Chapter 23 Electric field Chapter 23 Electric field 23.1 Properties of Electric Charges 23.2...

Chapter 23

Electric field

23.1 Properties of Electric Charges23.2 Charging Objects By Induction23.3 Coulomb’s Law23.4 The Electric Field23.6 Electric Field Lines23.7 Motion of Charged Particles in a Uniform Electric Field

1 Norah Ali AL.Moneef

23.1 properties of electric 23.1 properties of electric chargecharge


Objects can be charged by rubbing or friction

(a) Rub a plastic ruler (b) bring it close to some tiny pieces of paper

Objects charged by this method will attract each other.


we conclude that charges of the same sign repel one another andcharges with opposite signs attract one another


When a glass rod is rubbed with silk, electrons aretransferred from the glass to the silk. Because of conservation of charge, each electron adds negative charge to the silk, and an equal positive charge is left behind on the rod. Also, because the charges are transferred in discrete bundles, the charges on the two objects are

In 1909, Robert Millikan discovered that electric charge always occurs as some integral multiple of a fundamental amount of charge e

the electric charge q is said to be quantized

and we can write q = N e, where N is some integer

the electron has a charge – e and the proton has a charge of equal magnitude but opposite sign +e . Some particles, such as the neutron, have no charge.

Note that the electrical charge is measured in coulombs).


A coulomb is the charge resulting from the transfer of 6.24 x 1018 of the charge carried by an electron

Norah Ali AL.Moneef66 Norah Ali AL.Moneef

• During any process, the net electric charge of an isolated system remains constant i.e. is conserved.How many electrons constitute 1 C?

N =1x10-

6/1.6x10-19

= 6x1012 electrons


23.2 Charging Objects By Induction

Electrical conductors are materials in which some of the electrons are freeelectrons 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

When electrical insulators such as glass, rubber are charged by rubbing, only the area rubbed becomes charged, and the charged particles are unable to move to other regions of the material

.When materials such as copper, aluminum, and silver are good electrical conductors are charged in some small region, the charge readily distributes itself over the entire surface of the material

Semiconductors are a third class of materials, and their electrical propertiesare somewhere between those of insulators• Conductor transfers charge on contact

• Insulator does not transfer charge on contact• Semiconductor might transfer charge on contact


(a) The charged object on the left induces a charge distribution on the surface of an insulator due to realignment of charges in the molecules.


(e) When the rod is removed, the remaining electrons redistribute

uniformly and there is a net uniform distribution of positive charge on the

sphere

Charging a metallic object by induction (that is, the two objects never touch each other)

(a) A neutral metallic sphere ,

with equal numbers of positive and

negative charges

(b) The electrons on the neutral sphere are

redistributed when a charged rubber rod is

placed near the sphere.

. (c) When the sphere is grounded, some of its electrons leave through the ground

wire.

(d) When the ground connection is removed, the sphere has excess positive charge that is nonuniformly

distributed10 Norah Ali AL.Moneef

Conceptual QuestionAssume that you have two uncharged, insulated metallic spheres A and B that are in contact with each other. If you bring a positively charged insulated rod near sphere A and hold it there while you move sphere B away, what charge will sphere B have? (A) Negative charge (D) No net charge (B) Positive charge (E) None of these (C) Either pos. or neg. charge


23.3 Coulomb’s LawCharles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects using the torsion balance

rF e 2

1

we can express Coulomb’s law as an equation giving the magnitude of the electric force (sometimes called the Coulomb force) between two point charges

Coulomb’s law


charges of the same sign repel one anothercharges with opposite signs attract one another


where rˆ is a unit vector directed from q1 toward q2,. the electric force obeys Newton’s third law, the electric force exerted by q2 on q1 is equal in magnitude to the force exerted by q1 on q2 and in the opposite direction; that is, FF21 = - F12.

magnitude

direction


The force between two charges gets stronger as the charges move closer together.

The force also gets stronger if the amount of charge becomes larger. The force between two charges is directed along the line connecting their centers.

Electric forces always occur in pairs according to Newton’s third law, like all forces.

(do not put signs on the charges when you use Coulomb’s law)


Example• Which charge exerts greater force?

Two positive point charges, Q1=50C and Q2=1C, are separated by a distance L. Which is larger in magnitude, the force that Q1 exerts on Q2 or the force that Q2 exerts on Q1?

