hp 16 win

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Transparencies Sample Problems

Visual Concepts

Standardized Test Prep

Resources



Electric Forces and FieldsChapter 16

Section 1 Electric Charge

Section 2 Electric Force

Section 3 The Electric Field

Table of Contents



Section 1 Electric ChargeChapter 16

Objectives

• Understand the basic properties of electric charge.

• Differentiate between conductors and insulators.

• Distinguish between charging by contact, charging by induction, and charging by polarization.



Chapter 16Section 1 Electric Charge

Properties of Electric Charge

• There are two kinds of electric charge.– like charges repel– unlike charges attract

• Electric charge is conserved.– Positively charged particles are called protons.– Uncharged particles are called neutrons.– Negatively charged particles are called electrons.



Chapter 16

Electric Charge





Properties of Electric Charge, continued

• Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge.

• Charge is measured in coulombs (C).

• The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton.

e = 1.602 176 x 10–19 C



Chapter 16

The Milikan Experiment




Chapter 16

Milikan’s Oil Drop Experiment





Transfer of Electric Charge

• An electrical conductor is a material in which charges can move freely.

• An electrical insulator is a material in which charges cannot move freely.




Transfer of Electric Charge, continued

• Insulators and conductors can be charged by contact.

• Conductors can be charged by induction.

• Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor.



Chapter 16

Charging by Induction





Transfer of Electric Charge, continued

• A surface charge can be induced on insulators by polarization.

• With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.



Section 2 Electric ForceChapter 16

Objectives

• Calculate electric force using Coulomb’s law.

• Compare electric force with gravitational force.

• Apply the superposition principle to find the resultant force on a charge and to find the position at which the net force on a charge is zero.



Chapter 16

Coulomb’s Law

• Two charges near one another exert a force on one another called the electric force.

• Coulomb’s law states that the electric force is propor-tional to the magnitude of each charge and inversely proportional to the square of the distance between them.


1 22

2

charge 1 charge 2electric force = Coulomb constant

distance

electric C

q qF k

r



Chapter 16

Coulomb’s Law, continued

• The resultant force on a charge is the vector sum of the individual forces on that charge.

• Adding forces this way is an example of the principle of superposition.

• When a body is in equilibrium, the net external force acting on that body is zero.




Chapter 16

Superposition Principle




Chapter 16

Sample Problem

The Superposition Principle

Consider three point charges at the corners of a triangle, as shown at right, where q1 = 6.00 10–9 C, q2 = –2.00 10–9 C, and q3 = 5.00 10–9 C. Find the magnitude and direction of the resultant force on q3.




Chapter 16Section 2 Electric Force

Sample Problem, continuedThe Superposition Principle1. Define the problem, and identify the known

variables.Given:

q1 = +6.00 10–9 C r2,1 = 3.00 m

q2 = –2.00 10–9 C r3,2 = 4.00 m

q3 = +5.00 10–9 C r3,1 = 5.00 m

= 37.0º

Unknown: F3,tot = ? Diagram:




Sample Problem, continuedThe Superposition PrincipleTip: According to the superposition principle, the resultant

force on the charge q3 is the vector sum of the forces exerted by q1 and q2 on q3. First, find the force exerted on q3 by each, and then add these two forces together vectorially to get the resultant force on q3.

2. Determine the direction of the forces by analyzing the charges.

The force F3,1 is repulsive because q1 and q3 have the same sign.

The force F3,2 is attractive because q2 and q3 have opposite signs.




Sample Problem, continuedThe Superposition Principle3. Calculate the magnitudes of the forces with

Coulomb’s law.

–9 –9293 1

3,1 22 2

–83,1

–9 –9293 2

3,2 22 2

–93,1

5.00 10 C 6.00 10 CN m 8.99 10

( 3,1) C 5.00 m

1.08 10 N

5.00 10 C 2.00 10 CN m8.99 10

( 3,2) C 4.00m

5.62 10 N

C

C

q qF k

r

F

q qF k

r

F




Sample Problem, continuedThe Superposition Principle

4. Find the x and y components of each force.

At this point, the direction each component must be taken into account.

