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MUATAN dan
MEDAN LISTRIK
Yohanes Edi Gunanto
Muatan Listrik

Two types of charges exist




They are called positive and negative
Named by Benjamin Franklin
Like charges repel and unlike
charges attract one another
Nature’s basic carrier of positive
charge is the proton

Protons do not move from one material
to another because they are held firmly
in the nucleus
Attractive force
Repulsive force
Asal muatan





Atoms consist of a nucleus
containing positively
charged protons.
The nucleus of an atom is
surrounded by an equal
number of negatively
charged electrons.
The net charge on an atom
is zero.
An atom may gain or lose
electrons, becoming an ion
with a net negative or
positive charge.
Polar molecules have zero
net charge but their charges
are unevenly distributed in
space (e.g. water).




Nuclear diameter ~ 10-15 m
(femtometer)
Atomic diameter ~ 10-9 m
(nanometer)
Classes of Materials
 CONDUCTORS are materials in which
charges may move freely (e.g. copper).
 INSULATORS are materials in which
charges cannot move freely (e.g. glass).
 SEMICONDUCTORS are materials in which
charges may move under some conditions
(e.g. silicon).
Bagaimana muatan diukur ?

elektroskope / elektrometer
Hukum Coulomb
 q1
F12
F21
q2
r12
k | q1 | | q2 |
| F12 |
2
r12
k 1
4 0
For charges in a
VACUUM
2
N
m
k = 8.99  109 2
C
with  0  8.85 10
12
2
C
2
Nm
Quantum of Electric Charge




Electric charge is quantized. The smallest possible unit
is the charge on one electron or one proton:
e= 1.602 x 10-19 Coulombs
No smaller charge has ever been detected in an
experiment.
Catatan kecil:
Ahli Fisika partikel elementer (juga ahli Fisika energi
tinggi, 1963) berteori bahwa ada partikel yang lebih
kecil, disebut quark, yang mempunyai muatan 2/3 e
atau 1/3 e.
Deteksi eksperimen secara langsung pada partikelpartikel ini sulit dimungkinkan karena secara teori tidak
ada quark bebas.
Principle of Superposition

For a system of N charges q1, q2, q3, …,
qN, the resultant force F1 on q1 exerted by
charges q2, q3, …, qN is:
 


F1  F12  F13   F1N
• Each charge may be considered to exert a force
on q1 that is independent of the other charges
present.
Contoh soal :
Hitung besar gaya listrik pada sebuah elektron dalam
atom hidrogen (karena tertarik oleh sebuah proton, Q2 =
e). Jarak rata-rata proton-elektron: 0,53 x 10-10 m
Jawab:
Q1 = - e = - 1,6 x 10-19 C
Q2 = e = 1,6 x 10-19 C
r = 0,53x10-10 m
tanda negatif berarti attraktif.
(Kemana arah gaya? gaya pada elektron
mengarah ke proton)
Hitung gaya (net gaya) yang bekerja pada Q3
dari gambar berikut karena dua muatan yang
lain.
Electric Field
Electric Field Line
Patterns



Point charge
The lines radiate
equally in all
directions
For a positive
source charge, the
lines will radiate
outward
Electric Field Line
Patterns

For a negative
source charge, the
lines will point
inward
Electric Field Line
Patterns


An electric dipole
consists of two
equal and opposite
charges
The high density of
lines between the
charges indicates
the strong electric
field in this region
Electric Field

The ELECTRIC FIELD E is defined in terms of
the electric force that would act on a positive
test charge q0 :

 Fe
E
in N/C
q0
The electric force on a positive test charge q0 at a
distance r from a single charge q:

qq 0
Fe  k 2 rˆ
r
• The electric field at a distance r from a single
charge q:

 Fe
q
E
 k 2 rˆ
q0
r


Fe  q0 E
Electric Field due to
a Group of Charges:

qi
E  k  2 rˆi
i ri
rˆi is a unit vector from qi to q0
Example Problem



Four point charges
are at the corners of a
square of side a as
shown.
Determine the
magnitude and
direction of the electric
field at the location of
q.
What is the resultant
force on q?
2q
a
a
a
3q
q
a
4q
Electric Field ON axis of dipole
-q
+q
a
P
x
Superposition : E  E  E
E 
kq
a

x 
2

2
E  
kq
a

x 
2

2




1
1

E  kq 

2
2

a 
a 
 x    x   
2 
2 

E  kq
2 xa
2 2
 2 a
x  


4



 kq
2kpx
2 2
 2 a
x  


4


2 xa
2 2
 2 a
x  


4


p = qa
“dipole moment”
-- VECTOR
-
+
What if x>> a? (i.e. very far away)
2kpx 2kp
E 4  3
x
x

E

p
r
3
E~p/r3 is actually true for ANY point far from a dipole
(not just on axis)
Electric Dipole in a Uniform Field



Net force on dipole = 0;
center of mass stays
where it is.
Net TORQUE t: INTO
page. Dipole rotates to
line up in direction of E.
| t | = 2(QE)(d/2)(sin q)
= (Qd)(E)sinq
 |p| E sinq

= |p x E|
The dipole tends to
“align” itself with the
field lines.
Distance between charges = d
Electric Field from Continuous Charge
Distributions

If a total charge Q is distributed continuously, it
may be sub-divided into elemental charges dQ,
each producing an electric field dE:
dQ
dE  k 2
r
dQ
E   dE  k  2
r

dQ
1
dQ
E  k  2 rˆ 
rˆ
2

r
4 0 r
1
k
4πε0
ε0 = permittivity of free space
Uniform Charge Distributions



Volume Charge Density: ρ=Q/V
Surface Charge Density: σ=Q/A
Linear Charge Density:
λ=Q/l
l  Q/L
s  Q/A
r  Q/V
Example: Field on Bisector of
Charged Rod





Uniform line of charge +Q
spread over length L
What is the direction of
the electric field at a point
P on the perpendicular
bisector?
(a) Field is 0.
(b) Along +y
(c) Along +x
• Choose symmetrically
located elements of length dx
• x components of E cancel
P
y
a
x
dx
q
o
L
dx
Line Of Charge: Field on bisector
Distance
dE
d  a x
2
2
q
Charge per unit length l 
L
P
a
dq
k ( dq)
dE 
2
d
dx
Q
x o
L
k (l dx)a
dE y  dE cosq  2
(a  x 2 )3 / 2
a
cosq  2
2 1/ 2
(a  x )
L/2
L/2
dx


x
E y  kl a 
 kl a 

2
2 3/ 2
2
2
2
(a  x )
 a x  a  L / 2
L / 2

2klL
a 4a  L
2
2
What is E very far away from the line (L<<a)?
What is E if the line is infinitely long (L >> a)?
2klL
2kl
Ey 

2
a
a L

A rod of length l has a uniform positive charge per unit length
λ and a total charge Q. Calculate the electric field at a point
P that is located along the long axis of the rod and a
distance a from one end.
Start with
dq  ldx
1 dq
1 ldx
dE 

2
2
4 0 x
4 0 x

Then :
l a
E

a
l dx
l

2
4 0 x
4 0
l a

a
dx
l

2
x
4 0
l a
 1
 x 
 a
1 Q1
1 
Q
E
 

4 0 l  a l  a  4 0 a(l  a)

Finalize
 l => 0 ?
 a >> l ?
TERIMA KASIH