TUTORIAL 6
Review of Electrostatic
Outline
★
★
★
★
★
★
★
★
Some mathematics
Coulomb’s Law
Gauss’’s Law
Postulations for electrostatics
Electric potential
Poisson’s equation
Boundary conditions
Capacitance
September 8, 2004
8.022 – Lecture 1
Gradient
Some mathematics
!
Let’s define the infinitesimal displacement dl " dxxˆ # d
★
Del operator
★
★
The “del” operator
An operator works as
Gradient of a scalar
& !f
!f
!f
!f
Definition:
df "
dx #
dy #
dz " )
!x
! y ! !! z#
#+ ! x
a vector.
$%'
xˆ &
#y
) #x
!f
,
!# y
!f '
,
* ( $ dx , dy , dz %
"! z !, # # # "
Divergence
yˆ &
zˆ %
, ,
( '
(
#z * ) #x #y #z *
!
Definition of Gradient:
Given a vector function v ( x, y, z )
Properties:
& !f !f !f '
!a fvector !f
!f
It
looks
like
!
How fast a function varies when
its
grad f . -f v.( x, y, zxˆ)#% v xˆ y&ˆ #
,v , v, ) *
zˆ .
)
ˆ
ˆ
(
,
&
%
v
y
v
z
v
!xy y !z z + !xx !y y z !z ,
components vary, and the direction is It works like!axvectorCurl
But it’s not a real vector because it’s meaningless by itself
the one of maximum rate of increase.we
It’s define
an operator.
its divergence as:
★
!
!
!
★
Divergence of a vector
How much a vector
a point.
★
★
Curl of a vector
How much a vector
around a point.
★
Conclusions:
How it works:
!
v ( x, y!, z )
Given a! vector
function
#functions:
v y x,
#vx when
#vyz and z va
! f(x,y,z)
! varies
how
fast
-f Itmeasures
can act on
both
scalar
and
vector
v
v
div
%
$
+
%
&
&
!
!
! Logical
of
the
function:
gradient
#ofx derivative!
#$yf (vector)
#z
ˆ lar
ˆconcept
( x, y!, zActing
)extension
" vx on
xˆ !avsca
y y ! vz z " (vx , v y , vz )
spreads!varound
Acting on
a vectorbut
function
product: divergence
f is !a scalar
is a dot
vector!
function
-f with
! !
Acting on a scalar function with cross product: curl $ , f
Observations:
we define
its 8,curl
September
2004as:
8.022
x̂ – Lecture
ŷ 1 ẑ
!
The divergence is a scalar
September 8, 2004
8.022 – Lecture 1
! interpretat
!
Geometrical
# ion: it#measures
# how much t
!
$around
% v "a point”.
“spreads
#x
#y
#z
circulates
September 8, 2004
– Lecture
vx
v8.022
vz 1
y
Observations:
!
!
!
The curl is a vector
Geometrical interpretation: it measures how much the function
!
v
Brief Review
Brief
Review
1. What
is field ? Scalar field and vector field
Brief Review
Some mathematics
1.2.
1.
2.
2.3.
★
★
Definitions
physical
grad
What
is fieldand
? Scalar
fieldmeanings
and vectoroffield
What
is field ?and
Scalar
field
and
vectorfield
field
divergence
curl
of
a
______
Definitions and physical meanings of gradie
Definitions
and
physical
of gradie
Divergence
Theorem:
divergence
and
curl of a meanings
______ field
Divergence Theorem
curl of a ______ field
3. divergence
Divergenceand
Theorem:
!
!
A
dv
#
A
!
d
s
Divergence
Theorem:
v
s
★
The surface integral of the curl of3.
a vector
field
!! Adv # A !d s
over an open surface is equal to the closed
line
integral
v
s
4. Stoke’s Theorem:
!
!