F3,1: Fx = (F3,1)(cos 37.0º) = (1.08 10–8 N)(cos 37.0º)

Fx = 8.63 10–9 N

Fy = (F3,1)(sin 37.0º) = (1.08 10–8 N)(sin 37.0º)

Fy = 6.50 10–9 N

F3,2: Fx = –F3,2 = –5.62 10–9 N

Fy = 0 N




Sample Problem, continued


5. Calculate the magnitude of the total force acting in both directions.

Fx,tot = 8.63 10–9 N – 5.62 10–9 N = 3.01 10–9 N

Fy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N






6. Use the Pythagorean theorem to find the magni-tude of the resultant force.

2 2 9 2 9 23, , ,

–93,

( ) ( ) (3.01 10 N) (6.50 10 N)

7.16 10 N

tot x tot y tot

tot

F F F

F






7. Use a suitable trigonometric function to find the direction of the resultant force.

In this case, you can use the inverse tangent function:

–9,

–9,

6.50 10 Ntan

3.01 10 N

65.2º

y tot

x tot

F

F



Coulomb’s Law, continued

• The Coulomb force is a field force.

• A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects.




Section 3 The Electric FieldChapter 16

Objectives

• Calculate electric field strength.

• Draw and interpret electric field lines.

• Identify the four properties associated with a conductor in electrostatic equilibrium.



Chapter 16Section 3 The Electric Field

Electric Field Strength

• An electric field is a region where an electric force on a test charge can be detected.

• The SI units of the electric field, E, are newtons per coulomb (N/C).

• The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge.



Chapter 16

Electric Fields and Test Charges





Electric Field Strength, continued

• Electric field strength depends on charge and distance. An electric field exists in the region around a charged object.

• Electric Field Strength Due to a Point Charge

2

2

charge producing the fieldelectric field strength = Coulomb constant

distance

C

qE k

r



Chapter 16

Calculating Net Electric Field





Sample Problem

Electric Field Strength

A charge q1 = +7.00 µC is at the origin, and a charge q2 = –5.00 µC is on the x-axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin.



Chapter 16


Electric Field Strength1. Define the problem, and identify the known

variables.Given:

q1 = +7.00 µC = 7.00 10–6 C r1 = 0.400 m

q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 m = 53.1º

Unknown:

E at P (y = 0.400 m)

Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.




Chapter 16

Sample Problem, continuedElectric Field Strength2. Calculate the electric field strength produced by

each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

–69 2 2 51

1 2 21

–69 2 2 52

2 2 22

7.00 10 C8.99 10 N m /C 3.93 10 N/C

(0.400 m)

5.00 10 C8.99 10 N m /C 1.80 10 N/C

(0.500 m)

C

C

qE k

r

qE k

r




Chapter 16

Sample Problem, continuedElectric Field Strength3. Analyze the signs of the

charges.

The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.




Chapter 16

Sample Problem, continuedElectric Field Strength4. Find the x and y components of each electric field

vector.

For E1: Ex,1 = 0 N/C

Ey,1 = 3.93 105 N/C

For E2: Ex,2= (1.80 105 N/C)(cos 53.1º) = 1.08 105 N/C

Ey,1= (1.80 105 N/C)(sin 53.1º)= –1.44 105 N/C




Chapter 16

Sample Problem, continuedElectric Field Strength5. Calculate the total electric field strength in both

directions.

Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08 105 N/C = 1.08 105 N/C

Ey,tot = Ey,1 + Ey,2 = 3.93 105 N/C – 1.44 105 N/C = 2.49 105 N/C




Chapter 16

Sample Problem, continuedElectric Field Strength6. Use the Pythagorean theorem to find the

magnitude of the resultant electric field strength vector.

22

, ,

2 25 5

5

1.08 10 N/C 2.49 10 N/C

2.71 10 N/C

tot x tot y tot

tot

tot

E E E

E

E




Chapter 16

Sample Problem, continuedElectric Field Strength7. Use a suitable trigonometric function to find the

direction of the resultant electric field strength vector.In this case, you can use the inverse tangent function:

5,

5,

2.49 10 N/Ctan

1.08 10 N/C

66.0

y tot

x tot

E

E




Chapter 16

Sample Problem, continuedElectric Field Strength8. Evaluate your answer.

The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.





Electric Field Lines

• The number of electric field lines is proportional to the electric field strength.

• Electric field lines are tangent to the electric field vector at any point.



Chapter 16

Rules for Drawing Electric Field Lines




Chapter 16

Rules for Sketching Fields Created by Several Charges





Conductors in Electrostatic Equilibrium• The electric field is zero everywhere inside the

conductor.

• Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface.

• The electric field just outside a charged conductor is perpendicular to the conductor’s surface.

• On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.



