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8/2/2019 PTP 102 [1]
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1-ELECTRIC CHARGES
After running a plastic comb through your hair, you will find that the
comb attracts bits of paper. The attractive force is often strong enough to
suspend the paper from the comb, defying the gravitational pull of theentire Earth. The same effect occurs with other rubbed materials, such as
glass and hard rubber.
The important aspect of electricity that arises from experimentalobservations is that electric charge is always conserved in an isolated
system. That is, when one object is rubbed against another, charge is not
created in the process. The electrified state is due to a transferof charge
from one object to the other. One object gains some amount of negative
charge while the other gains an equal amount of positive charge. For
example, when a glass rod is rubbed with silk the silk obtains a negative
charge that is equal in magnitude to the positive charge on the glass rod.
We now know from our understanding of atomic structure that electronsare transferred from the glass to the silk in the rubbing process. Similarly,
when rubber is rubbed with fur, electrons are transferred from the fur to
the rubber, giving the rubber a net negative charge and the fur a net
positive charge. This process is consistent with the fact that neutral,
uncharged matter contains as many positive charges (protons within
atomic nuclei) as negative charges (electrons).
Thus, there are two kinds of charges in nature, with the property that
unlike charges attract one another and like charges repel one another. The
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force between charges varies as the inverse square of their separation and
charge is conserved and quantized.
2-Charging Objects by Induction
It is convenient to classify materials in terms of the ability of electrons
to move through the material:
Electrical conductors are materials in which some of the electrons are
free electrons1 that are not bound to atoms and can move relatively freely
through the material; electrical insulators are materials in which all
electrons are bound to atoms and cannot move freely through the
material.
1 2 3
4 5In the process of inducing a charge on the sphere, the charged rubber rod doesnt lose any of its negative charge
because it never comes in contact with the sphere. Furthermore, the sphere is left with a charge opposite that of the
rubber rod. Charging an object by induction requires no contact with the object inducing the charge.
Materials such as glass, rubber, and wood fall into the category ofelectrical insulators. When such materials are charged by rubbing, only
the area rubbed becomes charged, and the charged particles are unable to
move to other regions of the material. In contrast, materials such as
copper, aluminum, and silver are good electrical conductors. When such
materials are charged in some small region, the charge readily distributes
itself over the entire surface of the material. If you hold a copper rod in
your hand and rub it with wool or fur, it will not attract a small piece of
paper. This might suggest that a metal cannot be charged. However, ifyou attach a wooden handle to the rod and then hold it by that handle as
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you rub the rod, the rod will remain charged and attract the piece of
paper. The explanation for this is as follows: without the insulating wood,
the electric charges produced by rubbing readily move from the copper
through your body, which is also a conductor, and into the Earth. The
insulating wooden handle prevents the flow of charge into your hand.
Semiconductors are a third class of materials, and their electrical
properties are somewhere between those of insulators and those of
conductors. Silicon and germanium are well-known examples of
semiconductors commonly used in the fabrication of a variety of
electronic chips used in computers, cellular telephones, and stereo
systems. The electrical properties of semiconductors can be changed over
many orders of magnitude by the addition of controlled amounts of
certain atoms to the materials.
3-COULOMBS LAW
In 1785 Charles Coulomb (17361806) experimentally established
the fundamental law of electric force between two stationary charged
particles.
An electric force has the following properties:
1. It is directed along a line joining the two particles and is inversely
proportional to the square of the separation distance r, between them.
2. It is proportional to the product of the magnitudes of the charges, |q1|
and |q2|, of the two particles.
3. It is attractive if the charges are of opposite sign and repulsive if the
charges have the same sign.