A
dv
#
A !d s
of the vector along the contour bounding the surface.
v
s
4. Stoke’s Theorem:
(! " A)!d s # A !d l
4. Stoke’s Theorem:s
Stoke’s Theorem
c
$
"$
"$
"$
$
$
$
"$
(! " A)!d s # "$ A !d l
$
5. Two Null Identities:
(! " A)!d s # "$ A !d l
The surface integral of the curl of a vector field
$
s
c
5. Two
Identities:
over an open surface is equal to the closed
line Null
integral
s
★
c
" !!V " # #$$$$$$$!!!! " A " # #
of the vector along the contour bounding
the surface.
5. Two
Null!Identities:
Two Null identities
★
! " !!V " # #$$$$$$$!!!! " A " # #
! " !!V " # #$$$$$$$!!!! " A " # #
A curl-free field can be expressed as a gradient of a
scalar field.
★
A divergenceless field can be expressed as a curl of
Postulates of
a vector field.
★
Electrostatics in Fre
Postulates of Electrostatics in Free
0*+(1),-$'.)/2
%&''()(*+&,-$'.)/ :
Postulates
of
Electrostatics
in
Free
0*+(1),-$'.)/2
%&''()(*+&,-$'.)/ : %
Static Electric field
★
Field
★
★
Static Field
★
★
Space distribution of a vector or scalar parameter.
Field do not change with time.
Static Electric field
Electric charges are at rest, and electric field do not change with
time.
★
★
What’s involved?
★
Charge
★
Electrical field intensity
★
Electrical potential
Coulomb’s law
Coulomb’s law
★
Force between two charges
!
q1q 2
ˆ
F2 ! k
r
21
2
Coulomb’s Law
| r2 1 |
• If set q the point charge in free space, and draw a
!
★
Where:
! spherical surface s in postulate (1):
ElectricFfield
due
to one
point
charge
is theintensity
force that the
charge
q feels
due to
q
!
!
2
q
rˆ2 1 is the unit $vector
s # from q1 to q2
"% E!dgoing
2
1
The Electric field
of a positive charge is in the outward
s intensity
!
!
! Consequences:
!
radical
direction and has! a magnitude
proportional
to the charge and
q
"
F
!
"
F
Newton’s third$law:
" Rthe
(a
)
!
a
#1 ER ( %of
) =distance from the charge
E
ds
2
R to
R the
R square
"
%
inversely
proportional
s
!!
Like signs repel, opposite signs attract
★
!
!
September 8, 2004
8.022 –qLecture 1
E # a R ER # a R
%"! ! R
"
#
q &R p $ R '(
%"! ! R p $ R '
25
#
%"!above
outward
radialofdirection
andpoint
has acharge
magnitude
R p electric
$ R ' thefield
%"!
!R
The
tell
us! the
intensity
a positive
is in
proportional
to the square
thecha
dis
the outward radial direction and
andinversely
has a magnitude
proportional
toofthe
andfield
inversely
proportional
to the
square
of the
The above tell us the electric
intensity
of a positive
point
charge
is indistance from the charge
Coulomb’s law
the outward radial direction and has a magnitude proportional to the charge
and inversely proportional to the square of the distance from the charge
!
FAQ: why do we use cgs?
★
!
A system of discrete charges
Honest answer: because Purcell does…
September 8, 2004
8.022 – Lecture 1
5
A system of continuous
charges
Gauss’s
Law
27
★
The force on charge Q
to calculate E:
★ Ways
Integration
Gauss’s
Law
due to all the other charges is
the vector sum ofWays
the force
to calculate E: E # ) % a R &v" dv ' $$$ E # ) % a R & s" ds '
Gauss’s
Law
%"! ! v ' R
%"! ! s ' R
created by
the individual The superposition
principle:
&v
&s
&l
)
)
charges.
Ways
to calculate E: E # )
a
dv ' $$$ EWe
# can also
a useds
a
' $$$$$ E #law to calculat
Gauss’s
★
%
R
R
continuous
distribution
of
charges
"
"
%
%
s
L
'
'
R
R
R"
%"!
%"!
1
qk (R − R )"! !
!
!
