Kuliah Medan_listrik Statis

8/18/2019 Kuliah Medan_listrik Statis

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MEDAN LISTRIK


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A Brief History

of Electricity


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Ancient Greeks – Static Electricity

Rub amber with wool.

Amber becomes negatively charged by

attracting negative charges (electrons) from

the wool.

The wool becomes positively charged.

The amber can then pick up a feather.

How?


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William Gilbert (1544-1603)

Coined the word “electricity ” from the Greek

word elektron meaning amber.

English scientist and physician to QueenElizabeth.

In 1600 published "De Magnete, Magneticisque

Corporibus, et de Magno Magnete Tellure" ("On

the Magnet, Magnetic Bodies, and the Great

Magnet of the Earth").

Showed that frictional (static) electricity occurs in many common

materials.


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Pieter van Musschenbroek (1692 – 1761)

Dutch physicist from Leiden, Netherlands, whodiscovered capacitance and invented the Leyden jar .

Ref: http://chem.ch.huji.ac.il/~eugeniik/history/musschenbroek.htm

Leyden jar (also called condenser )

http://chem.ch.huji.ac.il/~eugeniik/history/musschenbroek.htmhttp://chem.ch.huji.ac.il/~eugeniik/history/musschenbroek.htm


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Leyden Jars

http://www.alaska.net/~natnkell/leyden.htm

http://home.earthlink.net/~lenyr/stat-gen.htm

Refs:

700 pF, 175 KV

Q = C x V

= 700 x 10-12 x 175 x 103

= 1.225 x 10-4 coulombs

No. of electrons =

1.225 x 10-4 coulombs / 1.6 x 10-19 coul/elec= 7.66 x 1014 electrons

http://www.alaska.net/~natnkell/leyden.htmhttp://home.earthlink.net/~lenyr/stat-gen.htmhttp://home.earthlink.net/~lenyr/stat-gen.htmhttp://home.earthlink.net/~lenyr/stat-gen.htmhttp://home.earthlink.net/~lenyr/stat-gen.htmhttp://www.alaska.net/~natnkell/leyden.htm


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Benjamin Franklin (1706 – 1790)

Conducted many experiments on staticelectricity from 1746 – 1751 (including his

lightning experiment) and became famous

throughout Europe by describing these

experiments in a series of letters to Peter

Collinson.


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Charles Coulomb (1736 – 1806)

Using a torsion balance Coulomb in 1784experimentally determined the law according to

which charged bodies attract or repel each other.

Coulomb’s Law

1 21 122

0 12

1

4

q q

r F e

7 2 9

0

110 9.0 10

4c

Unit: Newton meter 2 / coulomb2

volt meter / coulomb


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Alessandro Volta (1745 – 1827)

Interpreted Galvani’s experiment with

decapitated frogs as involving thegeneration of current flowing through

the moist flesh of the frog’s leg between

two dissimilar metals.

Argued with Galvani that the frog was

unnecessary.

In 1799 he developed the first battery (voltaic pile) that

generated current from the chemical reaction of zinc andcopper discs separated from each other with cardboard discs

soaked in a salt solution.

The energy in joules required to move a charge of one coulomb

through an element is 1 volt .


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Hans Christian Oersted (1777 – 1851)

Ref: http://chem.ch.huji.ac.il/~eugeniik/history/oersted.htm

1822

In 1820 he showed that a current produces amagnetic field.

X

http://chem.ch.huji.ac.il/~eugeniik/history/oersted.htmhttp://chem.ch.huji.ac.il/~eugeniik/history/oersted.htm


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André-Marie Ampère (1775 – 1836)

French mathematics professor who only a weekafter learning of Oersted’s discoveries in Sept.

1820 demonstrated that parallel wires carrying

currents attract and repel each other.

attract

repel

A moving charge of 1 coulomb per

second is a current of

1 ampere (amp).


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Michael Faraday (1791 – 1867)Self-taught English chemist and physicist

discovered electromagnetic induction in 1831 by

which a changing magnetic field induces an electric

field.

Faraday’s electromagnetic

induction ring

A capacitance of 1 coulomb per volt

is called a farad (F)


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Joseph Henry (1797 – 1878) American scientist, Princeton University

professor, and first Secretary of the SmithsonianInstitution.

Discovered self-

induction

Built the largest

electromagnets of his

day

Unit of inductance, L, is the “Henry”


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James Clerk Maxwell (1831 – 1879)Born in Edinburgh, Scotland;

Taught at King’s College in London (1860-1865) and was the first Cavendish Professor

of Physics at Cambridge (1871-1879).

