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Transcript of electricitye and magnetisme
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Topics:
8.022 (E&M) Lecture 4
More applications of vector calculus to electrostatics:
Laplacian: Poisson and Laplace equation
Curl: concept and applications to electrostatics
Introduction to conductors
2
Last time
Electric potential:
i it
( ) -r
r E = = i
21( )
2 8
EU
= =
4E =iG. Sciolla MIT 8.022 Lecture 4
Work done to move a unit charge from infinity to the point P(x,y,z)
Its a scalar!
Energy associated with an electric field:
Work done to assemble system of charges is stored in E
Gausss law in differential form:
Easy way to go from E to charge d stribution that created
w ithE ds
Volume Entirewith space
charges
rdV dV
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3
Laplacian operator
2f f i
2 2 2 2 2 22
2 2 2 2 2 2
( ) ( )f f f
f x y z x y zx y z x y z
f f ff f
x y z x y z
= + + + +
= + + = + +
i i
G. Sciolla MIT 8.022 Lecture 4
What if we combine gradient and divergence?
Lets calculate the div grad f (Q: difference wrt grad div f ?)
Laplacian Operator
4
Interpretation of Laplacian
2+y2
2 2 22
2 2 2
(2 2)4
f fx y z
aa
= + + =
= + =
gives the
G. Sciolla MIT 8.022 Lecture 4
Given a 2d function (x,y)=a(x )/4 calculate the Laplacian
As the second derivative, the Laplaciancurvature of the function
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Poisson equationLets apply the concept of Laplacian to electrostatics.
Rewrite Gausss law in terms of the potential
i = 4 i( 2i = ) =
2
= 4Poisson Equation
G. Sciolla MIT 8.022 Lecture 4 5
E
E
Laplace equation and Earnshaws Theorem
What happens to Poissons equation in vacuum?2 2 = 4 = 0 Laplace Equation
What does this teach us?In a region where satisfies Laplaces equation, then its curvaturemust be 0 everywhere in the region
The potential has no local maxima or minima in that region Important consequence for physics:
Earnshaws Theorem: It is impossible to hold a charge in stable equilibrium with electrostatic fields (no minima)
G. Sciolla MIT 8.022 Lecture 4 6
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G. Sciolla MIT 8.022 Lecture 4
C
7
Is the equilibrium stable? No!
(does the question sound familiar?)
G. Sciolla MIT 8.022 Lecture 4
Application of Earnshaws Theorem
8 charges on a cube and one free in the middle.
8
The circulation F
1 and C2
1and C
2
CF ds = i
C2
C1
C
C
1 2
1 2
1 2
' '
' '
C
C C C C
C C
F ds
F ds s
F ds F ds
= = + = + +
= +
i i i
i i i i i i
1 2
Consider the line integral of a vector function over a closed path C:
Lets now cut C into 2 smaller loops: C
Lets write the circulation C in terms of the integral on C
Circulation
C C C CF ds F ds
F d F ds F ds
= +
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9
The curl ofF
If we repeat the procedure N times:
F as circulation ofF in the limit A0
the
1
i
i
i
=
= =
C
0 C
A
F ds
F nA
ii
G. Sciolla MIT 8.022 Lecture 4
Define the curl of per unit area
where A is the area inside C
The curl is a vector normal to the surface A with direction given byright hand rule
LargeN
limcurl
Stokes Theorem
F over a closed line C and thesurface integral of the curl losed by C
1 1 1
1
L
1 1 1
A
curl curl ( )
i
i
i
i
C
i iC
i i i i
i
C
i
ii A
i i i
i i i
F ds
F ds AA
F ds
F n A dAA
A n
=
= = ==
=
= = =
= =
= = =
i
i i
i
i i i
A=
A
C
C
F
F dA
F ds
ds
=
i
i i i
CA
G. Sciolla MIT 8.022 Lecture 4 10
NB: Stokes relates the line integral of a functionof the function over the area enc
LargeN LargeN LargeN
LargeN
LargeN LargeN
In the limit 0: curl and
curlF n F A F A
=
argeN
curl
(definition of circulation)
curl Stokes TheoremF dA
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field is zero.
=C A
F ds i i= 0
CF ds i
Acurl = 0E dA i
curlE
G. Sciolla MIT 8.022 Lecture 4 11
Application of Stokes TheoremStokes theorem:
The Electrostatics Force is conservative:
The curl of an electrostatic
curl F dA
for any surface A
= 0
Curl in cartesian coordinates (1)
F
F
( , , ) ~ ( , , )2 2
( , , ) ~ ( , , )2 2
( , , )( ) ~ ( , , )2 2
( , , )( ) ~ ( , , )2
by
y y
a
c
zz z
b
dy
y y
c
a
z z
d
Fz zF d y
z
Fy yF d z
y
Fz zF d y F yz
yF d z z
=
= + +
= + +
=
i
i
i
i
2z
yz
zy
FFF ds
y z
= i
P
xy
z yz
a b
cd
G. Sciolla MIT 8.022 Lecture 4 12
Consider infinitesimal rectangle in yz plane
centered at P=(x,y,z) in a vector filed
Calculate circulation of around the square:
s F x y z y F x y z
s F x y z z F x y z
s F x y z x y z
s F x y F x y z
squareYZ
Adding the 4 components:
Fy
y z
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G. Sciolla MIT 8.022 Lecture 4
14
Curl in cartesian coordinates (2) Combining this result with definition of curl:
0
00
)
curl CA square yz
xx
y yz
F dsF n F ds FFA
Fx y y zFF
F dsy z
= = =
i
i i i
curl y yx xz z
x y z
x y z
F FF FF FF x y z F
y z z x x y x y z
F F F
= + +
l: easy to calculate!13
Similar results orienting the rectangles in // (xz) and (xy) planes
square
lim
(curl lim
y z
This is the usable expression for the cur
Summary of vector calculus in
Gradient:
Divergence:
l i
Curl:
, ,x y z
E
yx zFF F
Fx y z
= + +
iS VE dA = i i
4E =i
Purcell Chapter 2
A=
CF ds F dA i i
F F=
0E = G. Sciolla MIT 8.022 Lecture 4
electrostatics (1)
In E&M:
Gausss theorem:
In E&M: Gauss aw in differental form
Stokes theorem:
In E&M:
=
EdV
curl
curl
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Summary of vector calculus in
Laplacian:
In E&M:
lwith electrostati
2 i
Purcell Chapter 2
2 4 2 0 =
G. Sciolla MIT 8.022 Lecture 4
electrostatics (2)
Poisson Equation:
Laplace Equation:
Earnshaws theorem: impossib e to hold a charge in stable equilibriumc fields (no local minima)
=
Comment:This may look like a lot of math: it is!Time and exercise will help you to learn how to use it in E&M
Conductors and Insulators
Conductor
2O
Insulator
Inorganic materials: quartz, glass,
AuFree
electrons
Na+
Cl
G. Sciolla MIT 8.022 Lecture 4 16
: a material with free electrons
Excellent conductors: metals such as Au, Ag, Cu, Al,
OK conductors: ionic solutions such as NaCl in H
: a material without free electrons
Organic materials: rubber, plastic,
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.
