25
1 8.022 (E&M) – Lecture 4 Topics: More applications of vector calculus to elect rostatics: Laplacian: Poisson and Laplace equati on Curl: concept and applications to ele ctrostatics Introduction to conductors

8.022 (E&M) – Lecture 4

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8.022 (E&M) – Lecture 4. Topics:.  More applications of vector calculus to electrostatics:  Laplacian: Poisson and Laplace equation  Curl: concept and applications to electrostatics  Introduction to conductors. Last time…. 􀂄  Electric potential:. - PowerPoint PPT Presentation

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Page 1: 8.022 (E&M) – Lecture 4

1

8022 (EampM) ndash Lecture 4

Topics

More applications of vector calculus to electrostatics Laplacian Poisson and Laplace equation Curl concept and applications to electrostatics Introduction to conductors

2

Last timehellip

1048708 Electric potential

Work done to move a unit charge from infinity to the point P(xyz)

Itrsquos a scalar

Energy associated with an electric field

Work done to assemble system of charges is stored in E

Gaussrsquos law in differential form

Easy way to go from E to charge d stribution that created it

G Sciolla ndash MIT 8022 ndash Lecture 4

3

Laplacian operator

What if we combine gradient and divergenceLetrsquos calculate the div grad f (Q difference wrt grad div f )

Laplacian Operator

G Sciolla ndash MIT 8022 ndash Lecture 4

4

Interpretation of Laplacian

Given a 2d function (xy)=a(x2+y2)4 calculate the Laplacian

As the second derivative the Laplaciangives the curvature of the function

G Sciolla ndash MIT 8022 ndash Lecture 4

5

Poisson equation

Letrsquos apply the concept of Laplacian to electrostatics

Rewrite Gaussrsquos law in terms of the potential

Poisson Equation

G Sciolla ndash MIT 8022 ndash Lecture 4

6

Laplace equation and Earnshawrsquos Theorem

What happens to Poissonrsquos equation in vacuum

What does this teach us

In a region where φ satisfies Laplacersquos 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

Earnshawrsquos TheoremIt is impossible to hold a charge in stable equilibrium withelectrostatic fields (no minima)

G Sciolla ndash MIT 8022 ndash Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 2: 8.022 (E&M) – Lecture 4

2

Last timehellip

1048708 Electric potential

Work done to move a unit charge from infinity to the point P(xyz)

Itrsquos a scalar

Energy associated with an electric field

Work done to assemble system of charges is stored in E

Gaussrsquos law in differential form

Easy way to go from E to charge d stribution that created it

G Sciolla ndash MIT 8022 ndash Lecture 4

3

Laplacian operator

What if we combine gradient and divergenceLetrsquos calculate the div grad f (Q difference wrt grad div f )

Laplacian Operator

G Sciolla ndash MIT 8022 ndash Lecture 4

4

Interpretation of Laplacian

Given a 2d function (xy)=a(x2+y2)4 calculate the Laplacian

As the second derivative the Laplaciangives the curvature of the function

G Sciolla ndash MIT 8022 ndash Lecture 4

5

Poisson equation

Letrsquos apply the concept of Laplacian to electrostatics

Rewrite Gaussrsquos law in terms of the potential

Poisson Equation

G Sciolla ndash MIT 8022 ndash Lecture 4

6

Laplace equation and Earnshawrsquos Theorem

What happens to Poissonrsquos equation in vacuum

What does this teach us

In a region where φ satisfies Laplacersquos 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

Earnshawrsquos TheoremIt is impossible to hold a charge in stable equilibrium withelectrostatic fields (no minima)

G Sciolla ndash MIT 8022 ndash Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 3: 8.022 (E&M) – Lecture 4

3

Laplacian operator

What if we combine gradient and divergenceLetrsquos calculate the div grad f (Q difference wrt grad div f )

Laplacian Operator

G Sciolla ndash MIT 8022 ndash Lecture 4

4

Interpretation of Laplacian

Given a 2d function (xy)=a(x2+y2)4 calculate the Laplacian

As the second derivative the Laplaciangives the curvature of the function

G Sciolla ndash MIT 8022 ndash Lecture 4

5

Poisson equation

Letrsquos apply the concept of Laplacian to electrostatics

Rewrite Gaussrsquos law in terms of the potential

Poisson Equation

G Sciolla ndash MIT 8022 ndash Lecture 4

6

Laplace equation and Earnshawrsquos Theorem

What happens to Poissonrsquos equation in vacuum

What does this teach us

In a region where φ satisfies Laplacersquos 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

Earnshawrsquos TheoremIt is impossible to hold a charge in stable equilibrium withelectrostatic fields (no minima)

