Transcript Electric fields

Medan Listrik (Electric fields)
• Hukum Coulomb  bagaimana muatan listrik
berinteraksi dengan muatan listrik yang lain.
Tetapi bagaimana muatan listrik tersebut ‘tahu’ ada
kehadiran muatan listrik lain dan juga bagaimana
mampu mendeteksi muatan tersebut negatif atau positip?
• Maka muncul konsep medan listrik
Secara ilustratif
• Satuan medan listrik (N/C)
• Arah gaya listrik mendefinisikan
juga arah medan listrik
Garis-garis gaya Listrik (Electric field lines):
• Michael Faraday (abad 19) : memperkenalkan ide
medan listrik dan dipikirkan sebagai ruang di
sekitar benda bermuatan listrik dipenuhi oleh garisgaris gaya
• Garis-garis gaya sebagai cara memvisualisasikan
adanya medan listrik
• Arah garis-garis gaya menunjukkan arah gaya
listrik yang bekerja pada tes muatan positip
• Kerapatan garis gaya sebanding dengan
besarnya medan listrik
Kekuatan medan terkait dengan
jumlah garis-garis yang melintasi
satuan luas yang tegak lurus
medan
Karena gaya listrik
adalah
Kekuatan
(besarnya)
medan listrik:
Examples:
Two positive point charges, q1=+16 μC and
q2=+4.0 μC, are separated in a vacuum by a
distance of 3.0 m, as Figure illustrates. Find
the spot on the line between the charges
where the net electric field is zero.
SUPERPOSISI MEDAN LISTRIK
Example 5. Vector properties of Electric Fields
Two point charges are lying on the y axis in
Figure a: q1=–4.00 μ C and q2=+4.00 μ C. They
are equidistant from the point P, which lies on
the x axis. (a) What is the net electric field
at P?
MEDAN LISTRIK dari distribusi muatan listrik
Electric dipole電偶極
• In nature, many molecules carry no net
charge, but there are still finite electric fields
around the molecules.
• it is because the charge are not uniformly
distributed, the simplest form of charge
distribution is a electric dipole, as shown in
the figure.
• one can calculate the electric field along the
z-axis as follows:
Rewrite the electric field in the following form:
Because we are usually interested in the case where z>>d,
therefore we can use an approximation to simplify E:
Throw away higher order terms of d/z,
Or with p=qd (electric
dipole moment),
Note
• the dipole electric field reduces as 1/r3, instead of 1/r of a single
charge.
• although we only calculate the fields along z-axis, it turns out that
this also applies to all direction.
• p is the basic property of an electric dipole, but not q or d. Only
the product qd is important.
E
p
E
Electric field due to a line charge
• we now consider charges uniformly
distributed on a ring, rather than just a few
charges.
• again we use the superposition principle
of electric fields, just as what we have done
on electric dipoles.
• assume the ring has a linear charge
density λ.
• then for a small line element ds, the
charge is
So it produces a field at point P of:
Or rewrite as
• Here we only consider the field along z-axis,
dEcosθ, because any fields perpendicular to
z direction will be cancelled out at the end.
• since
Sum over all fields produced by other elements on the
ring:
An infinite long line charge
dE 
r
dl
40 r 2
R
cos 
and l  R tan 
dl  R sec 2 d
 d
dE 
40 R
For an infinite line,






E
cos

d


sin


40 R 
40 R

sin  2  sin 1 

40 R
2
1
2
1  2   / 2
1
E
2
40 R
A point charge in an electric field
By the definition of electric field, a point charge
will experience a force equal to:
The motion of a point charge can now be
described by Newton’s law.
Robert A. Millikan, in 1909, made use of this
equation to discover that charge is quantized and
he was even able to find the value of fundamental
charge e=1.6x10-19C.
The electric field between the
plates are adjusted so that the oil
drop doesn’t move: qE  mg
q  mg / E
He found that q=ne, always a multiple of a
fundamental charge e.
F  QE  ma
QE
a
m
1 2 1  QE  L 
y  at  
 
2
2  m  v x 
 0.66mm
2
Dipole in an electric field
• The response of an
electric dipole in an electric
field is different from a
charge in an electric field.
• Since a dipole has no net
charge, so the net force
acting on it must be zero.
• however, each charge in
the dipole does experience
forces from the field.
• the forces are equal in
magnitude but opposite in
direction, and so there is
net torque
  F  d  Fd sin 
  qEd sin 
 pE sin 
p is the dipole moment 偶極
矩
Or in vector form
  p E
Therefore, a dipole in an uniform electric field experiences a
torque that is proportional to the dipole moment, and does not
depends on any detail of the dipole.
  p E
Under the effect of the field, the dipole will
now oscillates back and forth, just like a
pendulum under the effect of gravitation field.
Potential energy
The motion of the dipole of cause
requires energy. The work done on
dipole by the torque is
W   d
Similar to gravitation potential, we can define a electric potential energy of
the dipole in an electric field :
In the vector form, the potential energy is simply a dot product of the dipole
moment and the field.
The potential energy is the lowest when the
angle = 0 (equilibrium position) and has the
largest values when the angle = 180°.
If there is energy loss to the surroundings,
the oscillation will die out gradually unless
energy is pumped in continuously from the
field (an oscillating field).
This is essentially the principle of a
microwave oven.
• Water molecule has large dipole moment of 6.2x10-30Cm.
• the dipoles vibrate in response to the field and generate thermal energy in
the surrounding medium
• materials such as paper and glass, which has no dipoles, do not become
warm
• The large dipole moment of
water molecule attracts Na and Cl
ions and breaks the ionic bond
between these ions.
• Therefore, salt dissolves easily
in water.
Note: A dipole in an uniform field experiences no net force, but it does in
a non-uniform field.
Induced
dipole
F+>F-
Charges in the
comb produce a
non-uniform field
E
2e
40 (a 2 / 2)
  p E
(a) τ=0
(b) τ=pE
(c) τ=0
the field produced by the rings is
E
QD
Q' D

40 ( D 2  R 2 )3 / 2 40 ( D 2  R'2 )3 / 2
0
 D  R'
Q '   2
2
D

R

2
 13 
  
5
2
3/ 2
Q



3/ 2
Q
(a)
4
4
mg   r 3 g  (1000)  (0.6 10 6 ) 3 (9.8)  8.87 10 15
3
3
qE  mg
(b)
q  8.87 10 15 / 462  1.9 10 17
 120e
U f  U i   p f  E  pi  E
 pE  cos(90   f )  cos(90   i ) 
電四極