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…
r
Electric potential: φ ( r ) = − ∫ E ids
∞
w ith E = -∇φ
Work done to move a unit charge from infinity to the point P(x,y,z)
It’s a scalar!
Energy associated with an electric field:
Work done to assemble system of charges is stored in E
U=
E2
1
dV
ρφ (r )dV = ∫
∫
2 Volume
8π
Entire
with
charges
space
Gauss’s law in differential form:
∇i E = 4πρ
Easy way to go from E to charge distribution that created it
G. Sciolla – MIT
8.022 – Lecture 4
2
1
Laplacian operator
What if we combine gradient and divergence?
Let’s calculate the div grad f (Q: difference wrt grad div f ?)
∇i∇f = (
=
∂
∂
∂
∂f
∂f
∂f
xˆ +
yˆ + zˆ )i( xˆ +
yˆ +
zˆ )
∂x
∂y
∂z
∂x
∂y
∂z
∂2 f ∂2 f ∂2 f ⎛ ∂2
∂2
∂2 ⎞
+ 2 + 2 = ⎜ 2 + 2 + 2 ⎟ f ≡ ∇2 f
2
∂x
∂y
∂z
⎝ ∂x ∂y ∂z ⎠
∇ 2 f ≡ ∇i∇f
G. Sciolla – MIT
Laplacian Operator
8.022 – Lecture 4
3
Interpretation of Laplacian
Given a 2d function φ(x,y)=a(x2+y2)/4 calculate the Laplacian
⎛ ∂2
∂2
∂2 ⎞
∇2 f = ⎜ 2 + 2 + 2 ⎟ f =
⎝ ∂x ∂y ∂z ⎠
a
= (2 + 2) = a
4
As the second derivative, the Laplacian
gives the curvature of the function
G. Sciolla – MIT
8.022 – Lecture 4
4
2
Poisson equation
Let’s apply the concept of Laplacian to electrostatics.
Rewrite Gauss’s law in terms of the potential
⎪⎧∇i E = 4πρ
⎨
2
⎪⎩∇i E = ∇i(−∇φ ) = −∇ φ
→ ∇ 2φ = −4πρ Poisson Equation
G. Sciolla – MIT
8.022 – Lecture 4
5
Laplace equation and Earnshaw’s Theorem
What happens to Poisson’s equation in vacuum?
∇ 2φ = −4πρ ⇒ ∇ 2φ = 0
Laplace Equation
What does this teach us?
In a region where φ satisfies Laplace’s equation, then its curvature
must be 0 everywhere in the region
The potential has no local maxima or minima in that region
Important consequence for physics:
Earnshaw’s Theorem:
It is impossible to hold a charge in stable equilibrium with
electrostatic fields (no minima)
G. Sciolla – MIT
8.022 – Lecture 4
6
3
Application of Earnshaw’s Theorem
8 charges on a cube and one free in the middle.
Is the equilibrium stable? No!
(does the question sound familiar?)
G. Sciolla – MIT
8.022 – Lecture 4
7
The circulation
Consider the line integral of a vector function F over a closed path C:
C’
∫
Γ=
C
F ids
Circulation
C2
C’
C1
Let’s now cut C into 2 smaller loops: C1 and C2
Let’s write the circulation C in terms of the integral on C1 and C2
Γ=
∫
=
C1
∫
∫
C
F ids =
∫
C1 −C '
F ids +
F ids − ∫ F ids +
C'
C
F ids +
1
G. Sciolla
– MIT
∫
C2
F ids
∫
C2
∫
C2 − C '
F ids +
F i ds =
∫
C'
⇒
F i ds
Γ = Γ1 + Γ 2
8.022 – Lecture 4
8
4
The curl of F
If we repeat the procedure N times:
i = LargeN
∑
Γ=
i =1
C
Γi
Define the curl of F as circulation of F per unit area in the limit A 0
curl F i nˆ ≡ lim
∫
C
A→ 0
F i ds
A
where A is the area inside C
The curl is a vector normal to the surface A with direction given by
the “right hand rule”
G. Sciolla – MIT
8.022 – Lecture 4
9
Stokes Theorem
Γ=
i = LargeN
∑
i =1
Γi =
LargeN
∑ ∫
i =1
Ci
∫
In the limit A → 0:
⎧
⎪Γ =
⎨
⎪
⎩
LargeN
∑
i =1
Ci
C
F ids
Ai
∫
∫ F ids = ∫
C
A
LargeN
∑
i =1
Ai curl F inˆ =
Γ=
⇒
F ids =
LargeN
∑
i =1
Ai
∫
Ci
F ids
A
Ai
→ curl F inˆ and
i = LargeN
∑
i =1
curl F i( Ai nˆ ) =
F ids
curl F idA
LargeN
∑
i =1
C
Ai → ∫ dA
A
curl F i Ai → ∫ curl F idA
A
(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 – MIT
8.022 – Lecture 4
10
5
Application of Stoke’s Theorem
Stoke’s theorem:
∫
C
F ids =
∫
A
curl F idA
The Electrostatics Force is conservative:
∫
∫
⇒
A
F ids = 0
C
curl E idA = 0 for any surface A
⇒
curl E = 0
The curl of an electrostatic field is zero.
