
density, ρ, can be defined. This is the amount of charge per unit volume of the object. If the
charge is uniformly distributed, this is
ρ=
Q
V
To add to the confusion, you must realize that the same object can be described as having two
different charge densities. For example, consider a plastic rod with charge distributed
throughout its volume. Obviously, the charge per unit volume, ρ, can be defined for this
object. However, you can also define the object as having a linear charge density, λ, reporting
the amount of charge present per meter of length. These two parameters will have different
values but refer to exactly the same object.
2
I will use lowercase q to designate the charge on a point particle and uppercase Q to designate the total
charge distributed on macroscopic objects.
4
Perfect Conductors and Perfect Insulators
Determining how electric charges in real materials respond to electric fields is incredibly
important but also incredibly complicated. In light of this, we will initially restrict ourselves
to two types of hypothetical materials.
In a perfect conductor, electric charges are free to move without any resistance to their
motion. Metals provide a reasonable approximation to perfect conductors, although, of course,
in a real metal a small amount of resistance to motion is present. When I refer to a material as
a metal, we will approximate the metal as a perfect conductor.
In a perfect insulator electric charges can not move, regardless of the amount of force applied
to them. Many materials act as insulators, but all real materials experience electrical
breakdown if the forces acting on charges become so great that the charges begin to move.
When I refer to an insulating material, like plastic, for example, we will approximate the
material as a perfect insulator.
Since electric fields create forces on electric charges, there can not be static electric fields
present inside perfect conductors. If a field was present inside a perfect conductor, the charges
inside the conductor would feel an electric force and hence move in response to that force.
They would continue to move until they redistributed themselves inside of the conductor in
such a way as to cancel the electric field. The system could not reach equilibrium as long as
an electric field was present. This re-arranging process would typically occur very quickly and
we will always assume our analysis takes place after it is completed.
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The Electric Field
Analysis Tools
Point Charges - Numerical
A sodium ion (Na+) and chlorine ion (Cl-) are
separated by 2.82 x 10-10 m. Find the electric
field 2.82 x 10-10 m to the right of the chlorine
ion.
Na+
Cl-
The electric field at this point will be the vector sum of the electric field from the sodium ion
(ENa) and the electric field from the chlorine ion (ECl).
Since the sodium is positively charged its
electric field points to the right, away
from it.
Since the chlorine is negative its electric
field points to the left, towards it, and is
larger in magnitude than the field from
the sodium since the point of interest is
closer to the chlorine. Thus, we should
expect the resultant field to point to the
left 3.
Na+
Cl-
ECl
ENa
Let’s calculate the electric field from each charge separately and then combine the results. For
the sodium ion:
E Na = +
kq
r2
(9.00 x10 9 )(1.6 x10 −19 )
(2.82 x10 −10 + 2.82 x10 −10 ) 2
= 4.53 x10 9 N
C
E Na = +
E Na
3
For the sake of consistency, we will use a common coordinate system with the +x-direction pointing to
the right and the +y-direction pointing to the top of the page. For three dimensional systems, the +zdirection will point directly out of the page. The coordinate axes will be indicated in the vast majority of
diagrams.
6
For the chlorine ion:
kq
r2
(9.00 x10 9 )(1.6 x10 −19 )
=−
(2.82 x10 −10 ) 2
= −18.1x10 9 N
C
ECl = −
ECl
ECl
So the resulting electric field at this location in space is:
r r
r
E = E Na + ECl
E x = +4.53 x10 9 − 18.1x10 9
E x = −13.6 x10 9 N
C
Numerical calculations involving electric fields typically involve extremely large- and smallvalued variables, giving your knowledge of scientific notation a good workout. One way to
avoid dealing with these extremely large and extremely small numbers is to perform the
calculations symbolically, determining a result in terms of general system parameters rather
than numerical values. You will have to do a mix of both numerical and symbolic calculations
in this course.
Point Charges - Symbolic
Find the electric field at the indicated point. The
charges are separated by a distance 4a.
+2q
-q
The electric field at this point will be the vector sum of the electric field from the left charge
and the electric field from the right charge. Let’s look at these two fields separately.
The electric field from the positive charge
has both +x- and +y-components. Therefore,
after calculating the magnitude of the field
we will have to break this field into separate
x- and y-components using trigonometry.
