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Chapter 19Electric Force and Electric Field

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Interaction of Electric Charge

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Charging an object A glass rod is rubbed with

silk

Electrons are transferredfrom the glass to the silk

Each electron adds a

negative charge to the silk

An equal positive charge isleft on the rod

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Charge

Unit:

C, Coulomb

+

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Electric ChargeElectric charge is one of the fundamental attributes of the

particles of which matter is made.

qproton e

qelectron e

e

1.6

10

19

C

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Electric Field

Like charges (++)

Electric dipole:

Opposite charges (+)

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Vectors are arrows

i

j

v1 2

i

j

v3 i j

v2

1.5i 0.5 j

v4 2i j

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What are these vectors?

i

jv1

2 i

v3 i 2 j

v2i 1.5 j

v4 3

i 2

j

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Magni tudeof a vector

= Length of the arrow

4

3

v 4 i 3 j

v 42 32 5

If v ai bj

v

v a2

b2

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What are the magnitudes?

i

jv1

2 i

v3 i 2 j

v2i 1.5 j

v4 3

i 2

j

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Magnitudes (solution)

i

jv1

2irv

1

2

v3 i 2 j

rv

3

(1)2 (2)2

12 22 5 2.24

v2i 1.5 j

rv2 1

2 1.52 1.80

v4 3i 2 j

r

v4 32

22

13 3.61

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Adding and subtracting vectors

u 2i 9

j

rv 5i 7.2 j

ru

rv 7i 16.2 j

ru

rv 3i 1.8 j

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Add and subtract

i

jv1 2i

v3 i 2 j

v2i 1.5 j

v4

3

i

2

j

Find:rv1

rv2 ,

rv1

rv2 ,

rv4

rv3

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v1 2i

v3 i 2 j

v4 3i 2 j

Find v1 v2 , v1 v2 , v4 v3

Solution

v2i 1.5 j

Very important!!!

rv1rv2

rv1 rv2

rv4

rv3

rv4

rv3

v1

v2 3i 1.5 j

r

v1

r

v2 (3)2

(1.5)2

32

1.52

3.35

rv4

rv3 (3i 2 j) (i 2 j)

3i 2 j i 2 j 2i 4 j

rv4

rv3 (2)

2 (4)2 22 42 4.47

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Notations

The following are all common notations of a vector:rv,v,v

The following are common notations of the magnitude

(i.e. the length of the arrow):

v,rv

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Vector Components

5

4

-3

i

j

v 4 i 3 j

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Terminology

v 4

i 3

j

Thex-component ofrv is 4, not 4 i .

They-component of

r

v is -3, not -3

j.

Usually written as:

vx 4,vy 3

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Decomposing a vector

v

Given v 7, 40o,

what are thex,ycomponents ofrv?

vx

vy

It means finding vx ,vy !

Hint: Once you know

one side of a right-

angle triangle and

one other angle, you

can find all the

lengths using cos,sin or tan.

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Trigonometry

cosb

h

sin ah

tana

b

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Solution

v 7

Given v 7, 40o,

what are thex,ycomponents ofrv?

vx v cos

7cos40o 5.36

vy v sin

7sin40o

4.50

v 5.36i 4.5j

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CheckOn the other hand, given

rv 5.36i 4.50 j,

you can deduce rv and .

v

vx

vy

v 5.362

4.52

7 (as expected)

tan

opposite

adjacent

vy

vx

4.5

5.36 0.8396

tan1(0.8396) 40o (as expected)

A l f

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Angles of a vector

x

Find the angles the four vectors make with the positivex-axis.

1 30o

2 180o 30o 150o

3 180o 30o 210oor 3 (180

o 30o) 150o

4 360o 30o 330oor 4 30

o

y

v1

v4

v2

v3

30

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Calculating the angles

v ai bj

tan1(b

a) , where

0oif a 0

180 oif a 0

Examples :

rv 2i 3j tan1(32

) 0o 56.3o

rv 2i 3j tan1(

3

2) 180o 123.7o

r

v 2

i 3

j tan1

(

3

2) 180o

236.3o

rv 2i 3j tan1(

3

2) 0o 56.3o( 360o 56.3o 303.7o)

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(-1) times a vector?

