First Beams at LHC!
PHY2049: Chapter 23
1
Gauss’ Law Says
qenc
∫SE ⋅ dA = ε0
ÎTotal
flux depends only on the amount of
enclosed charge, not on shape and size of the
Gaussian surface.
ÎCharges outside have no effect on the total
flux.
ÎThis does not mean they do not contribute to
E.
Remember:
‹Outward E field, flux > 0
‹Inward E field, flux < 0
PHY2049: Chapter 23
2
Conductors with No Current
ÎE
is zero everywhere inside
Why? Conductors are full of mobile charges (e.g., conduction electrons in a
background formed by immobile positive ions). If there were E, then the
charges must be moving because of the force F=qE. This would contradict
“no current.”
Note: even if there is an externally imposed E, it cannot go inside.
ÎAll
excess charge and induced charge must be on
surfaces
Why? Since E=0 everywhere inside, qenc enclosed by any Gaussian surface is
also zero everywhere inside.
Note: distribution of surface charge must be such to make E=0 everywhere
inside.
ÎE
is always normal to surface on conductor
Why? E component parallel to surface would cause surface charge to move.
This would contradict “no current.”
Note: distribution of surface charge must be such to make E normal.
PHY2049: Chapter 23
3
Use Gauss’ Law to Calculate E Fields
ÎSpherical
symmetry
‹E
field vs r inside uniformly charged sphere
‹Charges on concentric spherical conducting shells
ÎCylindrical
symmetry
‹E
field vs r for line charge
‹E field vs r inside uniformly charged cylinder
ÎRectangular
symmetry
‹E
field for charged plane
‹E field between conductors, e.g. capacitors
PHY2049: Chapter 23
4
Spherical Symmetry (1)
Uniformly Charged Sphere
ÎInsulator
or conductor?
‹ Conducting
sphere cannot be uniformly charged
ÎInside
symmetry, E must be radially symmetric
‹ E field has constant mag., ⊥ to Gaussian surface
+
+
+
+
‹ By
Gauss’ Law
qenc
∫SE ⋅ dA = E (4πr ) = ε0
2
3
Qr / R
1 Qr
E=
=
2
4πε 0 R 3
4πε 0 r
ÎOutside
Solve for E
1 Q
E=
4πε 0 r 2
Q
PHY2049: Chapter 23
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Gaussian surface
r 3 (sphere)
4 3
ρ ( πr ) = Q 3
3
R
3
+
+
+
+
+
+
ρ=
Q
4 3
πR
3
Volume charge
density
5
Spherical Symmetry (2)
Concentric Conducting Spherical Shell
ÎInside conductor
‹E
Must be 0
must be 0
qenc
∫SE ⋅ dA = 0 = ε0
+
–
Gauss’ Law
‹ Charge can be only on surfaces
+Q uniformly distributed on inner wall
ÎOutside
‹ By
symmetry, E must be radially symmetric
‹ E field has constant mag., ⊥ to Gaussian surface
q enc
∫SE ⋅ dA = E (4πr ) = ε0
2
–
–
−Q
Gauss’ Law
– Q uniformly distributed on outer surface
PHY2049: Chapter 23
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–
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-Q
+
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–
+
–
–
+
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–
–
+
+
–
–
Gaussian surface
(sphere)
Points toward shell
1 Q
E=−
4πε 0 r 2
6
Axial Symmetry (1): Line Charge
ÎInfinitely
long line, uniformly charged
‹ By
symmetry, E must be axially symmetric
‹ On curved surface, E field has constant
mag., ⊥ to Gaussian surface
‹ Through top and bottom surfaces, no ΦE
since E is ||
λh
qenc
∫SE ⋅ dA = E (h2πr ) + 0 + 0 = ε0
Gauss’ Law
1 λ
E=
2πε 0 r
Solve for E
PHY2049: Chapter 23
λ: linear charge density
7
Axial Symmetry (2): Uniformly Charged Cylinder
ÎInfinitely
tall cylinder, uniformly charged
R
‹ By
symmetry, E must be axially symmetric
‹ On curved surface, E field has constant mag.,
⊥ to Gaussian surface
‹ Through top and bottom surfaces, no ΦE since
E is ||
2
ρ
r
ρ ( hπr )
qenc
∫SE ⋅ dA = E (h2πr ) + 0 + 0 = ε0
1
E=
ρr Solve for E
2ε 0
h
Gauss’ law
ρ: volume
charge density
PHY2049: Chapter 23
Gaussian surface
(cylindrical)
8
Rectangular Symmetry (1):
ÎInfinitely
Uniformly Charged Sheet
wide and tall
symmetry, E must be ⊥ , same on both
sides
‹ By
σ (πr 2 )
σ: surface
charge density
q enc
∫SE ⋅ dA = E (πr ) + E (πr ) + 0 + 0 = ε0
2
σ
E=
2ε 0
ÎParallel
23-8
2
Solve for E
Gauss’ law
Constant!
conducting plates: read Section
PHY2049: Chapter 23
9
Some Comparisons
ÎSpherically
symmetric charge distribution
1 qtotal
E=
4πε0 r 2
ÎUniformly
long) line
ÎUniformly
charged, infinitely long (i.e., very
1 λ
E=
2πε 0 r
charged plane
σ
E=
2ε 0
No distance dependence!
PHY2049: Chapter 23
10