7
+2q
-q
The magnitude of the electric field from the positive charge is:
E2q =
kq
r2
k ( 2q)
( 2a) 2 + ( a) 2
2kq
= 2
5a
E2q =
E2q
To determine the direction of this field, notice that
the field vector points in the same direction as a line
connecting the source charge and the point of
interest. This line forms a right triangle with θ given
by:
+2q
a
θ
2a
a
2a
1
tan θ =
2
−1
θ = tan (0.5)
tan θ =
θ = 26.6 0
Thus,
2kq
cos 26.6
5a 2
2kq
= 2 (0.89)
5a
kq
= 0.36 2
a
2kq
sin 26.6
5a 2
2kq
= 2 (0.45)
5a
kq
= 0.18 2
a
E 2 qx =
E 2 qy =
E 2 qx
E 2 qx
E 2 qx
E 2 qx
8
Repeating the analysis for the magnitude of the electric field from the negative charge:
E−q =
kq
r2
k ( q)
( 2a) 2 + ( a ) 2
kq
= 2
5a
E−q =
E−q
This field points toward the negative charge, and thus has
+x- and –y-components. This field is also oriented at
26.6°.
a
θ
-q
2a
Thus,
kq
cos 26.6
5a 2
kq
= 2 (0.89)
5a
kq
= 0.18 2
a
kq
sin 26.6
5a 2
kq
= − 2 (0.45)
5a
kq
= −0.09 2
a
E − qx =
E − qy = −
E − qx
E − qx
E − qx
E − qx
The resultant electric field at this point is the sum of the fields from the two source charges:
E x = E 2qx + E− qx
E y = E2 qy + E − qy
kq
kq
+ 0.18 2
2
a
a
kq
E x = 0.54 2
a
E y = 0.18
kq
kq
− 0.09 2
2
a
a
kq
E y = 0.09 2
a
E x = 0.36
Thus, the electric field is predominately in the +x-direction with a slight upward (+y)
component. The magnitude of this field could be determined by Pythagoras’ theorem, and the
exact angle of the resulting field determined by the tangent function.
9
Continuous Charge Distribution
Electric charge is discrete, meaning it comes in integer multiples of electron and proton
charge. Therefore, the electric field can always be calculated by summing the electric field
from each of the electrons and protons that make up an object. However, macroscopic objects
contain a lot of electrons and protons, so this summation has many, many terms:
E=
∑
every electron and proton
kq
r2
As described earlier, we will assume that the charge on macroscopic objects is continuous,
and distributed throughout the object. Mathematically, this means that the summation over a
very large number of discrete charges gets replaced with an integral over a hypothetically
continuous distribution of charge.
Since this course does not require you to know calculus, instead of performing the integrals
necessary to determine the electric field from charge distributions I will tabulate the results
for three different types of charge distributions.
Uniformly Charged Sphere
Imagine a sphere of radius R with a uniform distribution of charge, either throughout its
volume or over its surface. For all points outside of the sphere, the electric field is given by:
E=
E
r
kQ
r2
where Q is the total charge on the sphere, and the
direction of the field is directly away from the center of
the sphere, assuming Q is positive.
If the charge is uniformly distributed throughout the
volume of a solid sphere,
Q = ρV
4
Q = ρ ( πR 3 )
3
R
If the charge is uniformly distributed over the surface
of the sphere,
Q = σA
Q = σ (4πR 2 )
10
Uniformly Charged Cylinder
Imagine a very long cylinder of radius R with a uniform distribution of charge along its
length. For all points outside of the cylinder, the electric field is given by:
E=
E
2kλ
r
where λ is the charge per unit length on the cylinder,
and the direction of the field is directly away from the
center of the cylinder, assuming λ is positive.
r
R
Note that the charge could be distributed throughout the
volume or over the surface of the cylinder.
Uniformly Charged Plate
Imagine a very large, very thin plate with a uniform distribution of charge along its surface.
For all points above and below the plate, the electric field is given by:
E
E = 4πkσ
where σ is the charge per unit area on the plate, and the
direction of the field is directly away from the plate,
assuming σ is positive.
Note that the electric field does not decrease in
magnitude the farther you are from the plate. This is
obviously an approximation. Basically, if the distance
from the plate is small compared to the size of the
plate, the electric field will have this constant value.