Provided a mathematical description of

Faraday’s lines of force.

Developed “Maxwell’s Equations” which describe

the interaction of electric and magnetic fields.

D

0 B

t

BE

t

DH J

Predicted that light was a form

of electromagnetic waves

D E

B H


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Albert Einstein (1879 – 1955)

In 1905 publishes his Special Theory of Relativitybased on two postulates:

1. Absolute uniform motion cannot be detected by any

means.

2. Light is propagated in empty space with a velocityc which is independent of the motion of the source.

This theory predicts seemingly unusual effects such as the measured lengthof moving bodies and time intervals being dependent on the frame of

reference being used for the measurement.


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Some Electrical Inventors

• Samuel F. B. Morse (Telegraph)

• Guglielmo Marconi (Wireless telegraph)

• Thomas Edison (Electric lights …..) • Nikola Tesla (A.C. generators, motors)

• John Bardeen and Walter Brattain

– Transistor

• Jack Kilby and Robert Noyce

– Integrated Circuit


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The TelegraphSamuel F. B. Morse

(1791 – 1872)


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Wireless Telegraph

Guglielmo Marconi Marconi Spark TransmitterBuilt at the Hall Street Chelmsford Factory

September, 1897


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Electric LightsThomas Edison

1847 - 1931

Replica of original lightbulb

Patent #223,898

Invented and developed

complete DC electric

generation and distribution

system for city lighting

systems

Carried on a major competition

with George Westinghouse who

developed an AC generation and

distribution system


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Alternating Current (AC) SystemsNikola Tesla

1856 - 1943Over 700 patents

Rotating magnetic field principle

Polyphase alternating-current system

Inducton motor

AC power transmission

Telephone repeaterTesla coil transfromer

Radio

Fluorescent lights

Bell Labs Museum


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The First

Point-ContactTransistor

1947

Bell Labs Museum


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The FirstJunction Transistor

1951

Bell Labs

Lab model

M1752

Outside the Lab


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Texas Instrument’s First IC -- 1958

Jack Kilby

Robert Noyce

Fairchild

Intel


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Properties of Electric Charges• There are 2 kinds of electric charge: positive (+) and

negative ( –)

– Carrier of positive charge in matter is the proton (charge = +e)

– Carrier of negative charge in matter is the electron (charge = – e)

– e = 1.602 10 – 19

C (typical “shock” experienced on a dryday transfers about 1 10 – 9 C)

– Charge is quantized (only comes in integer multiples of e)

• An object becomes electrically charged through

transfer of negative charge (movement of electrons) – Protons don’t move because they are tightly bound to

atomic nuclei

– Charge is conserved

– Neutral objects have equal amounts of + and – charge


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Properties of Electric Charges• Rubbing a rubber rod with wool transfers negative

charge to rod

– Wool has excess positive charge due to loss of negative

charge

• Rubbing a glass rod with silk transfers negative

charge to silk – Glass rod has excess positive charge

• Experiments show that:

– Negatively charged rubber rod is attracted to positively

charged glass rod

– Negatively charged rubber rod is repelled by another

negatively charged rubber rod

• Opposite charges attract, like charges repel

C f


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Conducting Properties of Materials• Insulators are materials in which electric charge

does not move easily

– They can be charged, but charge doesn’t move well – Glass, rubber, plastic, wood, and paper are examples

• Conductors are materials in which electric chargemoves easily – When an area becomes charged, charge distributes itself

over entire surface

– Copper, aluminum, and silver are examples

– Charge will remain on conductor if you hold it with an

insulator• Semiconductors are materials that have electrical

properties somewhere between conductors andinsulators

– Silicon and germanium are examples

M th d f Ch i /Di h i


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Methods of Charging/Discharging• Charging by rubbing

– Increases surface area of contact and enhances charge

transfer – Works for insulators but not for conductors

• Charging by conduction – Charged object brought in contact with a neutral object

– Neutral object becomes charged with same sign of chargeas object doing the charging

– Works when (originally) neutral object is insulated

• Discharging by grounding

– Negative charge leaves (or enters) object throughconducting path to Earth or other limitless reservoir ofcharge

– Third opening of electrical outlets is the ground (connectedto ground by wire and prevents static charge from building)

M th d f Ch i


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Methods of Charging• Charging by induction (no contact)

– Repulsive force between like charges in

charged rod and (insulated) neutral

conducting sphere causes redistribution of

charge on sphere (figure (b))

– Opposite (like) charges move closer to

(farther from) each other

– Rod would attract sphere

– Induced charge on sphere can remain

if some electrons leave through grounding

– + charge becomes equally distributed

because of high mobility of remaining

electrons

• In insulators, induced surface charges can occur

due to polarization (alignment of molecular charge)


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Application: Photocopiers

(from College Physics,Giambattista et al .)