G Sciolla ndash MIT 8022 ndash Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 4: 8.022 (E&M) – Lecture 4

4

Interpretation of Laplacian

Given a 2d function (xy)=a(x2+y2)4 calculate the Laplacian

As the second derivative the Laplaciangives the curvature of the function

G Sciolla ndash MIT 8022 ndash Lecture 4

5

Poisson equation

Letrsquos apply the concept of Laplacian to electrostatics

Rewrite Gaussrsquos law in terms of the potential

Poisson Equation

G Sciolla ndash MIT 8022 ndash Lecture 4

6

Laplace equation and Earnshawrsquos Theorem

What happens to Poissonrsquos equation in vacuum

What does this teach us

In a region where φ satisfies Laplacersquos 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

Earnshawrsquos TheoremIt is impossible to hold a charge in stable equilibrium withelectrostatic fields (no minima)

G Sciolla ndash MIT 8022 ndash Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 5: 8.022 (E&M) – Lecture 4

5

Poisson equation

Letrsquos apply the concept of Laplacian to electrostatics

Rewrite Gaussrsquos law in terms of the potential

Poisson Equation

G Sciolla ndash MIT 8022 ndash Lecture 4

6

Laplace equation and Earnshawrsquos Theorem

What happens to Poissonrsquos equation in vacuum

What does this teach us

In a region where φ satisfies Laplacersquos 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

Earnshawrsquos TheoremIt is impossible to hold a charge in stable equilibrium withelectrostatic fields (no minima)

G Sciolla ndash MIT 8022 ndash Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 6: 8.022 (E&M) – Lecture 4

6

Laplace equation and Earnshawrsquos Theorem

What happens to Poissonrsquos equation in vacuum

What does this teach us

In a region where φ satisfies Laplacersquos 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

Earnshawrsquos TheoremIt is impossible to hold a charge in stable equilibrium withelectrostatic fields (no minima)

G Sciolla ndash MIT 8022 ndash Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 7: 8.022 (E&M) – Lecture 4

7

Application of Earnshawrsquos Theorem

8 charges on a cube and one free in the middle

Is the equilibrium stable No

(does the question sound familiar)

G Sciolla ndash MIT 8022 ndash Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 8: 8.022 (E&M) – Lecture 4

8

The circulation

Consider the line integral of a vector function over a closed path C

Letrsquos now cut C into 2 smaller loops C1 and C2

Letrsquos write the circulation C in terms of the integral on C1 and C2

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 9: 8.022 (E&M) – Lecture 4

9

The curl of F

If we repeat the procedure N times

Define the curl of F as circulation of F per unit area in the limit A0

where A is the area inside C

The curl is a vector normal to the surface A with direction given by the ldquoright hand rulerdquo

G Sciolla ndash MIT 8022 ndash Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 10: 8.022 (E&M) – Lecture 4

10

Stokes Theorem

(definition of circulation)

Stokes Theorem

NB Stokes relates the line integral of a function F over a closed line C and the surface integral of the curl of the function over the area enclosed by C

G Sciolla ndash MIT 8022 ndash Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 11: 8.022 (E&M) – Lecture 4

11

Application of Stokersquos Theorem

Stokersquos theorem

The Electrostatics Force is conservative

The curl of an electrostatic field is zero

G Sciolla ndash MIT 8022 ndash Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 12: 8.022 (E&M) – Lecture 4

12

Curl in cartesian coordinates (1)

Consider infinitesimal rectangle in yz plane centered at P=(xyz) in a vector filed F Calculate circulation of around the square

G Sciolla ndash MIT 8022 ndash Lecture 4

Adding the 4 compone nts

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 13: 8.022 (E&M) – Lecture 4

13

Curl in cartesian coordinates (2)

Combining this result with definition of curl

Similar results orienting the rectangles in (xz) and (xy) planes

This is the usable expression for the curl easy to calculate

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 14: 8.022 (E&M) – Lecture 4

14

Summary of vector calculus in

electrostatics (1)

Gradient

In EampM

Divergence

Gaussrsquos theorem

In EampM Gaussrsquo aw in different al form

Curl

Stokersquos theorem

In EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 15: 8.022 (E&M) – Lecture 4

15

Summary of vector calculus in

electrostatics (2)

Laplacian

In EampM

Poisson Equation

Laplace Equation

Earnshawrsquos theorem impossib e to hold a charge in stable equilibrium with electrostatic fields (no local minima)

CommentThis may look like a lot of math it isTime and exercise will help you to learn how to use it in EampM

G Sciolla ndash MIT 8022 ndash Lecture 4

Purcell Chapter 2

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 16: 8.022 (E&M) – Lecture 4