11
Two Cylinders
A long coaxial cable has an inner wire of
negligible thickness with linear charge density
+λ and an outer conducting cylindrical shell of
radius R and linear charge density charge -λ.
a. Find the electric field halfway between the
wire and the shell.
b. Find the electric field outside of the shell.
From above we know that the electric field from a long cylinder is given by:
E=
2kλ
r
but this situation involves two nested cylinders. In the space between the wire and the shell
we can use this result to account for the wire, but how do we account for the cylindrical shell?
Amazingly enough, the mathematics that allowed us to calculate the result above shows that
the shell has no effect on the electric field in the space between the wire and the shell4. This
general result, known as Gauss’ Law, claims that only the electric charges “enclosed within”
the location of interest (the charges with r < R/2) can contribute to the electric field at r = R/2.
Therefore, at r = R/2, the electric field is given by:
E = E wire
2kλ
r
2kλ
E=
R
( )
2
4kλ
E=
R
E=
Outside of the shell we have to take into account that both of the cylinders that are “enclosed
within” the location of interest. Therefore,
E = E wire + E shell
2kλ 2kλ
−
r
r
E=0
E=
The resultant electric field is zero since the field from the wire and the field from the shell
have the same magnitude and point in opposite directions.
4
This assumes the wire and shell have uniform charge densities.
12
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The Electric Field
Activities
13
Determine the direction of the net electric field at each of the indicated points.
a.
+q
-q
+q
+q
b.
14
Determine the direction of the net electric field at each of the indicated points.
a.
-q
-q
-2q
-q
b.
15
Determine the direction of the net electric field at each of the indicated points.
a.
+q
+q
+q
b.
+q
+q
-q
-q
16
Determine the direction of the net electric field at each of the indicated points.
a.
+q
-q
-q
+q
b.
+q
-q
-q
17
For each of the charge distributions below, indicate the approximate location(s), if any, where the electric
field is zero.
a.
+q
+q
+q
-q
+2q
+2q
+2q
-4q
+8q
-2q
b.
c.
d.
e.
18
For each of the charge distributions below, indicate the approximate location(s), if any, where the electric
field is zero.
a.
-q
-q
-q
-q
b.
+q
-q
-q
+q
19
Determine the direction of the net electric field at each of the indicated points. The shaded region
represents a plastic rod with a uniform positive charge distribution.
a.
b.
20
Determine the direction of the net electric field at each of the indicated points. Each shaded region
represents a plastic rod with a uniform positive charge distribution and the unshaded region represents a
plastic rod with an equal-magnitude, uniform negative charge distribution.
a.
b.
21
Determine the direction of the net electric field at each of the indicated points. Each shaded region
represents a semi-circular plastic rod with a uniform positive charge distribution and the unshaded region
represents an identical plastic rod with an equal-magnitude, uniform negative charge distribution.
a.
b.
22
Determine the direction of the net electric field at each of the indicated points. In (a), the shaded region
represents a circular plastic hoop with a uniform positive charge distribution. In (b), the shaded region
represents a slice through the center of a thin plastic spherical shell with a uniform positive charge
distribution.
a.
b.
23
Determine the direction of the net electric field at each of the indicated points. In (a), the shaded region
represents a slice through the center of a plastic sphere with a uniform positive charge distribution. In (b),
the shaded region represents a slice through the center of a metal sphere with a uniform positive charge
distribution.
a.
b.
24
Determine the direction of the net electric field at each of the indicated points. The points are in the
perpendicular plane passing through the center of each circular hoop. Each plastic hoop has the same
uniform positive charge distribution.
a.
b.
25
a. For each of the scenarios described below, rank the objects on the basis of their net charge. All charge
densities are uniform
A
B
C
D
E
F
A 0.5 m long copper wire with charge density 1.0 µC/m.
A 0.5 m radius plastic sphere with charge density 1.0 µC/m3.
A 0.5 m radius copper sphere with charge density 1.0 µC/m2.
A 2.0 m long, 0.1 m radius plastic cylinder with charge density 2.0 µC/m3.
A 1.0 m long copper wire with charge density 0.5 µC/m.
A 1.0 m radius plastic sphere with charge density 0.5 µC/m3.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
b. A charge Q is distributed uniformly throughout the volume of each of the plastic objects described
below. Rank the objects on the basis of their charge density.