E=
∑
&
&charges
&l
)
)
)
Q
The E
superposition
v principle: discrete
s
4
πε
#
a
dv
E
#
a
ds
E
#
a
dl
'
$$$
'
$$$$$
'
$
R
−
R
0
Eis!dcontinuous?
s # $$
What
when thelaw
distribution
R
Rhappens
R of "charges
"k=1
" Gauss’s
%
%
%
We
can
also
use
to
calculate
E:
"
%
v
s
L
'
'
'
s
"! !& ! integral:
R
R
%"! !
%"! !Take theRlimit for q !dq%and
!!
'%
k
' 3
k
n
v'
R
i
q1
qN
Qasserts that the total outward flux of
law
2
We can qalso
use Gauss’s law to calculate E:Gauss’s
E!d s # $$
€
"% surfaceq in free
! space is equal to the total charge e
s
Q
q3
Q
q4
i
divided by ! !!
r
E! d s #
$$
law asserts
that the total outward flux of the E-field over any clo
"%Gauss’s
!
Q
s
! space is equal to the totalVcharge enclosed in the surface
surface
q5 in free
divided by ! !
lawcharge
asserts
that
total
outward
of the
E-field over any close
TheGauss’s
force on the
Q due
to the
all the
other
charges flux
is equal
! to i= N q i Q
dq Q
! dV Q
in of
free
is equal
total charge
in
the
surface
ˆ
ˆ
r
r
rˆ
Fenclosed
"
#
"
thesurface
vector sum
thespace
forces created
by to
thethe
individual
charges:
$ 2 i %
Q
2
2
%
Gauss’s
law
! through a generic surface
Simple example:
!E of charge
at center of sphere
E
Gauss’s Law
★
What if the surface is not spherical S?
Ways
Impossible integral?
Use intuition and interpretation of flux!
+Q
!
Version 1:
Surface Integration
!
★
Gauss’s law
!
Consider the sphere S1
Field lines are always continuous
E#
to calculate E:S
)
%"! !
%
v'
aR
&v
R
+Q
dv ' $$$S1 E #
"
" !S1=!S=4"Q
We can also use Gauss’s
Problem: ! Version 2:
1.10
or next+Q
lecture
!
Calculate !! E Purcell
for point
charge
at the center of a sphere of radius R
)
%"! !
%
s'
aR
&
R
law to calc
Gauss’s law asserts that the total outward flux of the
Q
Conclusion:
Solution:
E-field over
any
close
surface
in
free
space
is
equal
to
E!d sQ # $$
! electric flux ! through any closed surface S containing a net"
!
%
The
charge
s
!!
! .E ##" d A
& ' & ()* + & (&
- ".+ & "/0
+ & (& " divided by ε0
the total charge
enclosed
in",the
surface
is proportional to the charge enclosed:
★
★
!
! ! Q!
.1 , 2- ."3 + 4 (5 & "4 ."6 2/ .4 - 3!
& "7#8 " E E
# " dA
r$" " Q
!" # %"
Applications
★
#$
Gauss’s that
law the total outward flu
Gauss’s
law
asserts
enc
S
!
! September
Q
Q– Lecture 1!"in free space is equal to
8,Q2004
8.022surface
22 the total cha
!
#
#
#
#
%"
#
%"
"
"
E
d
A
d
A
d
A
R
Q
"
"
!" $ S
!"
$S
$ S R !"
divided
R
R
Find electric field distribution given chargeby ! !
R
8, 2004
8.022 – Lecture 1
distributionSeptember
especially
for symmetric
conditions.
Special case : Normal component of the electric field
intensity is a constant over a closed surface.
★
21
Where and how to apply Gauss’s Law?
Where
to
use
Gauss’s
law
Gauss’s Law can be best utilized when:
the E-field of charge distributions with some symmetry conditions,
such that the normal component of the electric field intensity is
constant over an enclosed surface
(
e.g. Determine the E-field intensity of an infinitely long,
straight, line charge of a uniform density
"' E!d∫sρ&l dl
"' =ρal LE !a ds & "'
side % S
s
%
)
r
r
side % S
Er ds &
!l L
dz &
="" LEaˆr #r E#r ⋅ $$$aˆ r ds =
'! '! EEr rd⋅$ds
!
s
!l side−s
$$$
E#a r Er #a r
""#=! r2πrLE r
L
€
r
""
∫
∫
!l
Q
#!