1 212 2

QQF k

LWhat is the force that Q1 exerts on Q2?

Therefore the magnitudes of the two forces are identical!!

Well then what is different? The direction.

What is the force that Q2 exerts on Q1? 2 121 2

Q QF k

L

What is this law?

Which direction? Opposite to each other!

Newton’s third law, the law of action and reaction!!


Two point charges separated by a distance rexert a force on each other that is given by Coulomb’s law. The force F21 exerted by q2 on q1 is equal inmagnitude and opposite in direction to the force F12 exerted by q1 on q2.(a) When the charges are of the same sign, the force is repulsive. (b) When the charges are ofopposite signs, the force is attractive

When more than two charges are present, the resultant force on any one of them equals the vector sum of the forces exerted by the various individual charges. For example, if four charges are present, then the resultant force exerted by particles 2, 3, and 4 on particle 1 is


• Double one of the charges– force doubles

• Change sign of one of the charges– force changes direction

• Change sign of both charges– force stays the same

• Double the distance between charges– force four times weaker

• Double both chargesforce four times stronger ( the force quadruples )


The electric force is very much like gravity. Both forces act at a distance. Charge is like mass, although mass is always positive and the force of gravity is always attractive. In fact, the general law of gravitation is very much like the Coulomb's Law. The force between two bodies of mass m1 and mass m2 a distance r apart is

Gravitation force is always attractive.


Two point charges repel each other with a force of 4×10 -5 N at a distance of 1 m. The two charges are:(a) both positive (b) both negative (c) alike (d) unlike

Example

ExampleTwo charges of- Q are 1 cm apart. If one of the charges is replaced by a

charge of +Q, the magnitude of the force between them is;(a) zero (b) smaller (c) the same (d) larger


Example :

Example :


find the magnitude of the force exerted between the proton and the electron in the hydrogen atom. The force between these tiny charges, each one of size 1.6 x 10-19 C (but opposite in sign, which we ignore here) separated by a distance of 5.29 x 10-11 m is

Example :


Example :


Two protons in a molecule are separated by 3.80 x 10-10 m. Find the electric force exerted by one proton on the other. (b) How does the magnitude of this force compare to the magnitude of the gravitational force between the two protons (mp=1.67 x10-27 Kg )

Thus, the gravitational force between charged atomic particles is negligible when compared with the electric force.

Example :


Example - Forces between Electrons

• What is relative strength of the electric force compared with the force of gravity for two electrons?(me=9.11x10-31 Kg)

422

2

102.4 n)calculatio the(do Gm

keFF

gravity

electric

2

2

2

2)(

r

mGF

r

ekF

egravity

electric


Example :


What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m

3.The magnitude of the Coulomb force is F = kq1q2/r2

F=(9 x109N·m2/C2)(26)(1.6x10–19 C)(1.6 x10–19C)/(1.5 x 10–12 m)2 = 2.7 x 10–3 N.

Example :


Zero Resultant Force, Example

Where is the resultant force equal to zero?

The magnitudes of the individual forces will be equalDirections will be opposite

This is another location at which the magnitudes of the forces on q3 are equal, but both forces are in the same direction at this location.29 Norah Ali AL.Moneef

Electric Force of 2 charges

Ex1) Find the total electric force on q1.

You must use Newton’s Laws for these problems

Step 2) Find the magnitudes of the forces individually

Step 1) Draw all forces

N4.8N7.2 F

)( N4.8

N4.8m15.0

C100.7C100.3CmN109

13

2

66229

2

3113

i

r

qqk

F

F

)( N7.2

N7.2m20.0

C100.4C100.3CmN109

12

2

66229

221

12

i

r

qqk

F

F


Example Calculate the net electrostatic force on charge Q3 shown in the figure due to the charges Q1 and Q2.

. (a) F32 is repulsive (the force on Q3 is in the direction away from Q2 because Q3 and Q2 are both positive) whereas F31 is attractive (Q3 and Q1 have opposite signs), so F31 points toward Q1.

(b) Adding F32 to F31 to obtain the net force.


The forces, components, and signs are as shown in the figure. Result: The magnitude of the force is 290 N, at an angle of 65° to the x axis.

Conceptual Example Make the force on Q3 zero.

In the figure, where could you place a fourth charge, Q4 = -50 μC, so that the net force on Q3 would be zero?