A
B
C
D
E
F
A solid cube with edge length L.
A solid cylinder of length L and radius 0.1L.
A solid sphere of radius 0.5L.
A solid half-sphere of radius L.
A spherical shell of inner radius 0.5L and outer radius L.
A cylindrical shell of length L, inner radius 0.1L and outer radius 0.2L
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
26
a. For each of the scenarios described below, rank the objects on the basis of their net charge. All charge
densities are uniform
A
B
C
D
E
F
A 1.0 m long copper wire with charge density 1.0 µC/m.
A 1.0 m long copper wire with charge density 1.0 µC/m2.
A 0.5 m radius copper spherical shell with charge density 1.0 µC/m2.
A 2.0 m long, 0.1 m radius plastic cylinder with charge density 2.0 µC/m3.
A 0.5 m radius copper sphere with charge density 1.0 µC/m2.
A 2.0 m long, 0.1 m radius copper cylindrical shell with charge density 2.0 µC/m2.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
b. A charge Q is distributed uniformly throughout the volume of each of the plastic objects described
below. Rank the objects on the basis of their charge density.
A
B
C
D
E
F
A solid cube with edge length L.
A solid cylinder of length 0.5L and radius 0.2L.
A solid sphere of radius 0.2L.
A solid half-sphere of radius 0.4L.
A spherical shell of inner radius 0.2L and outer radius 0.4L.
A cylindrical shell of length 0.5L, inner radius 0.1L and outer radius 0.2L
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
27
Two neutral metal spheres are insulated from the tabletop they rest on. A positively charged rod is initially
far from the spheres. Before each action below, the system is returned to this initial condition.
A
++++
++++
B
a. The spheres are near each other but do not touch, and the rod is brought near sphere A, but does not
touch. The rod is held at rest near sphere A.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
b. The spheres touch each other, and the rod is brought near sphere A, but does not touch. The rod is held at
rest near sphere A.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
c. The spheres are near each other but do not touch, and the rod touches sphere A. The rod is then removed.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
d. The spheres touch each other, and the rod touches sphere A. The rod is then removed.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
28
Two neutral metal spheres are insulated from the tabletop they rest on, but sphere B is grounded (connected
to a infinite reservoir of mobile charges). A positively charged rod is initially far from the spheres. Before
each action below, the system is returned to this initial condition.
A
++++
++++
B
a. The spheres are near each other but do not touch, and the rod is brought near sphere A, but does not
touch. The rod is held at rest near sphere A.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
b. The spheres touch each other, and the rod is brought near sphere A, but does not touch. The rod is held at
rest near sphere A.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
c. The spheres are near each other but do not touch, and the rod touches sphere A. The rod is then removed.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
d. The spheres touch each other, and the rod touches sphere A. The rod is then removed.
Is the net charge on sphere A positive, negative, or zero?
Is the net charge on sphere B positive, negative, or zero?
Explain the reason for your response:
29
A solid metal sphere is centered between two metal spherical shells. The sphere has a charge +2Q and the
shells are neutral.
A
B
C
D
E
The outer surface of the sphere.
The inner surface of the inner spherical shell.
The outer surface of the inner spherical shell.
The inner surface of the outer spherical shell.
The outer surface of the outer spherical shell.
a. Rank the surfaces on the basis of the charge on each surface.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
b. Rank the surfaces on the basis of their charge density.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
30
A solid metal sphere is centered between two metal spherical shells. Each of the three objects has charge
+2Q.
A
B
C
D
E
The outer surface of the sphere.
The inner surface of the inner spherical shell.
The outer surface of the inner spherical shell.
The inner surface of the outer spherical shell.
The outer surface of the outer spherical shell.
Rank the surfaces on the basis of the charge on each surface.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
31
A solid metal sphere is centered between two metal spherical shells. The sphere has charge –3Q, the inner
shell –2Q, and the outer shell +Q.
A
B
C
D
E
The outer surface of the sphere.
The inner surface of the inner spherical shell.
The outer surface of the inner spherical shell.
The inner surface of the outer spherical shell.
The outer surface of the outer spherical shell.