5

4

3 i

jv 4

i 3

j

i

j

What is - v?

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v 4i 3

j vv 4i 3

j

Points in the OPPOSITE direction!

What is - v?

5

4

3

v 4

i 3

j5

4

3v

v

4

i

3

j

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In General

If this is v

This is -v

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Adding Vectors Diagrammatically

u

vu v

v

You are allowed to move an arrow around as

long as you do not change its direction andlength.

Method for adding vectors:

1. Move the arrows until

the tail of one arrow is at

the tip of the other

arrow.

2. Trace out the resultant

arrow.

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Addition of vectors

You are allowed to move an arrow around as

long as you do not change its direction.

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Adding in a different order

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Order does not matter

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Subtracting Vectors Diagrammatically

u

v

u

v

v

u v

u v

What about uv?

v

u v u (v)

If we know (rv) we know

ru

rv

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Example

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Example

What isA B C?

A

C

B

D

A B C

AB

C

D

Addi t 1

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Adding vectors 1Add the three vectors to find the

total displacement.

s1 2.6 jrs2 4i

s3

3.1cos(45

o

)

i

3.1sin(45

o

)

j

2.19

i

2.19

j

stotal

s1

s2

s3 (4 2.19)i (2.6 2.19) j 6.19i 4.79j

or more precisely:r

stotal (6.19

i 4.79

j)km

Adding Vectors 2

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Adding Vectors 2

FindA 2

B.

A 2.8cos(60o

)

i 2.8sin(60o

)

j 1.40

i 2.42

jrB 1.9cos(60o)i 1.9sin(60o) j 0.95i 1.65j

2

B 2(0.95i 1.65 j) 1.90i 3.30 j

r

A 2r

B (1.40 1.90)

i (2.42 3.30)

j

0.50i 5.72 j

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Vector Notation of E field

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Vector Notation of E field

r: Unit vector pointing from the charge to the observer.

r 1

q

r

rE

q

40r2r

Charges produce electric field. The closer you are to the

charge, the stronger is the electric field.

Unit: V/m = N/C

0

8.85

10

12

C

2

N

1

m

2

: Permittivity

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Store k in your calculator

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Store kin your calculator

Type:

8.99E9 then STO

then ALPHA then K

then ENTER

If q=2C, r=1.3m, to

find the Efield, type:K*2/1.32

Electric Field (Magnitude)

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Electric Field (Magnitude)The magnitudeof the electric field produced by a single

point charge qis give by:

Eq

40

r2

Dont forget the absolute value!

Magnitude is alwayspositive.

Warnings

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WarningsDo not confuse the displacement vector rwithE.

r(blue arrow) points from the charge to the observer.rE(red arrow) is ALWAYS drawn with its tail at the obeserver,

and can point either away from OR toward the charge.

+Observerr E

r E

rec on o one

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rec on o onecharge)

If q 0,Eis in the SAME direction as r.

If q 0,r

Eis in the OPPOSITE direction as r.

rE

q

40r2r

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What is

r?

Example

q1 1nC

Observer atP.rrP1is the displacement vector from q1toP.

rP1is the corresponding unit vector.

rP1 1i 0.5 jrrP1 1

2 0.52 1.12

rP11i 0.5 j

1.12 0.89i 0.45 j

rP1

rP1rrP1

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Example (Continued)

The electric field vector is given by the red

arrow.

q1 1nCr

E1 q1

40rP12 rP1

109

40(1.12)2

(0.89i 0.45 j)

(6.38i 3.22j)V/m

f f

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The strategy in finding the electric field vector