We will use the term point charge to mean a particle of zero size that
carries an electric charge. The electrical behavior of electrons and protons
is very well described by modeling them as point charges. From these
observations, Coulomb proposed the following mathematical form for the
electric force between two charges:
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where ke is a constant called the Coulomb constant. In his experiments,
Coulomb was able to show that the value of the exponent of rwas 2 to
within an uncertainty of a few percent. Modern experiments have shown
that the exponent is 2 to within an uncertainty of a few parts in 1016
. The
value of the Coulomb constant depends on the choice of units. The SI unit
of charge is the coulomb (C). The Coulomb constant kein SI units has the
value
When using Coulombs force law, remember that force is a vector
quantity and must be treated accordingly. Active Figure below shows the
electric force of repulsion between two positively charged particles. Like
other forces, electric forces obey Newtons third law; hence, the forces
and are equal in magnitude but opposite in direction. (The
notation denotes the force exerted by particle 1 on particle 2;
likewise, is the force exerted by particle 2 on particle 1.) From
Newtons third law, F12 and F21 are always equal regardless of whether q1
and q2 have the same magnitude.
Example [1].The Hydrogen Atom
The electron and proton of a hydrogen atom are separated (on the
average) by a distance of approximately 5.3x10-11
m. Find the magnitudes
of the electric force and the gravitational force between the two particles.
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Solution: From Coulombs law, we find that the magnitude of the electric
force is
The gravitational force is
The ratio Fe/Fg~ 2x1039
. Thus, the gravitational force between charged
atomic particles is negligible when compared with the electric force. Note
the similarity of form of Newtons law of universal gravitation and
Coulombs law of electric forces. Other than magnitude, what is a
fundamental difference between the two forces?
Example [2]: Where Is the Resultant Force Zero?
Three point charges lie along the x axis as
shown in Figure. The positive charge q1 = 15.0
C is at x = 2.00 m, the positive charge q2 =
6.00 C is at the origin, and the resultant force
acting on q3 is zero. What is thex coordinate of
q3?
Solution: Because q3 is negative and q1 and q2 are positive, the forces F13
and F23are both attractive, as indicated in Figure. From Coulombs law,
F13 and F23 have magnitudes
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Only the positive root makes sense
Example [3]:Find the Resultant Force
Consider three point charges
at the corners of a triangle, as
shown in Figure, where q1 =
6.00x10-9
C, q2 = -2.00x10-9
C, and q3 = 5.00x10-9
C. (a)
Find the components of the
force exerted by q2 on q3. (b)
Find the components of the
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force exerted by q1 on q3. (c) Find the resultant force on q3, in terms of
components and also in terms of magnitude and direction.
Solution:
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Example [3]:Find the Charge on the Spheres
Two identical small charged spheres, each having a mass of 3.0x10-2
kg,
hang in equilibrium as shown in Figure. The length of each string is 0.15
m, and the angle is 5.0. Find the magnitude of the charge on each
sphere.
Solution: The two spheres exert repulsive forces on each other. If they
are held close to each other and released, they will move outward from
the center and settle into the configuration in Fig. after the damped
oscillations due to air resistance have vanished. The key phrase in
equilibrium helps us categorize this as an equilibrium problem , that one
of the forces on a sphere is an electric force. We analyze this problem by
drawing the free-body diagram for the left-hand sphere in Figure. The
sphere is in equilibrium under the application of the forces T from the
string, the electric force Fe from the other sphere, and the gravitational
force mg. Because the sphere is in equilibrium, the forces in the
horizontal and vertical directions must separately add up to zero:
,
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Considering the geometry of the right triangle in Figure, we see that sin = a/L. Therefore,
To finalize the problem, note that we found only the magnitude of the
charge |q | on the spheres. There is no way we could find the sign of the
charge from the information given. In fact, the sign of the charge is notimportant. The situation will be exactly the same whether both spheres
are positively charged or negatively charged.