16

Conductors and Insulators

Conductor a material with free electrons

Excellent conductors metals such as Au Ag Cu Alhellip

OK conductors ionic solutions such as NaCl in H2O

Insulator a material without free electrons

Organic materials rubber plastichellip

Inorganic materials quartz glasshellip

G Sciolla ndash MIT 8022 ndash Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 17: 8.022 (E&M) – Lecture 4

17

Electric Fields in Conductors (1)

A conductor is assumed to have an infinite supply of electric charges

Pretty good assumptionhellip

Inside a conductor E=0

Why If E 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 10-17 ndash 10-16 s (typical resistivity of metals)

G Sciolla ndash MIT 8022 ndash Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 18: 8.022 (E&M) – Lecture 4

18

Electric Fields in Conductors (2)

Electric potential inside a conductor is constant

Given 2 points inside the conductor P1 and P2 the Δφ would be

since E=0 inside the conductor

Net charge can only reside on the surface

If net charge inside the conductor Electric Field ne0 (Gaussrsquos law)

External field lines are perpendicular to surface

E component would cause charge flow on the surface until Δφ=0

Conductorrsquos surface is an equipotential

Because itrsquos perpendicular to field lines

G Sciolla ndash MIT 8022 ndash Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 19: 8.022 (E&M) – Lecture 4

19

Corollary 1

In a hollow region inside conductor =const and E=0 if there arenrsquot anycharges in the cavity

Why

Surface of conductor is equipotential

If no charge inside the cavity Laplace holds Φcavity cannot have max or minima Φ must be constant E = 0

Consequence

Shielding of external electric fields Faradayrsquos cage

G Sciolla ndash MIT 8022 ndash Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Slide 16
  • Slide 17
  • Slide 18
  • Slide 19
  • Slide 20
  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 20: 8.022 (E&M) – Lecture 4

20

Corollary 2

A charge +Q in the cavity wil induce a charge +Q on the outside of the conductor

Apply Gaussrsquos aw to surface - - - ins de the conductor

Why

G Sciolla ndash MIT 8022 ndash Lecture 4

because E=0 inside a conductor

Gausss law

( Conductor is overall neutral )

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
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Page 21: 8.022 (E&M) – Lecture 4

21

Corollary 3

The induced charge density on the surface of a conductor

caused by a charge Q inside it is

For surface charge layer Gauss tells us that ΔE=4πσ

Why

Since

G Sciolla ndash MIT 8022 ndash Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
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  • Slide 21
  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 22: 8.022 (E&M) – Lecture 4

22

Uniqueness theorem

Given the charge density (xyz) in a region and the value of the electrostatic

potential φ(xycz) on the boundaries there is only one function φ(xycz) which

describes the potential in that regionProve

Assume there are 2 solutions φ1 and φ2 they w ll satisfy Poisson

Both φ1 and φ2 satisfy boundary conditions on the boundary φ1 = φ2 =φ

Superposition any combination of φ1 and φ2 will be solution including

Φ3 satisfies Laplace no local maxima or minima inside the boundaries

On the boundaries φ3=0 φ3 = 0 everywhere inside region

φ1 = φ2 everywhere inside regionWhy do I careA solution is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
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  • Slide 22
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  • Slide 24
  • Slide 25
Page 23: 8.022 (E&M) – Lecture 4

23

Uniqueness theorem application 1

A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0

What is the potential inside the cavity

Solution

φ=φ0 everywhere inside the conductorrsquos surface including the cavityWhy φ=φ0 satisfies boundary conditions and Laplace equation The uniqueness theorem tells me that is THE solution

G Sciolla ndash MIT 8022 ndash Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
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  • Slide 25
Page 24: 8.022 (E&M) – Lecture 4

24

Uniqueness theorem application 2

Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively

What is the potential of the outer sphere (φinfinity=0)

What is the potential on the inner sphere

What at r=0

Outer sphere φ1=(Q1+Q2)R1

Solution

Inner sphere

Because of uniqueness

G Sciolla ndash MIT 8022 ndash Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
  • Slide 11
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  • Slide 22
  • Slide 23
  • Slide 24
  • Slide 25
Page 25: 8.022 (E&M) – Lecture 4

25

Next timehellip

More on Conductors in ElectrostaticsCapacitors

NB All these topics are included in Quiz 1 scheduled for Tue October 5 just 2 weeks from now

Reminders Lab 1 is scheduled for Tomorrow 5-8 pm Pset 2 is due THIS Fri Sep 24

G Sciolla ndash MIT 8022 ndash Lecture 4

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
  • Slide 10
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