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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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46
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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51
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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