Multiple Choice

1. In which way is the electric force similar to the gravitational force?

A. Electric force is proportional to the mass of the object.

B. Electric force is similar in strength to gravitational force.

C. Electric force is both attractive and repulsive.

D. Electric force decreases in strength as the distance between the charges increases.

Standardized Test PrepChapter 16



Multiple Choice, continued

1. In which way is the electric force similar to the gravitational force?

A. Electric force is proportional to the mass of the object.

B. Electric force is similar in strength to gravitational force.

C. Electric force is both attractive and repulsive.

D. Electric force decreases in strength as the distance between the charges increases.





2. What must the charges be for A and B in the figure so that they produce the electric field lines shown?

F. A and B must both be positive.

G. A and B must both be negative.

H. A must be negative, and B must be positive.

J. A must be positive, and B must be negative.





3. Which activity does not produce the same results as the other three?

A. sliding over a plastic-covered automobile seat

B. walking across a woolen carpet

C. scraping food from a metal bowl with a metal spoon

D. brushing dry hair with a plastic comb





4. By how much does the electric force between two charges change when the distance between them is doubled?


4

2

1 21

4

F.

G.

H.

J.




4. By how much does the electric force between two charges change when the distance between them is doubled?


4

2

1 4

1

2

F.

G.

H.

J.




Use the passage below to answer questions 5–6.

A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded.

5. What is this process of charging called?

A. charging by contact

B. charging by induction

C. charging by conduction

D. charging by polarization





Use the passage below to answer questions 5–6.

A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded.

6. What kind of charge is left on the conductor’s surface?

F. neutral

G. negative

H. positive

J. both positive and negative





7. What is the electric field strength 2.0 m from the center of the conducting sphere?

A. 0 N/C

B. 5.0 102 N/C

C. 5.0 103 N/C

D. 7.2 103 N/C


Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator.




8. What is the strength of the electric field at the surface of the conducting sphere?

F. 0 N/C

G. 1.5 102 N/C

H. 2.0 102 N/C

J. 7.2 103 N/C






9. What is the strength of the electric field inside the conducting sphere?

A. 0 N/C

B. 1.5 102 N/C

C. 2.0 102 N/C

D. 7.2 103 N/C






10. What is the radius of the conducting sphere?

F. 0.5 m

G. 1.0 m

H. 1.5 m

J. 2.0 m





Short Response

11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center?




Short Response, continued

11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center?

Answer: 0.0 N/C






12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer.





12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer.

Answer: not necessarily; The suspended object might have a charge induced on it, but its overall charge could be neutral.





13. One gram of hydrogen contains 6.02 1023 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38 106 m, what is the magnitude of the resulting compressional force on Earth?





13. One gram of hydrogen contains 6.02 1023 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38 106 m, what is the magnitude of the resulting compressional force on Earth?

Answer: 5.12 105 N





14. Air becomes a conductor when the electric field strength exceeds 3.0 106 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius.





14. Air becomes a conductor when the electric field strength exceeds 3.0 106 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius.

Answer: 1.3 10–3 C




Extended Response

Use the information below to answer questions 15–18.

A proton, which has a mass of 1.673 10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2 106 m/s.

15. What is the magnitude of the acceleration of the proton?




Extended Response, continued



15. What is the magnitude of the acceleration of the proton?

Answer: 6.1 1010 m/s2







16. How long does it take the proton to reach this speed?







16. How long does it take the proton to reach this speed?

Answer: 2.0 10–5 s







17. How far has it moved in this time interval?







17. How far has it moved in this time interval?

Answer: 12 m







18. What is its kinetic energy at the later time?







18. What is its kinetic energy at the later time?

Answer: 1.2 10–15 J





19. A student standing on a piece of insulating material places her hand on a Van de Graaff generator. She then turns on the generator. Shortly thereafter, her hairs stand on end. Explain how charge is or is not transferred in this situation, why the student is not shocked, and what causes her hairs to stand up after the generator is started.





19. (See previous slide for question.)Answer: The charge on the sphere of the Van de Graaff

generator is transferred to the student by means of conduction. This charge remains on the student because she is insulated from the ground. As there is no path between the student and the generator and the student and the ground by which charge can escape, the student is not shocked. The accumulation of charges of the same sign on the strands of the student’s hair causes the strands to repel each other and so stand on end.




Charging By Induction




Transfer of Electric Charge




Electric Field Lines

hp 16 win

Education

electric charge section

electric force section

electric force chapter16

transfer of electric

charge q

surface charge

c chapter16 section

induction chapter16