4-The Electric Field
The gravitational force and the electrostatic force are both capable of
acting through space, producing an effect even when there isnt any
physical contact between the objects involved. Field forces can be
discussed in a variety of ways, but an
approach developed by Michael Faraday
(17911867) is the most practical. In this
approach, an electric field is said to exist
in the region of space around a charged
object. The electric field exerts an electric
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force on any other charged object within the field. This differs from the
Coulombs law concept of a force exerted at a distance, in that the force
is now exerted by something -the field- that is in the same location as the
charged object. As an example, consider Figure, which shows a small
positive test charge q0 placed near a second object carrying a much
greater positive charge Q. We define the electric field due to the source
charge at the location of the test charge to be the electric force on the test
chargeper unit charge, or to be more specific the electric field vector E at
a point in space is defined as the electric force Feacting on a positive test
charge q0 placed at that point divided by the test charge:
Note that E is the field produced by some charge or charge distribution
separate from the test charge -it is not the field produced by the test
charge itself. Also, note that the existence of an electric field is a property
of its source - the presence of the test charge is not necessary for the field
to exist. The test charge serves as a detectorof the electric field.
To determine the direction of an electric field, consider a point charge q
as a source charge. This charge creates an electric field at all points in
space surrounding it. A test charge q 0 is placed at point P, a distance r
from the source charge, as in Figure. We imagine using the test charge to
determine the direction of the electric force and therefore that of theelectric field. However, the electric field does not depend on the existence
of the test charge, it is established solely by the source charge. According
to Coulombs law, the force exerted by q on the test charge is
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(a) The electric field atA due to the negatively charged sphere is downward, toward the negative charge.
(b) The electric field at P due to the positively charged conducting sphere is upward, away from the positive charge.
(c) A test charge q0 placed at P will cause a rearrangement of charge on the sphere, unless q0 is very small compared with the
charge on the sphere.
These points out an important property of electric fields that makes them
useful quantities for describing electrical phenomena. As the equationindicates, an electric field at a given point depends only on the charge q
on the object setting up the field and the distance rfrom that object to a
specific point in space. As a result, we can say that an electric field exists
at point P in Active Figure whether or not there is a test charge at P.
5-Electric Field of a Continuous Charge Distribution
Very often the distances between charges in a
group of charges are much smaller than the
distance from the group to some point of interest
(for example, a point where the electric field is
to be calculated). In such situations, the system
of charges is smeared out, or continuous. That is,
the system of closely spaced charges is
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equivalent to a total charge that is continuously distributed along some
line, over some surface, or throughout some volume.
To evaluate the electric field created by a continuous charge distribution,
we use the following procedure: first, we divide the charge distribution
into small elements, each of which contains a small charge q, as shown
in the figure. Next, we use equation
to calculate the electric field due to one of these elements at a point P.
Finally, we evaluate the total electric field at P due to the chargedistribution by summing the contributions of all the charge elements (that
is, by applying the superposition principle).
The electric field at P due to one charge element carrying charge The
electric field at P due to one charge element carrying charge q is
where ris the distance from the charge element to point P and r is a unitvector directed from the element toward P. The total electric field at P
due to all elements in the charge distribution is approximately
where the index i refers to the i th element in the distribution. Because the
charge distribution is modeled as continuous, the total field at P in thelimit qi= 0 is
where the integration is over the entire charge distribution. This is a
vector operation and must be treated appropriately.
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If a charge Q is uniformly distributed throughout a volume V, the
volume charge density is defined by
where has units of coulombs per cubic meter (C/m3).
If a charge Q is uniformly distributed on a surface of area A, the surface
charge density (lowercase Greek sigma) is defined by
where has units of coulombs per square meter (C/m2).
If a charge Q is uniformly distributed along a line of length !, the linear
charge density is defined by
where has units of coulombs per meter (C/m).
If the charge is nonuniformly distributed over a volume, surface, or line,the amounts of charge dq in a small volume, surface, or length element
are
Example [4]: Electric Field Due to Two
Point Charges
Charge q1 = 7.00 C is at the origin, and charge
q2 = -5.00 C is on thex-axis, 0.300 m from the
origin. (a) Find the magnitude and direction of
the electric field at point P, which has
coordinates (0, 0.400) m. (b) Find the force on a
charge of 2.00x10-8
C placed at P.
