TV Electric Charge (Notes) - TGB PHYSICStgbphysics.weebly.com/uploads/2/5/6/0/25608529/electric_charges_ppt.pdf · Electric Charge and Electric Field 16 A large electroscope is made

Electric Charge and

Electric Field

Electric Charge and Electric Field 16

AP Physics 4

Lecture Notes

� Coulomb’s Law

� The Electric Field

� Field Lines

� Electric Fields and Conductors


Coulomb’s Law

Coulomb’s law:

221

r

QQkF ====

r1Q 2Q

Charge (Q)

Coulombs (C)

1 C = 6.2421 x 1018 e

e = 1.602 x 10−−−−11119 C

Distance (m)

Coulomb’s constant (k)

k = 8.988 x 109 N⋅⋅⋅⋅m2/C2

Force (N)

12F 21F

2112 FFF ========


Coulomb’s Law

Coulomb’s law:

221

o r

QQ

πε4

1F ====

r1Q 2Q

oπε4

1k ====

2mN

2C12o 10x 8.85

kπ4

1ε

⋅⋅⋅⋅

−−−−========

Permittivity of free space

221

r

QQkF ====


The force is along the line connecting the charges,

and is attractive if the charges are opposite, and

repulsive if they are the same.

Coulomb’s Law


Vector addition review:

Coulomb’s Law

Two forces acting

on an object


tail-to-tip method

Parallelogram method

Components method

Coulomb’s law strictly applies only to point charges.

Superposition: for multiple point charges, the forces on

each charge from every other charge can be calculated

and then added as vectors.

Coulomb’s Law

1Q2Q

3Q

32F

31F

3Q

32F

31F

netF

Net force on Q3


Two charged dust particles exert a force of .32 N on each

other. What will be the force if they are moved so they

are only one-eighth as far apart?

Problem 16-06

Coulomb’s law is an inverse square law.

( ) N 48.20 32. 64 == NF


A person scuffing her feet on a wool rug on a dry day

accumulates a net charge of −−−−42 µµµµC. How many excess

electrons does she get,

Problem 16-07

By how much does her mass increase?

lectronse 10 x 622.2C 10 x 602.1

electron 1 C 10 x 42 14

196 ====

−−−−−−−−

−−−−−−−−

kg 10 x 389.2e 1

kg 10 x .119e 10 x 622.2 16

3114 −−−−

−−−−

−−−−−−−− ====


Compare the electric force holding the electron in orbit

(r = 0.53 × 10 l0 m) around the proton nucleus of the

hydrogen atom, with the gravitational force between the

same electron and proton.

What is the ratio of these two forces?

Problem 16-08

21

21

221

221

G

E

mGm

QkQ

r

mGmr

QkQ

F

F========

(((( ))))(((( ))))(((( ))))kg10 x .671kg10 x .119/kgmN 10 x .676

C 10 x 602.1 /CmN 10 x .988827312211

219229

−−−−−−−−−−−−

−−−−

⋅⋅⋅⋅

⋅⋅⋅⋅====

3910 x 3.2====


At each corner of a square of

side L there are four point

charges. Determine the force

on the charge 2Q.

Problem 16-16

Q2Q

3Q4QL

(((( )))) (((( )))) (((( ))))2

2

2

2o

22xL

kQ 8284.4222

L

kQ45cos

L2

Q4Q2k

L

QQ2kF ====++++====++++====

(((( )))) (((( )))) (((( ))))2

2

2

2o

22yL

kQ 8284.8226

L

kQ45sin

L2

Q4Q2k

L

Q3Q2kF ====++++====++++====

2

22y

2xQ2

L

kQ 0625.10 FFF ====++++====


A +4.75 µµµµC and a −−−−3.55 µµµµC charge are placed 18.5 cm apart.

Where can a third charge be placed so that it experiences no

net force?

Problem 16-20Q1 Q2

d x

Q

21 FF ====

(((( )))) 22

21

x

QQk

xd

QQk ====

++++

21

2

QQ

Qdx

−−−−====

C 10 x 5.3C 10 x .74

C 10 x 5.3cm 5.18x

66

6

−−−−−−−−

−−−−

−−−−==== cm 116====


The Electric Field

The electric field is defined as the force on a test charge

(small positive charge), divided by the charge:

q

FE ====

E E

F F

qEF ==== qEF −−−−====


Electric field of a point charge

+Q ••••

r

q++++

q

FE ====

qr

Qqk

2====

The Electric Field

2r

QkE ====

2r

QqkF ====


The Electric Field

Calculate forces using Coulomb’s law

Add forces vectorially to get result

Draw a diagram; show all charges, with signs,

and electric fields and forces with directions

Problem solving in electrostatics:


A proton is released in a uniform electric field, and it

experiences an electric force of 3.75 × 10−−−−14 N toward

the south. What are the magnitude and direction of the

electric field?