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Hukum Coulomb


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• Gaya yg dilakukan oleh satu muatan titikpada muatan titik lainnya bekerjasepanjang garis yang menghubungkankedua muatan tersebut. Besarnya gayaberbanding terbalik kuadrat jarak

keduanya, berbanding lurus dgnperkalian kedua muatan.

• Gaya tolak menolak muatan sama

• Gaya tarik menarik muatan beda

• Dengan k = 8,99 109 N.m2/C2

Hukum Coulomb

122

12

2112 r ˆF

r qkq

12

10854,8

xo


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•

Dua muatan titik masing-masing sebesar 0,05 Cdipisahkan pada jarak 10 cm. Carilah (a) besarnyagaya yang dilakukan oleh satu muatan pada muatanlainnya dan (b) Jumlah satuan muatan dasar pada

masing-masing muatan.

Contoh Soal


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0, 05C

0, 05C

10 cm

q1 q2 F21 F12

N

m

C xC xC m N

3-

2

66229

2

21

10x2,25

)1,0(

1005,01005,0/.8,99x10

r

qkq F

11

19

6

1012,3106,1

1005,0 x

C x

C x

e

q N

Neq

Solusi Soal no.1


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2 m 1,5 m

q1 = 25nC q2 = -10nC q0 = 20nCF20

F10

i N)(0,367

i)5,3(

)1020)(1025)(/.1099,8(

ˆ

2

99229

102

10

0110

m

C m N

r r

qkq F

N)i0,799(-

i)5,1(

)1020)(1010)(/.10(8,99

ˆ

2

99229

202

20

0220

m

C C C m N

r r

qkq F

N)i(-0,432i)799,0(i)367,0(2010 N N F F F total

Solusi Soal no.2

Soal


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q1 q

2

q3

Soal

Tiga muatan titik terletak pada sumbu x; q1 = 10 μCterletak pada titik asal, q2 = +4 μC berada pada x=80

cm, dan q0 = 20 μC berada pada x = 60 cm sepertiterlihat pada gambar di bawah. Carilah gaya total padaq0 akibat q1 dan q2.


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N ,m

C Nm F

N m

C C Nm F

81)1(

)1020)(1010)(/109(

2)6,0(

)1020)(104)(/109(

2

66229

13

2

66229

23

N N F

N N F

o

y

o

x

1,137sin)8,1(

4,137cos)8,1(

13

13

o

y x

dan

N F

N N F N F

664,1

1,3arctan

4,31,34,1

1,10,2dan4,1

22

Solusi Soal


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•

Untuk menghindari kesalahan yang mungkinterjadi dalam konsep gaya maka diperkenalkanlah

konsep medan listrik. Dimana:

)( 0 kecil qq

F E

o

Medan Listrik


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• Hukum Coulomb untuk E

akibat satu muatan titik.

• Hukum Coulomb untuk E

akibat suatu sistem

muatan titik.

02

0

î

i

ii r

r

kq E

i

i

i

ii r

r

kq E E 02

0

ˆ

Contoh Soal


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• Sebuah muatan positif q1=+8nC berada pada titik asal dan muatan

kedua positif q2=+12nC berada pada sumbu x = 4m dari titik asal.

Carilah medan lisriknya di sumbu x untuk:

– P1 yang berjarak x=7m dari titik asal.

– P2 yang berjarak x=3m dari titik asal.

– Mungkinkah kita menemukan suatu daerah bebas medan listrik

yang diakibatkan oleh kedua muatan listrik tersebut

Contoh Soal

+ +P1 P2

4 m

7 m

q1=8nC

q2=12nC


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3 m

+ + P1 P2

4 m

7 m

q1=8nC

q2=12nC

)P(di )/5,13()/0,12()/47,1(

3

1012/1099,8

7

108/1099,8

1

2

9229

2

9229

22

2

21

1

iC N iC N iC N

im

C C Nmi

m

C C Nm

i x

kq

i x

kq

E

)P(di )/100()/108()/99,7(

)(1

1012/1099,8

3

108/1099,8

2

2

9229

2

9229

22

2

2

1

1

iC N iC N iC N

im

C C Nmi

m

C C Nm

i x

kqi x

kq E

Solusi soal

Latihan Soal!


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• Hitunglah nilai E di P3 !

