Preview Objectives Properties of Electric Charge Transfer of Electric Charge Chapter 16 Section 1 Electric Charge.

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• Objectives

• Properties of Electric Charge

• Transfer of Electric Charge

Chapter 16 Section 1 Electric Charge

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.


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.

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Visual Concept


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

Section 1 Electric Charge


Visual Concept


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.


Visual Concepts

Visual Concept

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.

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• Objectives

• Coulomb’s Law

• Sample Problem

Chapter 16 Section 2 Electric Force

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.

Section 2 Electric Force

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.



Visual Concept


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.



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.


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• Objectives

• Electric Field Strength

• Sample Problem

• Electric Field Lines

• Conductors in Electrostatic Equilibrium

Chapter 16 Section 3 The Electric Field

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.


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.


Visual Concept


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


Visual Concept


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.

Section 3 The Electric Field

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.


Visual Concept


Rules for Drawing Electric Field Lines


Visual Concept


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.

Preview Objectives Properties of Electric Charge Transfer of Electric Charge Chapter 16 Section 1 Electric Charge.

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kinds of electric charge

electric force slide

surface charge

fundamental unit of

slight charge separation

induction slide

c slide

net force