Problem 16-24

+ south

E

F

q

FE ==== N/C 10x 2.34

C 10x 1.602

N 10x 3.75 591

14========

−−−−

−−−−


What are the magnitude and direction of the electric

field at a point midway between a +7.0 µµµµC and a −−−−8.0 µµµµC

charge 8.0 cm apart? Assume no other charges are

nearby.

Problem 16-28

+ -••••

d

q1 q2

E1

E2

21 EEE ++++====

(((( )))) (((( ))))22

21

2d

kq

2d

kqE ++++====

(((( ))))212qq

d

k4E ++++====

(((( ))))(((( ))))

(((( ))))N x10 8.0N x10 7.0m 08.0

m/CN x10 8.9884 662

9−−−−−−−− ++++

⋅⋅⋅⋅====

N/C x10 8.4E 7====


Two point charges, −−−−Q and +2Q, are separated by a distance

of d = 12 cm. The electric field at the point is zero.

How far from −−−−Q is P?

Problem 16-36

+-Q−−−− Q2++++

d = 12 cm

P

x••••

E1E2

21 EE ====

(((( ))))22

21

dx

kQ

x

kQ

++++====

12

1

QQ

Qdx

−−−−====

11

1

QQ2

Qdx

−−−−====

12

dx

−−−−====

12

cm 12

−−−−====

cm 29x ====


A proton (m = 1.67 x 10−−−−27 kg) is suspended at rest in a

uniform electric field E. Take into account gravity at the

Earth's surface, and determine E.

Problem 16-54

+q

E

GF

EF GE FF ====

mgqE ====

q

mgE ====

(((( ))))C 10x 602.1

m/s 9.80 kg 10x 67.1E

91

227

−−−−

−−−−====

N/C 10x 02.1E 7−−−−====


The electric field can be represented by field lines. These

lines start on a positive charge and end on a negative

charge.

The Electric Field


The number of field lines starting (ending) on a positive

(negative) charge is proportional to the magnitude of the

charge.

The Electric Field

The electric field is stronger where the field lines are closer

together.

••••A

••••B ••••C

E

E


Electric field of two charges:

The Electric Field


The electric field between two closely spaced, oppositely

charged parallel plates is constant.

The Electric Field


Field Lines

Field lines indicate the direction of the field; the field is

tangent to the line.

The magnitude of the field is proportional to the density

of the lines.

Field lines start on positive charges and end on negative

charges; the number is proportional to the magnitude of

the charge.

Summary of field lines:


Electric Fields and Conductors

When finger is removed

When charged rod is removed

Electrically neutral

Net positive charge

( on outer surface)

Charging by induction:





The electric field is

perpendicular to the

surface of a conductor

The electric field is zero

inside a conductor


Summary

Insulators: nonconductors

Conductors: electrons free to move

C 10x 1.602e 19−−−−====Charge on electron:

Charge is conserved

Two kinds of electric charge – positive and negative

Objects can be charged by conduction or induction

Charge is quantized in units of e

Coulomb’s law:2

11

r

QQkF ====

Electric field is force per unit charge:q

FE ====


A large electroscope is made with

"leaves" that are 78 cm long wires

with tiny 24 g spheres at the ends.

When charged, nearly all the

charge resides on the spheres. If the

wires each make a 30° angle with

the vertical what total charge Q

must have been applied to the

electroscope? Ignore the mass of the

wires.

Problem 16-64

2

Q

2

Q

LL

m m

θθθθ θθθθ


Problem 16-64

TF

mg

EFθ

0mgθcosFF Ty ====−−−−====∑∑∑∑0θsinFFF TEx ====−−−−====∑∑∑∑

ET FθsinF ==== mgθcosFT ====

mgθcosF

FθsinF

T

ET

====

====θtanmgF E ====⇒⇒⇒⇒

(((( ))))θtanmg

d

2Qk

F2

2

E ======== k

θtanmgd2Q ====⇒⇒⇒⇒

(((( ))))(((( ))))(((( ))))

229

o

C/mN 10x 988.8

30tangk 024.0m 78.02Q

⋅⋅⋅⋅==== C 10x 1.6 6−−−−====


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TV Electric Charge (Notes) - TGB PHYSICStgbphysics.weebly.com/uploads/2/5/6/0/25608529/electric_charges_ppt.pdf · Electric Charge and Electric Field 16 A large electroscope is made