∫
E r ds
side−s
If the E• ds = constant condition doesn’t exist, then the
Gauss’ law is not useful. See the following example
7
Ez
E1
Er
Er
E2
E1
Example 1
Determine the Electric field caused by a spherical
cloud of electrons with radium b and a volume charge
density ρ = −ρ0 for 0 ≤ R ≤ b and ρ = 0 for R > b .
€
€
€
b
R
R
Postulations
of Electrostatics
Postulates
of Electrostatics
in Free Space
%&''()(*+&,-$'.)/ :
0*+(1),-$'.)/2
%
(1) !!E # $$$$$$$$$$
&#
(2) ! " E # #
•
Q
"$ E!d s # &
s
"$ E!d l # #
$$$$$$$$$$
#
c
& # is the permittivity of free space
• Electric fields mentioned here only due to static charges in free space
• Physical meaning of them:
(1) Implies that a static electric field is not solenoidal
(2) Asserts that static electric fields are irrotational.
Q
"$
&
means the scalar line integral of the static electric field
E
!
d
l
#
#
"$
intensity around any close path vanishes. Kichhoff Voltage’s Law
s
E! d s #
is
$ a form of Gauss’Law
#
c
4
7
Electric potential
★
ential
Potential
Scalar fieldElectric
Electric
Potential
A curl-free
vector
fieldalways
can
always
be expressed
as the gradient
A
curl-free
vector
field
could
be
expressed
as
the
gradient
of
A
curl-free
vector
field
can
always
be
expressed
asathe
d can always be expressed as the gradient of a
scalar field scalarscalar
field: field:
★
P"
E%!d l E!d l E# % ,E
V# % ,V
& &%%V
'
E# % ,V V" % VV
&
"
&P&
P"
'
It is more common to calculate V first, then apply the equation ab
E! d l
& & %'
P&
P&
calculate V first, then applyIt the
equation
above to
to calculate V first, then apply the eq
is more
common
find E out
★
Due to one chargefind E out
For For
discrete
charges:
continuous
s:
R
P"
(!a R dR &
"
Rk *
q
For
R discrete
q charges:
'
V & %&
q
For cont
distributed charge
dR &
q
q
R
distribute
!
*
&
!
"#
R
"#
R
)
)
dR
'
a
(
!
a
&
v
)"# ! R
v
!
!
R
R
"
'
VR &
* R
V & n *'
dv
&
'
"#
"#
)
)
v
*
!
!
)"#V! & R
& )"# ! v * qRk
n
V&
)
"#
+
q
&
!
&
k
!
&
R
R
"#
)
%
*
s
s
&
k
&
V
&
!
k
&
V
+
V&
ds *
&
'
'
s
*
s
*
R
R
"#
)
%
*
)"#V! & R
)"# ! ! kR&&
k
V
& distributed
% 'a Rcharges:(!a
For continuous
R
"
!
q
q
E #qa&R
#Ra R'(
#
q
$ "% (a R ER )!a R ds
#" ER ( %" R ) =
R EpR $
"
7 P
%#"! ! R
s
#
R! #
V" % V& & % E!d l E # a R E!E
# a%R,%V
"
"! ! R
P&
%"! ! R p $ R '
&
R
$
R
'(
q
qV first, then apply
p
It is more common
to
calculate
the The
equation
above
to electric field inten
E # a R ER # a R
#
above
tell
us
the
#
"
%
R
"!
R p electric
$ R ' thefield
%"!
!
outward
radialofdirection
andpoint
has ac
The above
tell
us! the
intensity
a positive
find E out
scalar field:
"
'
Electric potential
proportional
to the squar
the outward radial direction and
andinversely
has a magnitude
proportional
For continuous
For discrete charges: and inversely proportional
to the
square
of the
distance from th
The above
tell
us
the
electric
field
intensity
of
a
positive
point
charge
is
in
R gradientqof a
q
n always
asof
the
distributed charges:
★ beAexpressed
system
discrete
charges
the
direction
has a magnitude proportional to the charge
V outward
dR &
& % 'aradial
(!a Rand
R
"
Similar
when
we
are
!
