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Chapter Presentation

Transparencies Sample Problems

Visual Concepts

Standardized Test Prep

Resources

(3)

Electric Forces and Fields

Chapter 16

Section 1 Electric Charge Section 2 Electric Force

Section 3 The Electric Field

Chapter 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.

(5)

Chapter 16 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.

(6)

Chapter 16 Electric Charge

(7)

Chapter 16 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

(8)

Chapter 16 The Milikan Experiment

(9)

Chapter 16 Milikan’s Oil Drop Experiment

(10)

Chapter 16 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.

(11)

Chapter 16 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.

(12)

Chapter 16 Charging by Induction

(13)

Chapter 16 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.

(14)

Section 2 Electric Force

Chapter 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.

(15)

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 proportional to the magnitude of each charge and inversely proportional to the square of the distance between them.

   

 

 

  

 



1 2 2

charge 1 charge 2 electric force = Coulomb constant

electric C

F k q q

r

(16)

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.

(17)

Chapter 16 Superposition Principle

(18)

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.

(19)

Chapter 16 Sample Problem, continued

The Superposition Principle

1. Define the problem, and identify the known variables.

Given:

q₁ = +6.00  10^–9 C r_2,1 = 3.00 m q₂ = –2.00  10^–9 C r_3,2 = 4.00 m q₃ = +5.00  10^–9 C r_3,1= 5.00 m

 = 37.0º

Unknown: F_3,tot= ? Diagram:

(20)

Chapter 16 Sample Problem, continued

Tip: 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 F_3,1 is repulsive because q₁ and q₃ have the same sign.

The force F_3,2 is attractive because q₂ and q₃ have opposite signs.

(21)

Chapter 16 Sample Problem, continued

3. Calculate the magnitudes of the forces with Coulomb’s law.

   

 

   

 

–9 –9

2 3 1 9

3,1 2 2 2

–8 3,1

–9 –9

2 3 2 9

3,2 2 2 2

–9

5.00 10 C 6.00 10 C 8.99 10 N m

( 3,1) C 5.00 m

1.08 10 N

5.00 10 C 2.00 10 C 8.99 10 N m

( 3,2) C 4.00m

5.62 10 N

C

F k q q r F

   

   

     

 

   

   

     

 

(22)

Chapter 16 Sample Problem, continued

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

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

F_3,1: F_x = (F_3,1)(cos 37.0º) = (1.08  10^–8 N)(cos 37.0º) F_x = 8.63  10^–9 N

F_y = (F_3,1)(sin 37.0º) = (1.08  10^–8 N)(sin 37.0º) F_y= 6.50  10^–9 N

F_3,2: F_x = –F_3,2 = –5.62  10^–9 N F_y = 0 N

(23)

Chapter 16 Sample Problem, continued

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

F_x,tot = 8.63  10^–9 N – 5.62  10^–9 N = 3.01  10^–9 N F_y,tot = 6.50  10^–9 N + 0 N = 6.50  10^–9 N

(24)

Chapter 16 Sample Problem, continued

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

     

 

2 2 9 2 9 2

3, , ,

–9 3,

( ) ( ) (3.01 10 N) (6.50 10 N) 7.16 10 N

tot x tot y tot

tot

F F F

F

(25)

Chapter 16 Sample Problem, continued

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 N

tan 3.01 10 N

65.2º

y tot x tot

F F

(26)

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.

Chapter 16

(27)

Chapter 16 Objectives

• Calculate electric field strength.

• Draw and interpret electric field lines.

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

(28)

Chapter 16 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.

(29)

Chapter 16 Electric Fields and Test Charges

(30)

Chapter 16 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

charge producing the field electric field strength = Coulomb constant

distance

C

E k q r

(31)

Chapter 16 Calculating Net Electric Field

(32)

Chapter 16 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.

(33)

Chapter 16 Sample Problem, continued

1. 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

(34)

Chapter 16 Sample Problem, continued

2. 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.

 

9 2 2 –6 5

1 21 2

1

–6

9 2 2 5

2 22 2

2

7.00 10 C

8.99 10 N m /C 3.93 10 N/C

(0.400 m) 5.00 10 C

8.99 10 N m /C 1.80 10 N/C

(0.500 m)

C

E k q r E k q

r

  

       

 

  

       

 

(35)

Chapter 16 Sample Problem, continued

3. 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.

(36)

Chapter 16 Sample Problem, continued

4. Find the x and y components of each electric field vector.

For E₁: E_x,1 = 0 N/C

E_y,1 = 3.93  10⁵ N/C

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

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

(37)

Chapter 16 Sample Problem, continued

5. Calculate the total electric field strength in both directions.

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

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

(38)

Chapter 16 Sample Problem, continued

6. Use the Pythagorean theorem to find the

magnitude of the resultant electric field strength vector.

