Chapter 23: Gauss’ Law
PHY2049: Chapter 23
1
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 around due to force F=qE.
This would contradict “no current.”
Note: even if there is an externally imposed E, it cannot go inside
ÎAll
excess charge must be on outer surface.
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.”
PHY2049: Chapter 23
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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
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Q
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Gaussian surface
r 3 (sphere)
4 3
ρ ( πr ) = Q 3
3
R
3
+
+
+
+
+
+
ρ=
Q
4 3
πR
3
Q
PHY2049: Chapter 23
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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
4
Spherical Symmetry (2)
Concentric Conducting Spherical Shell
ÎNote:
Problem 23-4 on page 614
‹ In
this problem, point charge –Q in cavity is not at center. This
makes the problem harder. In particular, you do not need to
understand explanation given in book for why charge on outer
surface is uniform. Rigorous explanation in fact requires a tool,
so-called uniqueness theorem, which is beyond introductory
physics.
PHY2049: Chapter 23
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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
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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
Gaussian surface
(cylindrical)
PHY2049: Chapter 23
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Rectangular Symmetry (1):
ÎInfinitely
Uniformly Charged Sheet
wide and tall
symmetry, E must be ⊥ , same on both
sides
‹ By
σ (πr 2 )
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
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Some Comparisons
ÎSpherically
symmetric charge distribution
1 qtotal
E=
4πε0 r 2
ÎUniformly
line
charged, infinitely long (i.e., very long)
1 λ
E=
2πε 0 r
ÎUniformly
charged plane
σ
E=
2ε 0
no distance dependence
PHY2049: Chapter 23
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How to calculate E field, if there is no
symmetry?
ÎNumerically
elements dq
integrate dE produced by all charge
ÎUsually
easier to numerically compute electric
potential V first and then E
Electrical potential is subject of Chapter 24
PHY2049: Chapter 23
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