*
&
"#
R
"#
R
)
)
v
and inversely proportional
to the square! of the distance
from the dv
charge
!
*
V&
P"
ial
'
'
using
Coulomb’s
law to
v* R
)
"#
q
!
k
P&
V&
calculate
the electric
+
!
&
5
ulate V first, then apply the
equation
above
to
)"# ! k && R % R k *
s
* to a
& intensity
V field
dsdue
'
s* R
)"#
!
Gauss’s
Law
continuous charge
For continuous
!
&
★
Aq system of continuous charges
Wayssource
to lcalculate
E:
*
&
V
dl
Gauss’s
Law
distributed charges:
)"# ! 'l * R
a R dR &
% ' E! d l
)"# ! R
E# % ,
&V
&
n
!v
Gauss’s
)"# ! 'v *Law
R
V&
dv *
Ways to calculate E:
E#
)
%"! !
%
v'
aR
&v
R
8
dv ' $$$ E #
"
)
%"! !
%
!s
&v
&s
)
)
)
Ways
E:
*
V & to calculate
ds
E#
a R " dv ' $$$ EWe
# can also
a Ruse
ds
' $$$$$ E #law to
Gauss’s
'
"
%
%
s
*
)"# ! R
%"! ! v ' R
%"! ! s ' R
%"! !
!l&v
&s
&l
)&
)
)
Q
*
dl
EV# &
a
dv
E
#
a
ds
E
#
a
dl
'
$$$
'
$$$$$
'
$
E! d s #
R
Ruse
R
"
" Gauss’s law to
"
'
%
%
%
We
can
also
calculate
E:
l
*
v
s
L
'
'
'
"
%
s
)"#
RR
R
R
%"!
%"! !
%"! !
!
! !
&
8
Qasserts that the total outw
law
We can also use Gauss’s law to calculate E:Gauss’s
E!d s # $$
"% surface in free
! space is equal to the tot
s
Q
divided by ! !!
Poisson’s equation
★
Laplacian
Laplacian stands for the divergence
of the gradient of
★
2
∇ V = ∇ ⋅ ∇V
r ∂ r ∂ r ∂
r ∂V r ∂V r ∂ V
∇ V = ( ax + ay + ay ) ⋅ ( ax
+ ay
+ ay
)
∂x
∂y
∂y
∂x
∂y
∂y
2
∂ 2V ∂ 2V ∂ 2V
= 2 + 2 + 2
∂x
∂y
∂y
★
€
★
€
Poisson Equation
is a second order differential
equation holds at every points in space.
★
Laplace Equation
is the governing equation for
€
problems involving a set of conductors
★
ρ
∇ V =−
ε
2
2
∇ V =0
Summary of vector calculus in
electrostatics
(1)
Summary
!
The potential !"is always continuous
!
E is not always continuous: it can “jump”
!
When we have surface charge distributions
! Remember problem #1 in Pset 2
" When solving problems always check for consistency!
!
!
' & & & (
5
%
6
,
,
Gradient:
.6
/ &x &y &z 0
!
!
In E&M: E * + 5 6
! &Fx &Fy &Fz
Divergence: 5" F *
3
3
&x
&y
&z
! !
!
!
Gauss’s theorem: 2 E "dA * 2 5 " EdV
S
!
!
In E&M: Gauss’ law in differential form
Curl:
!
!
V
G. Sciolla – MIT
In E&M:
Summary
$2
!
54 E * 0
G. Sciolla – MIT
21
!
5" E * 478
!
!
curl F * 5 4 F
Stoke’s theorem:
8.022 – Lecture 6
C
! !