Solution: The force on Q3 due to Q4 must exactly cancel the net force on Q3 from Q1 and Q2. Therefore, the force must equal 290 N and be directed opposite to the net force

cm 8.31318.0290

10501065109

66943

mrF

QQKr


23 .4 The Electric FieldThe electric field at any point in space is

defined as the force exerted on a tiny positive test charge divide by the test chargeElectric force per unit charge

What kind of quantity is the electric field?Vector quantity.

What is the unit of the electric field?N/C

The magnitude of the electric field at a distance r from a single point charge Q is

FE

q

FE

q

2kQq r

q

2

kQ

r 2

0

1

4

Q

r


Find the electric field

Example :


Example: Electric field of a single point charge.

Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C.

. Electric field at point P (a) due to a negative charge Q, and (b) due to a positive charge Q, each 30 cm from P.Solution: Substitution gives E = 3.0 x 105 N/C. The field points away from the positive charge and towards the negative one.

CNE /10310 230

10 63 1099

r

QK 5

22


A map of the electrical field can be made by bringing a positive test charge into an electrical field.

•This is represented by drawing lines of force or electrical field lines.

•You can draw vector arrows to indicate the direction of the electrical field.

•When brought near a negative charge the test charge is attracted to the unlike charge and when brought near a positive charge the test charge is repelled.


Point negative charge

E= kq1/r2

q1

q1

r


(a) If q is positive, then the force on the test charge is directed away from q.

A test chargeq0 at point P is a distance r from apoint charge q.

(b) For the positive source charge, the electric field at P points radially outward from q.

(c) If q is negative, then the force on the test charge is directed toward q.

(d) For the negative source charge, the electric field at P points radially inward toward q.


The electric field vector 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 direction of the line is that of the force on a positive test charge placed in the field.

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.

Electric field lines penetrating two surfaces. The magnitude of the field is greater on surface A than on surface B.

23.6 Electric Field Lines


23.6 Electric Field Lines (Point Charge)

Electric Field (vector)

Field Lines(Lines of

force)Electric field lines (lines of force) are continuous lines whose direction is everywhere that of the electric field


The charge on the right is twice the magnitude of the charge on the left (and opposite in sign), so there are twice as many field lines, and they point towards the charge rather than away from it.


Equal charges:

same number density

Unequal charges

–Ve > + Ve

Because the charges are of equal magnitude, the number of lines that begin atthe positive charge must equal the number that terminate at the negative charge. At points very near the charges, the lines are nearly radial. The high density of linesbetween the charges indicates a region of strong electric field

Like charges (++) Opposite charges (+ -)


N2/N1=Q2/Q1.

if object 1 has charge Q1 and object 2 has charge Q2, then the ratio of number of lines is

The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge

QN linessince


Electric Field Lines:.

The rules for drawing electric field lines are :

• 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.


For discrete point charges, we can use the superpositionprinciple and sum the fields due to each point charge:

q1

q2

q3

p

E(r) Ei

i

at any point P, the total electric field due to a group of source charges equals thevector sum of the electric fields of all the charges.


What are the signs of the

charges whose electric

fields are shown at below?

1)

2)

3)

4)

5) no way to tell

Electric field lines originate on originate on

positive chargespositive charges and terminate terminate

on negative chargeson negative charges.


Which of the charges has

the greater magnitude?1)

2)

3) Both the same

The field lines are denser around denser around

the red chargethe red charge, so the red one red one

has the greater magnitudehas the greater magnitude.

Follow-up:Follow-up: What is the red/brown ratio of What is the red/brown ratio of magnitudes for the two charges?magnitudes for the two charges?


Which of the following statements about electric field lines associated with electric charges is false? (a) Electric field lines can be either straight or curved. (b) Electric field lines can form closed loops. (c) Electric field lines begin on positive charges and end on negative charges. (d) Electric field lines can never intersect with one another.

Answer: (b). Electric field lines begin and end on charges and cannot close on themselves to form loops.

