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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

Section 1 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.



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 2

2

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, continued


1. 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

θ = 37.0º

Unknown: F3,tot = ? Diagram:






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 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.






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

( ) ( )( )

( ) ( )( )

–9 –9293 1

3,1 22 2

–8

3,1

–9 –9293 2

3,2 22 2

–9

3,1

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

× × ⋅ = = ×

= ×

× × ⋅ = = ×

= ×






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






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 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






7. Use a suitable trigonometric function to find the

direction of the resultant force.

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

ϕ

ϕ

×= =

×

=

–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.




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

Section 3 The Electric Field




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


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



1. Define the problem, and identify the knownvariables.

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

θ = 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



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.

( )

( )

–69 2 2 51

1 2 2

1

–69 2 2 52

2 2 2

2

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



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.




Chapter 16



4. 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



5. 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



6. 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



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:

ϕ

ϕ

×= =

×

= °

5,

5

,

2.49 10 N/Ctan

1.08 10 N/C

66.0

y tot

x tot

E

E




Chapter 16



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.





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 isperpendicular 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.

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