1. Draw an arrow from the charge to the observer

2. Write down the vector rrand its magnitude rr

3. Calculate r

rrrr

4. CalculaterE q

40r2r

5. If there are more than one charge, repeat for each one

6.rEtotal

rE1

rE2L

rEN

Fi d th it t

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Find the unit vectors

A

B

C

D

E

q1

Find :r

rA1rrA1

rA1

then do the same

for the other point

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Warnings

Do not confuse the displacement vector rwithE.

rr(blue arrow) points from the charge to the observer, and is

used to calculate

r.rE(red arrow) is ALWAYS drawn with its tail at the obeserver.

q1 1nC

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Example - Two Charges

See supplementary notes

-1nC

+1nC

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Example

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Example

4 cm 3 cm

q1 q2P

q1 1 108C, q2 2 10

8C

Find the E field at point P

S l ti Pr1 r2

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Solution

4 cm 3 cm

q1 q2P

q1 1 108C, q2 2 10

8CE1

q1

40r12r1

q1

40r12i

(1 108 )

4(8.85 1012 )(0.04)2

i (56.2

i )kV/ m

E1E2

E2q2

40r22r2

q2

40r22

(i )

(2 10

8 )

4(8.85 1012 )(0.03)2(i ) (199.8i )kV/ m

1 r2

r1i , r2

i

EtotalE1E2 (56.2i 199.8i )kV/ m (256.0i )kV/ m

Example

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Example

x 7-x

q1 q2P

q1 1 108C, q2 2 10

8C

Find the point P such that E = 0.

7 cm

xamp q1 1 108C, q2 2 108C

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xampe

x 7-x

q1 q2P

q1 , q2

q1

40x2

q2

40 (7 x)2

q1

x2

q2

(7 x)2

(7 x)2

x2 q

2

q1 7 x

x q

2

q1 2

7 x 2x

x 71 2

16.9cm(rejected) or 2.9cm

Th diff b t

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The difference between

field vectors and field lines

Field vectors Field lines

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Properties of field lines

Field lines never cross each other

Field lines never terminate in vacuum

Field lines originate from positive charges andterminate at negative charge

Field lines may go off to infinity

The tangent of a field line gives the direction of

the E field at that particular point

Dipole

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Dipole

Field vectors Field lines

Similar to this

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Similar to this

You connects the field vectors tofind the field lines.

Electric Field and Electric Force

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Electric Field and Electric Force

Electric field can be used to calculate the electric forc

F q E

FandEare parallel when qis positive.

FandEare opposite when qis negative.

Example :

A chargeq 2Cin an electric fieldrE (2i 3j)N/Cwill feel a force:

rF (2C)(2i 3j)N/C (4i 6 j)N

Two point charges

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Two point charges

q1 q2r

E21E12

The E field (magnitude) at

point 2 due to charge 1:

E21q1

40r2

The force (magnitude) on

charge 2 due to charge 1:

F21 q2E21 q1q2

40r2

The E field (magnitude) at

point 1 due to charge 2:

E12q2

40r2

The force (magnitude) on

charge 1 due to charge 2:

F12 q1E12 q1q2

40r2

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Another Notation

Fq1q2

40r2

F kq1q2

r2

k1

40 8.99 109Nm2C2

Finding the Electric Force

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Finding the Electric Force

q1

q2

q3

There are two (equivalent) methods of

finding the force on a charge (say, q1).

Method 1 (usingEfield - recommended) :

Find the electric field at point 1 due to the other two chargesrE1

rE12

rE13 (Field at point 1 due to q2 and q3)

rF1 q1

rE1

Method 2 (using Coulomb's Law):

Find the forcevectorson charge 1 due to the other two charges using Coulomb's LawrF1

rF12

rF13 (Forces at point 1 due toq2 and q3)

Note that you must first write the forces as vectors,cannotadd the magnitude instearF1

rF12

rF13

n ng e orce on a

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n ng e orce on acharge

At pointP:rE (9.60i 3.16 j)N/C

P

-1nC

+1nCIf a charge q3 2nCis placed at point P,

the force on it will be given by :rF3 q3

rE (2nC)(9.60i 3.16 j)N/C

(19.20i 6.32 j) 109N