Electric Fields in Conductors (1)
E=0
Why? IfE ii
-17 -16 s (typi iE E
+
-
E
---
++
-
--
+
+
+
+
G. Sciolla MIT 8.022 Lecture 4 17
A conductor is assumed to have an infinite supply of electric chargesPretty good assumption
Inside a conductor,is not 0 charges w ll move from where the potential is higher to where
the potential is lower; m gration will stop only when E=0.
How long does it take? 10 -10 cal resist vity of metals)
Electric Fields in Conductors (2)
Electric potential inside a conductor is constant 1 and P2 the would be:
2
10
P
P = = i
Net charge can only reside on the surface
External field li =0
G. Sciolla MIT 8.022 Lecture 4 18
Given 2 points inside the conductor P
since E=0 inside the conductor.E ds
If net charge inside the conductor Electric Field .ne.0 (Gausss law)
nes are perpendicular to surfaceE// component would cause charge flow on the surface until
Conductors surface is an equipotentialBecause its perpendicular to field lines
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Corollary 1
Why? If no charge inside the cavity lds
or minima E=0
Shielding of external electric fi
E=0
G. Sciolla MIT 8.022 Lecture 4 19
In a hollow region inside conductor, =const and E=0 if there arent anycharges in the cavity
Surface of conductor is equipotential
Laplace ho cavity cannot have max
must be constant
Consequence:elds: Faradays cage
Corollary 2
lconductor
Why?
l l i
+Q
-
--
--
-
-
++
+
+
++
+
+
4 ( )
( )Q Q Q Q Q
=
= + = = =
i
i
G. Sciolla MIT 8.022 Lecture 4 20
A charge +Q in the cavity wil induce a charge +Q on the outside of the
App y Gausss aw to surface - - - ins de the conductor
0 because E=0 inside a conductor
Gauss's law
conductor is overall neutral
inside
inside outside inside
E dA
E dA Q Q
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Corollary 3
=E /4
Why? E=4 Since Einside=0E =4
+Q
-
--
--
--
++
+
+
++
+
+
G. Sciolla MIT 8.022 Lecture 4 21
The induced charge density on the surface of a conductorcaused by a charge Q inside it is induced surface
For surface charge layer, Gauss tells us that
surface induced
Uniqueness theorem
potential ion i
1 and 2 i :
1 and 2 1= 2= 1and 2 will i
3= 2 1:
3 l 3=0 3
1= 2
2 2 2
3 2 1( ) ( )r r r r r = =
2
1
2
2
( )r r
r r
= =
Why do I care?
G. Sciolla MIT 8.022 Lecture 4
Given the charge density (x,y,z) in a region and the value of the electrostatic(x,y,z) on the boundaries, there is only one funct (x,y,z) wh ch
describes the potential in that region.
Prove:
Assume there are 2 solutions: ; they w ll satisfy Poisson
Both satisfy boundary conditions: on the boundary,
Superposition: any combination of be solution, includ ng
satisfies Lapace: no local maxima or minima inside the boundaries
On the boundaries =0 everywhere inside region
everywhere inside region
( ) 4 ( ) 4 ( ) 0=
( ) 4
( ) 4 ( )
A solution is THE solution!
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G. Sciolla MIT 8.022 Lecture 4 23
Uniqueness theorem: application 1 A hollow conductor is charged until its external surface reaches a
potential (relative to infinity) =0.
What is the potential inside the cavity?
++
+
+
++
+
+
0=?
Solution
=0 everywhere inside the conductors surface, including the cavity.Why? =0 satisfies boundary conditions and Laplace equation
The uniqueness theorem tells me that is THE solution.
G. Sciolla MIT 8.022 Lecture 4 24
Uniqueness theorem: application 2
Two concentric thin conductive spherical shells or radii R1 and R2carry charges Q1 and Q2 respectively.
What is the potential of the outer sphere? (infinity=0)
What is the potential on the inner sphere?
What at r=0?
R1
R2Q1
Q2
1 1
2 2 22 1 2
1 22 2
2 12 2 2
2 1
Because of uniqueness: ( )
R R
R R
Q Q QE ds dr
r R R
Q Qr r R
R R
= = =
= + =