   

 

   

 

2 2

, ,

2 2

5 5

5

1.08 10 N/C 2.49 10 N/C 2.71 10 N/C

tot x tot y tot

tot

E E E

E E

(39)

Chapter 16 Sample Problem, continued

7. 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/C

tan 1.08 10 N/C

66.0

y tot x tot

E E

(40)

Chapter 16 Sample Problem, continued

Electric Field Strength 8. 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.

(41)

Chapter 16 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.

(42)

Chapter 16 Rules for Drawing Electric Field Lines

(43)

Chapter 16 Rules for Sketching Fields Created by Several Charges

(44)

Chapter 16 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.

(45)

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.

Chapter 16

(46)

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.

Chapter 16

(47)

Multiple Choice, continued

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.

Chapter 16

(48)

Multiple Choice, continued

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.

Chapter 16

(49)

Multiple Choice, continued

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

Chapter 16

(50)

Multiple Choice, continued

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

Chapter 16

(51)

Multiple Choice, continued

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

Chapter 16

4 2

1 2 1 4 F.

G.

H.

J.

(52)

Multiple Choice, continued

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

Chapter 16

4 2

1 4 1 2 F.

G.

H.

J.

(53)

Multiple Choice, continued

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

Chapter 16

(54)

Multiple Choice, continued

5. What is this process of charging called?

A. charging by contact B. charging by induction C. charging by conduction D. charging by polarization

Chapter 16

(55)

Multiple Choice, continued

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

F. neutral G. negative H. positive

J. both positive and negative

Chapter 16

(56)

Multiple Choice, continued

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

F. neutral G. negative H. positive

J. both positive and negative

Chapter 16

(57)

Multiple Choice, continued

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

Chapter 16

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.

(58)

Multiple Choice, continued

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

Chapter 16

(59)

Multiple Choice, continued

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

Chapter 16

(60)

Multiple Choice, continued

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

Chapter 16

(61)

Multiple Choice, continued

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

Chapter 16

(62)

Multiple Choice, continued

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

Chapter 16

(63)

Multiple Choice, continued

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

Chapter 16

(64)

Multiple Choice, continued

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

Chapter 16

(65)

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?

Chapter 16

(66)

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

Chapter 16

(67)

Chapter 16 Short Response, continued

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.

(68)

Chapter 16 Short Response, continued

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.

(69)

Chapter 16 Short Response, continued

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?

(70)

Chapter 16 Short Response, continued

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

(71)

Chapter 16 Short Response, continued

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.

(72)

Chapter 16 Short Response, continued

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

(73)

Chapter 16 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?

(74)

Chapter 16 Extended Response, continued

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

Answer: 6.1  10¹⁰ m/s²

(75)

Chapter 16 Extended Response, continued

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

(76)

Chapter 16 Extended Response, continued

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

Answer: 2.0  10^–5 s

(77)

Chapter 16 Extended Response, continued

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

(78)

Chapter 16 Extended Response, continued

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

Answer: 12 m

(79)

Chapter 16 Extended Response, continued

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

(80)

Chapter 16 Extended Response, continued

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

Answer: 1.2  10^–15 J

(81)

Chapter 16 Extended Response, continued

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.

(82)

Chapter 16 Extended Response, continued

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.

(83)

Chapter 16 Charging By Induction

(84)

Chapter 16 Transfer of Electric Charge

(85)

How to Use This Presentation

How to Use This Presentation

Resources

Chapter 16

Table of Contents

Chapter 16

Objectives

Chapter 16

Properties of Electric Charge

Chapter 16

Electric Charge

Chapter 16

Properties of Electric Charge, continued

Chapter 16

The Milikan Experiment

Chapter 16

Milikan’s Oil Drop Experiment

Chapter 16

Transfer of Electric Charge

Chapter 16

Transfer of Electric Charge, continued

Chapter 16

Charging by Induction

Chapter 16

Transfer of Electric Charge, continued

Chapter 16

Objectives

Chapter 16

Coulomb’s Law

   

 

Chapter 16

Coulomb’s Law, continued

Chapter 16

Superposition Principle

Chapter 16

Sample Problem

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued

   

 

   

 

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued

Coulomb’s Law, continued

Chapter 16

Chapter 16

Objectives

Chapter 16

Electric Field Strength

Chapter 16

Electric Fields and Test Charges

Chapter 16

Electric Field Strength, continued

 

Chapter 16

Calculating Net Electric Field

Chapter 16

Sample Problem

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued

 

 

Chapter 16

Sample Problem, continued

Chapter 16

Sample Problem, continued