8/2/2019 PTP 102 [1]

1/20

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

8/2/2019 PTP 102 [1]

2/20

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

8/2/2019 PTP 102 [1]

3/20

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:

8/2/2019 PTP 102 [1]

4/20

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.

8/2/2019 PTP 102 [1]

5/20

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

8/2/2019 PTP 102 [1]

6/20

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

8/2/2019 PTP 102 [1]

7/20

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:

8/2/2019 PTP 102 [1]

8/20

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:

,

8/2/2019 PTP 102 [1]

9/20

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

8/2/2019 PTP 102 [1]

10/20

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

8/2/2019 PTP 102 [1]

11/20

(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

8/2/2019 PTP 102 [1]

12/20

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.

8/2/2019 PTP 102 [1]

13/20

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.

8/2/2019 PTP 102 [1]

14/20

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

8/2/2019 PTP 102 [1]

15/20

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

8/2/2019 PTP 102 [1]

16/20

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.

8/2/2019 PTP 102 [1]

17/20

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

8/2/2019 PTP 102 [1]

18/20

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.

8/2/2019 PTP 102 [1]

19/20

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.

8/2/2019 PTP 102 [1]

20/20

SUMMARY