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VE ∇−=
Calculation of E - previous two lectures
1) Coulomb’s Law
Calculate E due to point charges
Use superposition principle to get total E
2) Calculate electric potential V and use
•V is a scalar so all we need to do is add up the numbers
•If we need to know the work done to move a charge between two points all we need to use is the difference in electric potential
Aim of lecture is to introduce another method to calculate E - (22-2 – 22-6 Tipler)
Lectures 5 & 6: Gauss’s Law
Johann Carl Friedrich GaussBorn: 30 April 1777 in Brunswick, Duchy of Brunswick
Died: 23 Feb 1855 in Göttingen, Hanover number theory,statistics ,analysis, differential geometry, electrostatics, astronomy, and optics
It’s all to do with FLUX
?
1) One can guess the charge polarity
2) Can we also work out the magnitude of the charge?
Coulomb’s law: Q to E
Gauss’ Law: from E to Q
q
r
For points on the spherical surface
204 r
qE
πε=
∫ AdE.
Note result is independent of r
Point charge q
∫= EdA0εq
=
€
=q
4πε0r2
× 4πr2∫= dAE
Imaginary spherical surface
€
dr A
€
εo
r E • d
r A ∫ = qencl
This is Gauss’ law.
RESULT HOLDS FOR ANY SHAPE OF CLOSED SURFACE AND FOR ANY CHARGE DISTRIBUTION INSIDE IT
€
εo
r E • d
r A ∫ = qencl
The closed surface is called a
Gaussian Surface
Can be proved, but not required for this module.
Concept of Electric Flux E
Since the density of field linesis proportional to E,
€
rE • d
r A ∫ Net number of lines
passing through the surface
Electric Flux E
€
rE .d
r A
surface∫
Unit: Nm2C-1
Electric Flux E: Flux of what?A useful concept, but no real materialflow.
The sun: Flux of photons
Water fall: flux of molecules
r
q
What is the net flux through the Gaussian Surface?
€
rE • d
r A ∫ = 0
€
rE = 0 ??
If the charge falls outsidethe Gaussian Surface, the net flux through thesurface is zero.
Plane Area A
Outward Normal E-field
Concept of Electric Flux E
Electric flux = EAcos
In terms of vectors:
Flux E = E.A
A (a vector normal to the area)
€
φE = E .d Asurface
∫
In general
∫= AdEE .surface
For a closed surface
Circle represents a closed surface
For a number of charges inside a volume:
∑∫ =i
i
surface
qAd.E0
1
ε
Gauss’s Law: the electric flux of E out of any closed surface enclosing a total charge Qenclosed situated in a vacuum (air for practical purposes) is
€
φE = E .dAsurface
∫ =Qenclosed
ε0
Gauss’ law and Coulomb’s law
Assume we know nothing about Coulomb’s law,we try to use Gauss’ law to get the E field from a point charge.
r Eq
€
rE • d
r A ∫ =
q
εo
€
E dA∫ = E(4πr2) =q
εo
€
E =q
4πε0r2
Example: Sphere of radius a uniformly charged throughout its volume. Total charge Q
Q
(a) E-field for r > a
Method: set up an (imaginary) Gaussian Surface and use symmetry
r0
24.ε
π QrEAdE =∫ =
204 r
QE
πε=∴
Coulomb’s Law
The E field at r from a uniformly charged sphere is equivalent to the field at that point produced by a point charge Q at r=0.
NB: 1) r > a,2) Uniformly charged sphere
€
E =Q
4πε0a2
For r = a
(b) E-field for r < a
24 rEE π×=
304 a
QrE
πε=∴
r3
3
0 34
341
a
rQ
π
π
ε××=
30
3
a
Qr
ε=
€
∴E =Qr
4πε0a3
r
The E field from charges within the shell from r to a is
ZeroThe E field from charges within the sphere r, is proportional to the charges enclosed.
€
∴E =Qencl
4πε0r2
=1
4πε0r2
(r3
a3Q)
For r = a
€
E =Q
4πε0a2
~1/r2
~r
E-fields in Conductors
• EXTERNAL E-FIELD. Electrons are free to move, and exist in vast quantities
E = 0 in a conductor
Thus Q=0E = 0
Application: protection of electronics
Field lines in the space surrounding a charged flat plate and cylinder made visible by small bits of thread suspended in oil
Aim
To illustrate how to apply Gauss’s law to known charge distributions
Applications of Gauss’s Law
•Spherical Shell
•Sheet of Charge
•Thin Wire
Three types:
(1) E-field due to a uniform spherical shell of charge
+Q
0
00
=∴
==∑∫
E
qAd.E
ε
Inside
+Q
0
00
=∴
==∑∫
E
qAd.E
ε
Inside
Outside
204 r
QE
πε=
(1) E-field due to a uniform spherical shell of charge
Fig. 22-23
(2) E-field due to an infinite thin plane sheet of charge
A
Symmetry: E is to the sheet; E has the same magnitude at any given distance on either side of the sheet
Use a cylindrical Gaussian surface - axis to the sheet
0
2εσA
EA=
Gauss’s Law:
Imaginary Gaussian Surface
02εσ
=E
The result agrees with the one we found by taking the infinite radius limit of a disk!
What is the E-field due to an infinite conducting sheet carrying a charge?
II. E-fields in Conductors – a reminder
•Charges move until E = 0 inside a conductor.
•Each point in a conductor is at the same V.
•No free charges exist inside a conductor.
•Free (extra) charges reside on surface.