F "ds =
2
A
! !
curl F "dA
w
la
#(x,y,z)
t)
i( n
’s
) ! % '4(#
s
s
"
u
a
G
)# E % 4(#
8.022 – Lecture 4
$(x,y,z)
2
2 !"
" 2
Purcell
Chapter
! ' ! % ' E #ds14
2
1
&
1
"
E % ') !
! %
&
V
dq
r
!(x,y,z)
P"
V& & % ' E!d l
P&
E # % ,V
Capacitance
to calculate V first, then apply the equation above to
Definition For continuous
q
distributed charges:
dR
(
!
a
&
R
"
!
&
★ )"#
R
R
The
potential
of
an
isolated
conductor is directly proportional to
v
!
*
V&
dv
'
v* R
)
"#
the
total
charge
on
it.
qk
!
!s
&
Rk *
V&
ds *
Q=CV
'
s
*
)"# ! R
★
The constant C& is the!lcapacitance of the isolated conductor.
V&
dl *
'
)"# ! l * R
★
Procedure
8
es:
★
★
★
★
★
★
Choose an appropriate coordinate system
Assume charge +Q and -Q on the conductors
By Gauss’s law, find E form Q
Find potential difference between conductors -Q and +Q
Find C by Q/V
!"#$%&'#
Example 2
#$%&'&%()*+$%*,-(-)-$*.$)/*$%*&0(&1$23)&11(%$%41(,5+(%&1$
-6+.&%3-$*.$&$13,7)8$"922$&,5$+&5(($:22$&,5$;22<$=83$
5(313%)+(%$2&)3+(&1$>3)/33,$)83$-6+.&%3-$8&-$&$+31&)(?3$
'3+2())(?()4$!"#$#%#&#'()"*+#,-.".#"#/0#1.203".4#/5#116#
7.8."1/5.#8-.#92:29/8259.#;<#8-.#92:29/8;"6
Solution to example 2
Solution to example 1
Q
=−
2πεL
€
1
Q
dr
=
−
(ln 7 − ln5)
∫7 r
2πεL
5
Some hints for project
★
Idea
★
Using numerical method to solve integration problems.
★
Physical meaning of matrix multiplication
Some hints for project
Numerical Analysis of Charged
Numerical Analysis of Charged Conducting Plates
How to calculate Electrical potential at a point
P ( x, y , z )
P ( x, y , z )
★
★
Point to Point
Q
φ = Ri
4 πεR
y
★
Surface to Point
€1
φ=
4 πε
x
σs
∫∫ R ds
s
Ri
y
x
1
The electric potential at point P(x,y,z)
( ( x, yat, zpoint
) $ 'P(x,y,z)
The electric potential
i
contributed by cell i can be expressed as
4%&asR i
"S i
contributed by cell i can be expressed
When charge
where
(x,y,z)
!!
dx#(dy( x# , y, z ) $
2
2
2
distribution
is
#
#
#
Ri $ ( x ) xi ) *where
( y ) yi ) *R( z$) z(i x) ) x#) 2 * ( y ) y#) 2 * ( z ) z #) 2
constant
i
i
i
i
Some hints for project
to calculate
Electrical
potential
at a point
If thereHow
are total
2 conducting
plates with
2 N number
of cells, the
electric potential
P(x,y,z)
contributed by all the cells can be
Discretionatofpoint
continues
surface
expressed as
★
★
★
2N
) ( x, y , z ) # !( i
1
" " 4&' R
ds%
Numerical Analysis iof
#1Charged
$S Conductingi Plates
i
P( x, y, Dept.
z ) of Elec. Eng., The Chinese University of2Hong Kong, Prof.
