electricitye and magnetisme

7/28/2019 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


2/13

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


Given a 2d function (x,y)=a(x )/4 calculate the Laplacian

As the second derivative, the Laplaciancurvature of the function


3/13

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)



4/13


C

7

Is the equilibrium stable? No!

(does the question sound familiar?)


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


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


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


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


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


7/13


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


8/13

15

Summary of vector calculus in

Laplacian:

In E&M:

lwith electrostati

2 i

Purcell Chapter 2

2 4 2 0 =


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


: 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,


9/13

.

Electric Fields in Conductors (1)

E=0

Why? IfE ii

-17 -16 s (typi iE E

+

-

E

---

++

-

--

+

+

+

+


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


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


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


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


11/13

22

Corollary 3

=E /4

Why? E=4 Since Einside=0E =4

+Q

-

--

--

--

++

+

+

++

+

+


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?


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!


12/13


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.


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

= = =

= + =

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electricitye and magnetisme