(3) E-field due to an infinite conducting sheet
Q distributes until there is a charge densityσ on both surfaces.
σ σ
E E
We now repeat the “pill box”
construction.
NO E-field
σ σ
(3) E-field due to an infinite conducting sheet
A
Apply Gauss’s Law
€
φE = EA =σA
ε0
0εσ
=∴E
Factor of 2 difference between the fields of insulator and conducting infinite planes. Why?
E
(4) E-field just outside a charged conductor
0εσ
=E
(5) E-field between parallel charged metal plates
+ + + + + + + + + + + + + +
- - - - - - - - - - - - - - - -
Charge density σ E
0εσ
=E
+ + + + + + + + + + + + + +
- - - - - - - - - - - - - - - -
Cylindrical Gaussian Surface
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Hyper-inflation: If you have a wheelbarrow of cash in the street, someone steals onlythe wheelbarrow.
Hyper-inflation:No accurate definition
∫ =×=0
2ε
λπ
lrlEAd.E
rE
02πελ
=
Gauss’s Law in the Differential Form
( )
∫∫
∫
=∴
=
VS
V
dVAd.E
dVrQ
ρε
ρ
0
1
( )dVF.Ad.FVS ∫∫ ∇=
Apply Gauss’s Mathematical theorem
( ) ∫∫∫ =∇=VVS
dVdVEAdE ρε 0
1..
Hence
This can only be true for any volume if
0ερ
=∇E.
€
∇.r E = (
∂
∂xˆ i +
∂
∂yˆ j +
∂
∂zˆ k ).(Ex
ˆ i + Eyˆ j + E z
ˆ k )
=∂Ex
∂x+
∂Ey
∂y+
∂E z
∂z
€
∇.r E ≡ Div
r E
The divergence of E
In cylindrical coordinates:
€
∇.r E =
1
r
∂
∂r(rE r) +
1
r
∂Eϕ
∂ϕ+
∂E z
∂z
rE
02πελ
= Has only radial component
€
∇.r E =
1
r
∂
∂r(rE r) =
1
r
∂
∂r(r
λ
2πεor) = 0
0ερ
=∇E.
In spherical coordinates:
€
∇.r E =
1
r2
∂
∂r(r2E r ) +
1
rsinθ
∂Eθ
∂θ(Eθ sinθ) +
1
rsinθ
∂Eϕ
∂ϕ
Q aShow that for r>a,
€
∇.r E = 0
for r<a
€
∇.r E = ??
Summary
•In situations of high symmetry (planar, spherical, cylindrical), Gauss’s law allows us to compute quantitatively the E-field in a straightforward manner
Next installment:
Electric Potential of line of charge and spherical distributions
dS2
dS1
Local charge gives E = σ/2ε0 in both directions
To ensure E = 0 inside, rest of charge must produce a field E = σ/2ε0 at dS1 and hence dS2 in direction dS2
Net field out of surface dS2 is Enet = σ/2ε0 + σ/2ε0 = σ/ε0
Conductor’s Surface
Gauss developed a theorem - originally for gravitational fields - that was put into its simplest form for E-fields, in terms of pictures, by Michael Faraday
Michael Faraday 1791 - 1867
Concept of Electric Flux E
More field lines exit than enter. The net number is the same as that for a single charge equal to the net charge within the surface
Concept of Electric Flux E
The net number of lines leaving any surface enclosing the charges is proportional to the net charge enclosed by the surface
The mathematical quantity that corresponds to the number of field lines crossing a surface is called the electric flux E
1p 1p
1 m
One pence 3 g copper
T&M lectures: Molar mass of Cu 63.5 g
One pence = (3/63.5) mols of Cu = 2.84 1022 Cu atoms
One Cu atoms contains 29 electrons. Hence one pence contains 8.25 1023 electrons
This corresponds to 131956 C. 0.0001% of that is 0.132 C
0.132 C at 1 m gives F = 157,000,000 N (17,662 Tons)
Gauss’s Law: the electric flux of E out of any closed surface enclosing a total charge Qenclosed situated in a vacuum (air for practical purposes) is
€
φE = E .dAsurface
∫ =Qenclosed
ε0
Concept of Electric Flux E
As long as the surface encloses both charges, the number of electric field lines leaving the surface is EXACTLY equal to the number of lines entering the surface no matter where the surface is drawn.
A closed surface
q
General Proof (see section 22-6)
Imaginary Closed Surface
204 r
qE
πε=
q
r
For points on the spherical surface
204 r
qE
πε=
∫ AdE.
Note result is independent of r
Point charge q - total flux through a spherical surface
∫= EdA0εq
=22
0
44
rr
q ππε
×=∫= dAE
Imaginary spherical surface
The E-field at dS is:24 r
qE
oπε=
Flux of E out of dS is
πε
ϕ cos4 2
0
dSr
qd =
Total flux is:
€
φE =q
4πε0
dΩ∫
Ω= dq
04πε
0εq
=
q
q
r
€
E .dA∫
Point charge q
0εq
=
€
=q
4πε0r2
× 4πr2
Imaginary spherical surface
€
dr A
The net flux through the closed surface is
€
q
ε0
Gauss’s Law is equivalent to Coulomb’s law, but its use greatly simplifies problems that have a high degree of symmetry. It states that the total electric flux through a closed surface enclosing a total charge Qenclosed situated in a vacuum (air for practical purposes) is:
SUMMARY
€
φE = E .dAsurface
∫ =Qenclosed
ε0