G. Sciolla – MIT
8.022 – Lecture 4
11
Curl in cartesian coordinates (1)
Consider infinitesimal rectangle in yz plane
centered at P=(x,y,z) in a vector filed F
Calculate circulation of F around the square:
⎧b
⎪ ∫ F ids
⎪a
⎪c
⎪ ∫ F ids
⎪b
⎨d
⎪ F ids
⎪∫
⎪c
⎪a
⎪ ∫ F ids
⎩d
=Fy ( x, y, z −
⎡
∆z
∆z ∂Fy ⎤
)∆y ~ ⎢ Fy ( x, y, z ) −
⎥ ∆y
2
2 ∂z ⎦
⎣
⎡
∆y
∆y ∂Fz ⎤
, z )∆z ~ ⎢ Fz ( x, y, z ) +
∆z
2
2 ∂y ⎥⎦
⎣
⎡
∆z
∆z ∂Fy ⎤
=Fy ( x, y, z + )(− ∆y ) ~ − ⎢ Fy ( x, y, z ) +
⎥ ∆y
2
2 ∂z ⎦
⎣
=Fz ( x, y +
=Fz ( x, y −
z
d ∆y c
P ∆z
a
b
y
x
⎡
∆y ∂Fz ⎤
∆y
∆z
, z )(− ∆z ) ~ − ⎢ Fz ( x, y, z ) −
2 ∂y ⎦⎥
2
⎣
Adding the 4 components: ⇒
G. Sciolla – MIT
⎛ ∂F ∂F ⎞
F ids = ⎜ z − y ⎟ ∆y∆z
∂z ⎠
⎝ ∂y
squareYZ
∫
8.022 – Lecture 4
12
6
Curl in cartesian coordinates (2)
Combining this result with definition of curl:
⎧
F ids
⎪ curl F inˆ ≡ lim ∫ C
A→ 0
⎪
A
⇒
⎨
⎛ ∂Fz ∂Fy ⎞
⎪
∆
∆
=
−
F
ds
y
z
i
⎜
⎟
⎪ ∫
∂z ⎠
⎝ ∂y
⎩square
(curl F ) x = lim
∆x → 0
∆y → 0
∫
square
F ids
∆x∆y
⎛ ∂F ∂Fy ⎞
=⎜ z −
⎟
∂z ⎠
⎝ ∂y
Similar results orienting the rectangles in // (xz) and (xy) planes
⎛ ∂F ∂F
curl F = ⎜ z − y
∂z
⎝ ∂y
⎞
⎛ ∂Fx ∂Fz
−
⎟ xˆ + ⎜
∂x
⎝ ∂z
⎠
xˆ
∂
F
⎛
⎞
∂Fx
∂
⎞
y
⎟ yˆ + ⎜ ∂x − ∂y ⎟ zˆ ≡ ∂x
⎠
⎝
⎠
Fx
yˆ
zˆ
∂
∂y
Fy
∂
≡ ∇× F
∂z
Fz
This
is the usable expression8.022
for–the
curl: easy to calculate!
G. Sciolla – MIT
Lecture 4
13
Summary of vector calculus in
electrostatics (1)
Gradient:
In E&M:
⎛ ∂ ∂ ∂ ⎞
∇φ ≡ ⎜
,
,
⎟φ
⎝ ∂x ∂y ∂z ⎠
E = −∇ φ
Divergence: ∇i F =
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
Gauss’s theorem:
∫
S
E idA =
∫
V
∇ i EdV
In E&M: Gauss’ law in differential form
Curl:
∇i E = 4πρ
curl F = ∇ × F
Stoke’s theorem:
In E&M:
∫
C
F ids =
∫
A
curl F idA
∇× E = 0
G. Sciolla – MIT
8.022 – Lecture 4
Purcell Chapter
14 2
7
Summary of vector calculus in
electrostatics (2)
Laplacian: ∇ 2φ ≡ ∇ i ∇ φ
In E&M:
Poisson Equation:
Laplace Equation:
∇ 2φ = −4πρ
∇ 2φ = 0
Earnshaw’s theorem: impossible to hold a charge in stable equilibrium
with electrostatic fields (no local minima)
Comment:
This may look like a lot of math: it is!