K.-L. Wu 2
2
Ri # ( x * xi%) + ( y * yi%) + ( z * zi%)
Ri
Basic Electromagnetics,
If there are total 2 conducting plates with 2 N number of cells, the
Lesson 10
Some hints for project
Numerical Analysis of Charged Condu
P ( x, y , z )
How to calculate Electrical potential at a point
★
Numerical Analysis
of Charged
Conducting
Matrix representation
of sets
of linear
equationsPlates
★
2N
%* n
n !1
Ri # ( x * xi%) 2 + ( y
Ri
& & 4()
'Sn
1
( x1 # xn" ) $ ( y1 # yn" ) $ ( z1 # zn" )
2
2
2
ds" ! V / 2
Numerical Analysis of C
:
If there are total 2 conducting plates with 2 N number of cells, the
electric potential at point P(x,y,z) contributed by all the cells can be
expressed as
6 V /2 3
In a matrix form, there is
1 0
0
#
x
y
z
ds%
)
(
,
,
)
(
!
1
"
"
!
4&' R
ds" ! V / 2
0
0
$ ( y # y" ) $ ( z # z" )
0 V /2 0
,
lmn -.+ n / % 5
2
1
"
#
V
/
2
ds ! #V / 2
0
0
$ ( y # y" ) $ ( z # z" )
0 ! 0
The matrix elements can be
:
0# V / 2 0
:
expressed as
4
1
1
:
2N
2N
%* n
n !1
& & 4()
'S n
( x N # xn" )
2
2N
%* & & 4()
n !1
n
'Sn
2N
%* & & 4()
n !1
n
'Sn
( x N $1 # xn" )2
2
N
( x2 N # xn" ) $ ( y2 N
2
2
n
N $1
i #1
N
2
n
N $1
n
Basic Electromagnetics,
lmn %
$Si
i
Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu
2
n
# yn" ) $ ( zx2 nN "#a zn" )2
2
i
!
dsy" !" a#V / 2
dxn$
n
!
dyn$
1
Some hints for project
Numerical Analysis of Charged Conducting Plates
The potential at the center of !Sm due to unit charge over !S n can be
similarly evaluated, but the formula is complicated. For most purposes it is
sufficiently
accurate to treat the charge on !S n as if it were a point charge,
mn
and use
What is l
★
lmn
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and how to find it
!Sn
a2
&
%
, m"n
2
2
2
4'(Rmn '( ( xm # xn ) $ ( ym # yn ) $ ( zm # zn )
Midpoint
of each
small
Q1:
In Itone
face,
x,y,
zmall,nthecomes
this
point,
we
havesurface
discussed
details of howfrom..
to analyze a static
electric field problem consisting of two conducting plates. The same
principle can be applied to other static electric field problems consisting of
y
any step
number of conducting2aobjects of anyb shapes.
1
starting pint
2
Please pay attention to (1) How the problem is approximated? and (2) How
in Matlab: the integral equation is converted into a matrix equation?
.
i=1;
.
for m=b-a:-2*a:a-b
for n=-b+a:2*a:b-a
.
y_coord(i)=m;
x_coord(i)=n;
.
z_coord(i)=b;
i=i+1;
.
end
2n
end
end point
b-a
Basic Electromagnetics,
O
Dept. of Elec.
Chinese
University
Prof. K.-L.
-bEng.,
-b+aThe
-b+3a
-b+5a
-b+7a of Hong Kong,b-3a
b-a Wu
-b+5a
-b+a
L=2b
Difficult, But not impossible. - Godfather II
x
Lesson 10
x
y
z
d
d
d
d
-d
-d
-d
-d
Some hints for project
★
What is lmn and how to find it
x y
z
1
d
2
d
.
d
.
d
i
xi yi zi
.
-d
.
-d
j
xj yj zj
.
-d
2n
-d
L1,1 L1,2
L1,2n
L2,1
Li,j
L2n,1
Some hints for project
Numerical Analysis of Charged Conducting Plates
6 V /2 3
re is
0 ! 0
0
0
0 V /2 0
,lmn -.+ n / % 5# V / 22
0
0
0 ! 0
an be
0# V / 2 0
4
1
z
y
    
    
σ
=
V
⋅
 i  i 
    
cell n
(xn, yn, zn)
a
a
a
a
x
1
&' ( xm # xn$ ) " ( ym # yn$ ) " ( zm # zn$ )
2
2
2
If ds$ % dxn$ 7 dyn$
potential at the center of )Sm due to a uniform charge
€
lij




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