Upload
devaki
View
68
Download
0
Tags:
Embed Size (px)
DESCRIPTION
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
Citation preview
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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