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Solution:
6-Electric Field Lines
We have defined the electric field mathematically. We now explore a
means of representing the electric field pictorially. A convenient way of
visualizing electric field patterns is to draw curved lines that are parallel
to the electric field vector at any point in space. These lines, called
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electric field lines and first introduced by Faraday, are related to the
electric field in a region of space in the following manner:
The electric field vector E is tangent to the electric field line at each
point. The line has a direction, indicated by an arrowhead that is the same
as that of the electric field vector.
The number of lines per unit area through a surface perpendicular to the
lines is proportional to the magnitude of the electric field in that region.
Thus, the field lines are close together where the electric field is strong
and far apart where the field is weak.
These properties are illustrated in
the figure. The density of lines
through surface A is greater than the
density of lines through surface B.
Therefore, the magnitude of the
electric field is larger on surface A
than on surface B. Furthermore, the
fact that the lines at different locations
point in different directions indicates that the field is nonuniform.
Representative electric field lines for the field due to a single positive
point charge are shown in the figure. This two-dimensional drawing
shows only the field lines that lie in the plane containing the point charge.
The lines are actually directed radially outward from the charge in all
directions; thus, instead of the flat wheel of lines shown, you should
picture an entire spherical distribution of lines. Because a positive test
charge placed in this field would be repelled by the positive source
charge, the lines are directed radially away from the source charge. The
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electric field lines representing the field due to a single negative point
charge are directed toward the charge. In either case, the lines are along
the radial direction and extend all the way to infinity. Note that the lines
become closer together as they approach the charge; this indicates that the
strength of the field increases as we move toward the source charge.
The rules for drawing electric field lines are as follows:
The lines must begin on a positive charge and terminate on a negative
charge. In the case of an excess of one type of charge, some lines will
begin or end infinitely far away.
The number of lines drawn leaving a positive charge or approaching a
negative charge is proportional to the magnitude of the charge.
No two field lines can cross.
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7-Motion of Charged Particles in a Uniform
Electric Field
When a particle of charge q and mass m is placed in an electric field
E, the electric force exerted on the charge is q E. If this is the only force
exerted on the particle, it must be the net force and causes the particle to
accelerate according to Newtons second law. Thus,
The acceleration of the particle is therefore
If E is uniform (that is, constant in magnitude and direction), then the
acceleration is constant. If the particle has a positive charge, its
acceleration is in the direction of the electric field. If the particle has a
negative charge, its acceleration is in the direction opposite the electric
field.
Example [5]:
A positive point charge q of mass m is released from
rest in a uniform electric field E directed along the x
axis, as shown in the figure. Describe its motion.
Solution:
The acceleration is constant and is given by qE/m.
The motion is simple linear motion along the x axis.
Therefore, we can apply the equations of kinematics
in one dimension
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Choosing the initial position of the charge as xi= 0 and assigning vi= 0
because the particle starts from rest, the position of the particle as a
function of time is
The speed of the particle is given by
The third kinematic equation gives us
from which we can find the kinetic energy of the charge after it has
moved a distance x =xf - xi:
We can also obtain this result from the workkinetic energy theorem
because the work done by the electric force is
Fex = qE
x and W=
K.
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PROBLEMS
1- The figure shows two positively chargedparticles fixed in place on an x axis. The
charges are q1 = 1.60 x10-19
C and q2 =3.20 X 10
-19C, and the particle separation
is R = 0.0200 m. What are the magnitude
and direction of the electrostatic force
on particle 1 from particle e2?
2- An object having a net charge of 24.0 C is placed in a uniformelectric field of 610 N/C directed vertically. What is the mass of
this object if it floats in the field?
3- In the figure determine thepoint (other than infinity) at
which the electric field is
zero.
4- Two point charges are located on the x axis. The first is a charge+Q atx = -a. The second is an unknown charge located at x = +3a.The net electric field these charges produce at the origin has a
magnitude of 2keQ/a2. What are the two possible values of the
unknown charge?
5- Three charges are at the corners of anequilateral triangle as shown in the figure. (a)
Calculate the electric field at the position of
the 2.00 C charge due to the 7.00 C and -4.00 C charges. (b) Use your answer to part
(a) to determine the force on the 2.00 C
charge.
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SUMMARY