GIANCOLI- CHAPTER 16 Electric Forces and Fields. Electric Charge Electric charge is conserved Electric Charge is quantized One unit of charge : e= 1.60219.

GIANCOLI- CHAPTER 16

Electric Forces and Fields

Electric Charge

Electric charge is conserved

Electric Charge is quantized One unit of charge : e= 1.60219 x 10-19 C

C stands for Coulomb, the unit of electric charge

A proton has a charge of +1.60 x 10-19 C

An electron has a charge of -1.60 x 10-19 C

Conductors & Insulators

Conductors: Materials in which electric charges move freely Examples: most metals

Insulators: Materials in which electric charges do not move freely Examples: Plastic, glass, silk, rubber

Charging by contact

The two objects are rubbed together and electrons are transferred from one to the other electrons from the fur

are transferred to the rod

Charging by Induction

To charge by induction, a charged object is brought close to (not touching!) a conductor and then a conducting wire connects the conductor to the ground and the electrons travel to the ground

Charging by Polarization

Charging by polarization creates a surface charge A charged object is brought close to an insulator and

the electrons and protons realign themselves to create one side that is more positive and one that is more negative

Coulomb’s Law

Coulomb’s Law describes the mathematical relationship between electric force, distance and electric charge for two objects

1 22electric C

q qF k

r

Electric force= Coulomb’s Constant x (charge 1)(charge 2)distance2

kC= 8.99 x 109 Nm2

C2

Coulomb’s Law

Remember that force is a vector!

For problems involving two charges, the direction is either “attractive” or “repulsive”

the direction of the force between a positive charge and negative charge is attractive a

the direction of the force between two negative charges is repulsive

Example Problem

Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of +12 x 10-9 C and the other is given a charge of -18 x 10-9 C A. Find the electric force exerted on one sphere by the

other B. The spheres are connected by a conducting wire.

After equilibrium has occurred, find the electric force between the two spheres.

Use Coulomb’s Law

1 22electric C

q qF k

r

9 929 5

22

12 10 18 108.99 10 2.2 10 Attractive

0.30

x C x CNmF x x N

C m

What does it mean “after equilibrium has occurred”? The charge on each sphere is the same

9 929 7

22

3 10 3 108.99 10 8.99 10 Repulsive

0.30

x C x CNmF x x N

C m

Three charges in a line

Three charges are arranged in a line as shown below. The distance between Q1 and Q2 is 0.35 m and the distance between Q2 and Q3 is 0.45 m.

What is the net electrostatic force acting on Q3? Q1

+4 µC

Q2-5 µC

Q3+8µC

Find the force acting on Q3 from Q1

Find the force acting on Q3 from Q2

Net force acting on Q3= 1.33 N Left

Right N 45.08.0

108104109

2

662

29

13

m

CxCxCNm

x

F

Left N 78.145.0

108105109

2

662

29

23

m

CxCxCNm

x

F

Two Dimensional Problem

Three charges are arranged in a triangular pattern as shown below. The distance between Q1 and Q2 is 0.35 m and the distance between Q2 and Q3 is 0.45 m.

Find the net electrostatic force on Q2Q1+4 µC

Q2-5 µC

Q3+8µC

Solving the Problem

We already have the numerical values for the force acting on Q2from Q3…but the direction is different F32= 1.78 N Right

Find the force from Q1 on Q2

UpN 47.1

35.0

105104109

2

662

29

12

m

CxCxCNm

x

F

Finish the Problem

Use vector addition (Pythagorean theorem) to find the net force acting on Q2.

Direction?

N 31.247.178.1 22 netF

EofN 6.3978.1

47.1tan 1

1.78 N

1.47 N

Electric Field

Electric force, like gravitational force, is a field force Remember: Field forces can act through space even

when there is no physical contact between the objects involved

A charged object has an electric field in the space around it

Electric Field Lines

Electric Field Lines point in the direction of the electric field

The number and spacing of field lines is proportional to the electric field strength The electric field is strong where the field lines are

close together and weaker when they are far apart

Electric Field Lines

The lines for a positive charge point away from the charge

The lines for a negative charge point towards the charge

Electric Field Lines

This diagram shows the electric field lines for two equal and opposite point charges Notice that the lines begin on the positive charge and

end on the negative charge

Electric Field Lines

This diagram shows the electric field lines for two positive point charges Notice that the same number of lines emerges from

each charge because they are equal in magnitude

Electric Field Lines

If the charges are unequal, then the number of lines emerging from them will be different Notice that the positive charge has twice as many

lines

Calculating Electric Field StrengthThe equation for the electric field

produced by a point charge is:

Kc=9x109 Nm2/C2 ,r is the distance from the charge and q is the charge producing the field

The unit for E is N/C

Electric field strength is a vector!! If q is positive, then E is directed away

from q If q is negative, then E is directed

toward q

2r

qkE c

Calculating the force from an electric field

If a charged object is placed in an electric field, we can calculate the force acting on it from the electric field

Remember that F is a vector!!

F qE

Sample Problem p. 647 #3

An electric field of 2.0 x 104 N/C is directed along the positive x-axis

a. What is the electric force on an electron in this field?

b. What is the electric force on a proton in this field?

Sample Problem p. 647 #3

E= 2.0 x 104 N/C , q= 1.6 x 10-19 CF=qE= 3.2 x 10-15 N for both the electron

and the proton

What about the direction? The electric field is pointing along the positive x

axis (to the right) which means there’s a positive charge to the left

E field+

For the proton

Since the electric field is pointing to the right, if you put a proton in it, the proton will want to move away towards the right and the direction of the force on it will be to the right

Answer: 3.2 x 10-15 N along the positive x axis (to the right)

+ +F

For the electron

Since there’s a positive charge causing the electric field to point towards the right, an electron would feel attracted to the positive charge. Therefore, the force acting on it is toward the left

Answer: 3.2 x 10-15 N along the negative x axis (to the left)+ -

F

Sample Problem

Find the electric field at a point midway between two charges of +30 nC and 60 nC separated by a distance of 30.0 cm

+30 nC +60 nC

For the 30 nC charge:

Direction of the E-field for both charges is “away” since they’re both positive

C

N

m

xx

r

qkE c 000,12

15.0

)1030(1092

99

2

For the 60 nC For the 60 nC charge:charge:

+30 nC +60 nC

C

N

m

xx

r

qkE c 000,24

15.0

)1060(1092

99

2

Which one will win?

At the midway point, the 30nC charge’s field strength is 12000 N/C toward the 60 nC charge and the 60 nC charge’s field strength is 24,000 N/C toward the 30 nC charge.

The 60 nC charge will win. Since the field’s point in opposite directions, you have to subtract

Answer: 12,000 N/C toward the 30 nC charge

Sample Problem

A constant electric field directed along the positive x-axis has a strength of 2.0 x 103 N/C. Find the electric force exerted on a proton by the field Find the acceleration of the proton

Answer

F=qE=(1.6x10-19 C)(2.0 x 103 N/C)= 3.2 x 10-16 N

Direction?

Answer: 3.2 x 10-16 N along the positive x-axis (to the right)

E field+ +

F

Answer

B. What is the acceleration?

Ask Newton! F=ma a = F/m= 3.2 x 10-16 N/1.6x10-27 kg a= 2 x 1011 m/s2 along the positive x axis