Example :


Find the electric field due to a point charge of 0.5 mC at a distance of 4 cm from it in vacuum

q =0.5×10–3 C, r = 4×10–2 m,

Recall E =kq/r2

and k=9 x 109 N.m2/C2

E = 9x109 N.m2/C2 0.5 X10-3 C/(4x 10-2 m)2 = 2.82×1010 N/C

Example :


3. Rank the magnitudes E of the electric field at points A, B, and C shown in the figure.

A) EC>EB>EA

B) EB>EC>EA

C) EA>EC>EB

D) EB>EA>EC

E) EA>EB>EC

.C

.A

.B

Example :


Rank the electric field strength in order from smallest to largest.

A: E1 < E2 < E3 = E4

B: E3 = E4 < E2 < E1

C: E2 = E3 < E4 < E1

D: E1 < E4 < E2 = E3

Example :


A test charge of +3 µC is at a point P where an external electric field is directed to the right and has a magnitude of 4×106 N/C. If the test charge is replaced with another test charge of –3 µC, what happens to the external electric field at P ?

A. It is unaffected. B. It reverses direction. C. It changes in a way that cannot be

determined.

Example :


1. Left.2. Down.3. Right.4. Up.5. The electric field is zero.

At the position of the dot, the electric field points

Example :


Example of finding electric field from two charges

Find electric field due to both

charges at point p

We have q1=+10 nC at the origin, q

2 = +15 nC at

x=4 m.

What is E at y=3 m and x=0? point P

x

y

q1=10 nc q2 =15 nc4m

3m

P


Example continued

Field due to q1

E = 9x109 N.m2/C2 10 X10-9 C/(3m)2 = 10 N/C (in the y direction).

E =kq/r2 k=9 x 109 N.m2/C2

x

y

q1=10 nc q2 =15 nc4

3

Ey= 10 N/C Ex= 0

Field due to q2

5

E = 9 x 109 N.m2/C2 15 X10-9 C/(5m)2 =5.4 N/Cat some angle Resolve into x and y components

E

Ey=E sin C

Ex=E cos C

Now add all components

Ey= 10 + 3.4 = 13.4 N/CEx= - 4.3 N/C

E Ex2 Ey

2 magnitude

j


Example continued

xq1=10 nc q2 =15 nc4

3

Ey= 10 + 3.4 = 13.4 N/C

Ex= - 4.3 N/C

E

Magnitude of electric field

E E x2 E y

2

tan -1 Ey/Ex= tan -1 (13.4/-4.3)= 72.8⁰

Using unit vector notation we canalso write the electric field vector as:

CNE /1.143.44.13 22

jiE 4.13 3.4


Three point charges are aligned along the x axis as shown in the Figure. Find the electric field at

(a) the position (2.00, 0) and

Example :


From the figure find the electric field at (0,0)

9 91 3

1 2 21

8.99 10 3.00 10ˆ ˆ ˆ2.70 10 N C

0.100

ek q

r

E j j j

9 92 2

2 2 22

2 32 1

8.99 10 6.00 10ˆ ˆ ˆ5.99 10 N C

0.300

ˆ ˆ5.99 10 N C 2.70 10 N C

ek q

r

E i i i

E E E i j

9 ˆ ˆ5.00 10 C 599 2700 N Cq F E i j

6 6ˆ ˆ ˆ ˆ3.00 10 13.5 10 N 3.00 13.5 N F i j i j

(a)

Example :


If we know the electric field, we can calculate the force on any charge:

The direction of the force depends on the sign of the charge – in the direction of the field for a positive charge, opposite to it for a negative one.

23.7 Motion of a Charged Particle in a Uniform Electric Field


Electric Field linesUniform Field

Equispaced parallel straight lines

From infinity To infinity


23.7 Motion of a Charged Particle in a Uniform Electric Field

EqF

m

Eqa

amEqF

If the electric field E is uniform (magnitude and direction), the electric force F on the particle is constant.

If the particle has a positive charge, its acceleration a and electric force F are in the direction of the electric field E.

If the particle has a negative charge, its acceleration a and electric force F are in the direction opposite the electric field E.

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 Newton’s second law


An Accelerating Positive ChargeA positive point charge q of mass m is released from rest in a uniform electric field E directed along the x axis,. Describe its motion.The acceleration is constant a = qE/m.

The motion is simple linear motion along the x axis.We apply the equations of kinematics in one dimension

The speed of the particle is

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

A positive point charge q in a uniform electric field E undergoes constant acceleration in thedirection of the field.


the kinetic energy of the charge after it has moved a distance ∆x = xf - xi :

from the work–kinetic energytheorem because the work done by the electric force isFe ∆ x = qE ∆ x and

W = ∆ K.