Rank the surfaces on the basis of the charge on each surface.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
32
Find the electric field directly between a sodium ion (Na+) and chlorine
ion (Cl-) separated by 2.82 x 10-10 m.
Na+
Cl-
Qualitative Analysis
On the graphic above, sketch the direction of the electric field at the requested point. Explain why the
electric field points in this direction.
Mathematical Analysis
33
Find the electric field at the center of the chain of sodium ions (Na+)
and chlorine ions (Cl-). The interionic separation is 2.82 x 10-10 m.
Cl-
Na+
Cl-
Qualitative Analysis
On the graphic above, sketch the direction of the electric field at the requested point. Explain why the
electric field points in this direction.
Mathematical Analysis
34
Na+
Find the electric field at the center of the array of sodium ions (Na+)
and chlorine ions (Cl-). The interionic separation is 2.82 x 10-10 m.
Cl-
Na+
Na+
Cl-
Qualitative Analysis
On the graphic above, sketch the direction of the electric field at the requested point. Explain why the
electric field points in this direction.
Mathematical Analysis
35
In the H2O molecule illustrated at right, the oxygen atom has an effective
charge of -2e and each hydrogen atom +e. The effective charge separation
between the O and each H is 4.9 x 10-12 m. The molecule forms an angle of
1050. Find the electric field directly between the hydrogen atoms.
H
H
O
Qualitative Analysis
On the graphic above, sketch the direction of the electric field at the requested point. Explain why the
electric field points in this direction.
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Find the electric field at each of the indicated points. The
charges are separated by a distance 4a.
a.
b.
+q
d.
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-q
c.
Find the electric field at each of the indicated points. The
charges are separated by a distance 4a.
a.
b.
+2q
d.
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+q
c.
The two charges at right are separated by a distance 2a. Determine the
location(s), if any, where the electric field is zero.
+q
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+2q
The two charges at right are separated by a distance 2a. Determine the
location(s), if any, where the electric field is zero.
-2q
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+q
The two charges at right are separated by a distance 2a. Determine the
location(s), if any, where the electric field is zero.
-2q
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+3q
The nested spheres at right consist of an inner metal sphere of radius a
and charge –Q, and an outer metal spherical shell of inner radius b, outer
radius c, and charge –3Q. Find the charge densities on the three surfaces.
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The coaxial cylinders at right consist of an inner metal cylinder of radius
a and charge +2Q per unit length, and an outer metal cylindrical shell of
inner radius b, outer radius c, and charge –3Q per unit length. Find the
charge densities on the three surfaces.
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The coaxial cylinders at right consist of an inner metal cylinder of radius
6 mm and linear charge density 3.0 nC/m, and an outer metal cylindrical
shell of inner radius 10 mm, outer radius 12 mm, and linear charge
density +5.0 nC/m. The cylinders are 2 m long. Find the charge densities
on the three surfaces.
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A solid plastic sphere of radius 2.0 cm has a uniform charge density of
+3.0 nC/m3.
a. Find the electric field 4.0 cm from the center of the sphere.
Mathematical Analysis
b. If the sphere was partially hollow, with a radius of 1.0 cm for the hollow portion, and everything else
was the same, find the electric field 4.0 cm from the center of the sphere.
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A solid metal sphere of radius 2.0 cm has a uniform charge density of
+3.0 nC/m2.
a. Find the electric field 4.0 cm from the center of the sphere.
Mathematical Analysis
b. If the sphere was partially hollow, with a radius of 1.0 cm for the hollow portion, and everything else
was the same, find the electric field 4.0 cm from the center of the sphere.
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A long, hollow plastic cylinder has inner radius a, outer radius b, and
uniform charge density ρ.
a. Find the electric field in the region r < a.
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b. Find the electric field in the region r > b.
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A long, hollow metal cylinder has inner radius 1.0 cm, outer radius 2.0
cm, and 4.0 nC distributed uniformly per unit length.
a. Find the electric field 0.5 cm from the center of the wire.
Mathematical Analysis
b. Find the electric field 4.0 cm from the center of the wire.
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A long coaxial cable has an inner wire of negligible thickness with linear
charge density 1.2 nC/m and an outer conducting cylindrical shell of
radius 0.6 cm. The net charge on the cable is zero.
a. Find the electric field halfway between the wire and the shell.
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b. Find the electric field outside of the cable.
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