• Berapa besar sudut yang diciptakan resultan E di P3 terhadap sumbux positif.

+ +q1=8nC

q2=12nC

4 m

3 m

E di P3 ?

Latihan Soal!

1. Dari gambar tersebut di atas

2 Hit l h ( ) d li t ik E di d d j k 30 d i b h t


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2. Hitunglah (a) medan listrik E di udara pada jarak 30 cm dari sebuah muatantitik q1 = 5x10

-9C, (b) gaya pada suatu muatan q2 4x10-10C yang ditempatkan

30 cm dari q1, dan (c) gaya pada muatan q3 = -4x10-10C yang ditempatkan 30

cm dari q1 (dimana q2 tidak ada).

3. Tiga muatan ditempatkan pada tiga sudut sebuah bujur sangkar seperti padagambar. Setiap sisi bujursangkar adalah 30 cm. Hitunglah E pada sudut keempat!

Berapakah gaya yang diberikan oleh muatan 6μC pada sudut yang kosongtersebut?

4. Terdapat dua buah bola kecil bermuatan, q1 = +20x10-8C dan q2 = -5x10

-8C.Tentukan (a) medan listrik E pada titik P, (b) gaya pada muatan -4x10-8C yangditempatkan pada P, dan (c) posisi dimana medan listrik nol (jika tidak adamuatan -4x10-8C).

q1 q2 P5 cm 5 cm

-5μC+8μC

-4μC

G i i d li t ik


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• Garis medan listrik bermula dari muatan + dan berakhir pada muatan – • Garis2 digambar simetris, meninggalkan atau masuk ke muatan

• Jumlah garis yang masuk/meninggalkan muatan sebanding dgn besar

muatan

• Kerapatan garis2 pada sebuah titik sebanding dgn besar medan listrik di

titik itu

• Tidak ada garis2 yang berpotongan

Garis-garis medan listrik


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• Muatan titik dalam medan listrik akan mengalami gaya qE.

• Sehingga percepatan partikel dalam medan listrik memenuhi:

• Didapatkan dari: Fmekanik = Flistrik

E m

qa

Gerak Muatan Titik di

Dalam Medan Listrik


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• Sebuah elektron ditembakkan memasuki medan

listrik homogen E = (1000 N/C)i dengan

kecepatan awal Vo=(2 x 106 m/s)i pada arah

medan listrik.Berapa jauh elektron akan bergerak sebelum

berhenti?

Soal 6

Di l Li t ik


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• Dipol listrik terjadi jika dua muatanberbeda tanda dipisahkan olehsuatu jarak kecil L.

• Suatu dipol listrik ditandai oleh

momen dipol listrik p, yangmerupakan sebuah vektor yangmempunyai arah dari muatannegatif ke positif.

• p=qL, untuk gambar kartesiandiatas maka p=2aqi

-q +q

L

p=qL +-

Dipol Listrik

C h Di l li ik


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Contoh Dipol listrik :

Sebuah muatan positif dan sebuah muatan negatifmempunyai besar q yang sama TERPISAH SEJAUH 2a. Berapakah

besar medan yang ditimbulkan oleh KEDUA muatan inipada titik P, sejarak r sepanjang garis pembagi tegaklurus dari garis yang menghubungkan muatan MUATANtsb ? Anggap r >> a.

1 E

+q

-q

a

ar

θ

θ

θ

P

2 E

E

E

2 4 Medan Listrik oleh Distribusi Muatan Kontinu


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2.4 Medan Listrik oleh Distribusi Muatan Kontinu

Jika distribusi muatan tersebut adalah kontinu, makamedan yang ditimbulkannya di setiap titik P dapat

dihitung dengan membagi elemen2 yang sangat kecil dq.Medan yang ditimbulkan oleh setiap elemen akandihitung, dengan memperlakukan elemen2 tsb sebagai

muatan titik. diberikan oleh

Dimana r adalah jarak dari elemen muatan dq ke titik P.medan resultan kemudian dicari dari prinsip superposisi

dengan menjumlahkan kontribusi2 medan yangditimbulkan oleh semua elemen muatan, atau

r r dqr E d ˆ

41)( 2

0

)()( r E d r E

)(r E d

)(r E d


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M d k b t b t


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Medan karena batang bermuatanLihat gambar berikut:Sebuah muatan listrik dengan rapat muatan terletak pada batang dengan

panjang L hitunglah besarnya medan listrik di titik P sejauh b dari salah satuujung batang bermuatan


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Bila panjang batang tak

berhingga maka

y E

r

dq E E

dE E

y

y

0

2

0

2

cos24

1

cos


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