The electric field in the region between two oppositely charged flat metallic platesis approximately uniform .Suppose an electron of charge "e is projected horizontally into this field from the origin with an initial velocity at time t = 0.Because the electric field E in the Figure is in the positive y direction, the acceleration of the electron is in the negative y direction. That is,

ivi


An electron is projected horizontally into a uniform electric field produced by two charged plates. The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.

the acceleration is constant, we can apply the equations of kinematics in two Dimensions with vxi = vi and vyi = 0.

Because the electric field E in the Figure is in the positive y direction, the acceleration of the electron is in the negative y direction. That is,


After the electron has been in the electric field for a time interval, the components of its velocity at time t are

Its position coordinates at time t are

t = xf / vi

After the electron leaves the field, the electric force vanishes and the electron continues to move in a straight line in the direction of v in with a speed v > vi


Electron Beams; Cathode Ray Tubes

• Televisions, Oscilloscopes, Monitors, etc. use an electron beam steered by electric fields to light up the (phosphorescent) screen at specified points

E-field

metal plates- - - - - - -

+ + + + + + +

electron beam

screen

cathode emitter


Which electric field is responsible for the trajectory of the proton?


An electron (mass m = 9.11×10-31kg) is accelerated in the uniform field E (E = 1.33×104 N/C) between two parallel charged plates. The separation of the plates is 1.25 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure. With what speed does it leave the hole?

F = qE = ma

a = qE/m

Vf 2 = vi2 + 2a (∆)

Vf 2 = 2ad = 2(qE/m) ∆

Vf 2 = 2a ∆ = 2(qE/m) ∆=2(1.9 x 10 -19C) (1.33×104 N/C) (1.25m) l 9.11×10-

31kgVf = 8.3 x 10 6 m/s


Example : Electron moving perpendicular to .

Suppose an electron traveling with speed v0 = 1.0 x 107 m/s enters a uniform electric field =2x104N/C , which is at right angles to v0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity.

E

E

Solution: The acceleration is in the vertical direction (perpendicular to the motion in -y -direction)a = –eE/m =1.6x1019x2x104/9.11x10-31. y = ½ ay

2

x = v0t; y = -(eE/2mv0

2)x2


Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling 1 cm.

Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec, so the distance the electron falls upward is y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m

V

E

d

•e

E


; The electrons in a particle beam each have a kinetic energy of 1.60 x 10-17 J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0 cm?

Example :


An electron and a proton are each placed at rest in an electric field of 520 N/C. Calculate the speed of each particle 48.0 ns after being released.

Example :


An object having a net charge of 24.0 C is placed in a uniform electric field of 610 N/C that is directed vertically. What is the mass of this object if it “floats” in the field?

Example :


Three point charges are located at the corners of an equilateral triangle. Calculate the net electric force on the 7.00 μC charge.

Example :


Electric charges have the following important properties:• Unlike charges attract one another, and like charges repel one another.• Charge is conserved.• Charge is quantized—that is, it exists in discrete packets that are some integralmultiple of the electronic charge.

Conductors are materials in which charges move freely. Insulators are materialsin which charges do not move freely.


where ˆr is a unit vector directed from the charge to the point in question.

The electric field is directed radially outward from a positive charge and radially inward toward a negative charge.

The electric field due to a group of point charges can be obtained by usingthe superposition principle. That is, the total electric field at some point equals the vector sum of the electric fields of all the charges:


Electric field lines describe an electric field in any region of space. The numberof lines per unit area through a surface perpendicular to the lines is proportionalto the magnitude of E in that region.

A charged particle of mass m and charge q moving in an electric field E has anacceleration


Unit Modifiers for ReferenceSmaller Centi =

10-2

Milli ( m ) = 10-3

Micro () = 10-6

Nano ( n ) = 10-9

Pico ( p ) = 10-

12

Largero Kilo (K )= 103

o Mega = 106

o Giga = 109

o Tera = 1015

Examples:5mC = .005C10k = 10000


Chapter 23 Electric field Chapter 23 Electric field 23.1 Properties of Electric Charges 23.2...

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Transcript of Chapter 23 Electric field Chapter 23 Electric field 23.1 Properties of Electric Charges 23.2...