Time and exercise will help you to learn how to use it in E&M
G. Sciolla – MIT
8.022 – Lecture 4
Purcell Chapter
15 2
Conductors and Insulators
Conductor: a material with free electrons
Excellent conductors: metals such as Au, Ag, Cu, Al,…
OK conductors: ionic solutions such as NaCl in H2O
Au
Free
electrons
ClNa+
Insulator: a material without free electrons
Organic materials: rubber, plastic,…
Inorganic materials: quartz, glass,…
G. Sciolla – MIT
8.022 – Lecture 4
16
8
Electric Fields in Conductors (1)
A conductor is assumed to have an infinite supply of electric charges
Pretty good assumption…
Inside a conductor, E=0
Why? If E is not 0
charges will move from where the potential is higher to where
the potential is lower; migration will stop only when E=0.
How long does it take? 10-17-10-16 s (typical resistivity of metals)
E
E
E
-
+
-
- +
-
- +
+
+
+
+
.
G. Sciolla – MIT
8.022 – Lecture 4
17
Electric Fields in Conductors (2)
Electric potential inside a conductor is constant
Given 2 points inside the conductor P1 and P2 the ∆φ would be:
P2
∆φ = ∫ Eids = 0 since E=0 inside the conductor.
P1
Net charge can only reside on the surface
If net charge inside the conductor
Electric Field .ne.0 (Gauss’s law)
External field lines are perpendicular to surface
E// component would cause charge flow on the surface until ∆φ=0
Conductor’s surface is an equipotential
Because it’s perpendicular to field lines
G. Sciolla – MIT
8.022 – Lecture 4
18
9
Corollary 1
In a hollow region inside conductor, φ=const and E=0 if there aren’t any
charges in the cavity
E=0
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: Faraday’s cage
G. Sciolla – MIT
8.022 – Lecture 4
19
Corollary 2
A charge +Q in the cavity will induce a charge +Q on the outside of the
conductor
+
+
+
+
+Q
+
+
+
+
Why?
Apply Gauss’s law to surface - - - inside the conductor
⎧ EidA = 0 because E=0 inside a conductor
⎪∫
⎨
⎪⎩ ∫ EidA = 4π (Q + Qinside ) Gauss's law
⇒ Qinside = −Q ⇒ Qoutside = −Qinside = Q (conductor is overall neutral)
G. Sciolla – MIT
8.022 – Lecture 4
20
10
Corollary 3
The induced charge density on the surface of a conductor
caused by a charge Q inside it is σinduced=Esurface/4π
+
+
+
+Q
-
+
-
+
-
+
σ
+
+
Why?
For surface charge layer, Gauss tells us that ∆E=4πσ
Esurface=4πσinduced
Since Einside=0
G. Sciolla – MIT
8.022 – Lecture 4
21
Uniqueness theorem
Given the charge density ρ(x,y,z) in a region and the value of the electrostatic
potential φ(x,y,z) on the boundaries, there is only one function φ(x,y,z) which
describes the potential in that region.
Prove:
Assume there are 2 solutions: φ1 and φ2; they will satisfy Poisson:
∇ 2φ1 (r ) = 4πρ (r )
∇ 2φ2 (r ) = 4πρ (r )
Both φ1 and φ2 satisfy boundary conditions: on the boundary, φ1= φ2=φ
Superposition: any combination of φ1and φ2 will be solution, including
φ3= φ2 – φ1:
∇ 2φ3 (r ) = ∇ 2φ2 (r ) − ∇ 2φ1 (r ) = 4πρ (r ) − 4πρ (r ) = 0
φ3 satisfies Laplace: no local maxima or minima inside the boundaries
φ3=0 everywhere inside region
On the boundaries φ3=0
φ1= φ2 everywhere inside region
Why do I care?
A solution is THE solution!
G. Sciolla – MIT
8.022 – Lecture 4
22
11
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?
+
+
+
+
φ0
φ=?
+
+
+
+
Solution
φ=φ0 everywhere inside the conductor’s surface, including the cavity.
Why? φ=φ0 satisfies boundary conditions and Laplace equation
The uniqueness theorem tells me that is THE solution.
G. Sciolla – MIT
8.022 – Lecture 4
23
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?
Solution
Q2
R2
R1
Q1
Outer sphere: φ1=(Q1+Q2)/R1
R1
Inner sphere φ2 − φ1 = − ∫ E ids = −
R2
R1
Q2
Q Q
dr = 2 − 2
2
r
R1 R2
R2
∫
Q Q
⇒ φ2 = 2 + 1 Because of uniqueness: φ (r ) = φ2∀r < R2
R2 R1
G. Sciolla – MIT
8.022 – Lecture 4
24
12
Next time…
More on Conductors in Electrostatics
Capacitors
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 – MIT
8.022 – Lecture 4
25
13