Chapter 23: Gauss’ Law
PHY2049: Chapter 23
1
Electric Flux
ÎSimple
‹E
definition of electric flux (E constant, flat surface)
at an angle θ to planar surface, area A
Φ E ≡ E ⋅ A = EA cos θ
‹ Units
ÎSimple
= N m2 / C (SI units)
Normal
E
example
‹ Let
E = 104 N/C pass through 2m x 5m rectangle, 30° to normal
‹ φE = 104 * 10 * cos(30) = 100,000 * 0.866 = 86,600
ÎMore
general ΦE definition (E variable, curved surface)
Φ E ≡ ∫ E ⋅ dA
S
PHY2049: Chapter 23
2
Example of Constant Field
(
)
A = 2iˆ + 3 ˆj m 2
E = 4iˆ
(
)
Φ E ≡ E ⋅ A = 2iˆ + 3 ˆj ⋅ 4iˆ = 8
PHY2049: Chapter 23
3
Flux Through Closed Surface
ÎSurface
elements dA
always point outward!
ÎSign
of ΦE
‹E
outward (+)
‹ E inward (−)
ΦE < 0
ΦE > 0
ΦE = 0
PHY2049: Chapter 23
4
Example: Flux Through Cube
G
ÎE field is constant: E = Ezˆ
‹ Flux
through
‹ Flux through
‹ Flux through
‹ Flux through
front face?
back face?
top face?
whole cube?
PHY2049: Chapter 23
5
Example: Flux Through Cylinder
ÎAssume
E is constant, to the right
‹ Flux
through left face?
‹ Flux through right face?
‹ Flux through curved side
‹ Total flux through cylinder?
PHY2049: Chapter 23
6
Example: Flux Through Sphere
ÎAssume
ÎE
point charge +Q
points radially outward (normal to surface!)
Φ E = ∫ E ⋅ dA
S
(
⎛ kQ ⎞
= ⎜ 2 ⎟ 4π r 2
⎝r ⎠
Q
= 4π kQ =
)
ε0
Foreshadowing of Gauss’ Law!
PHY2049: Chapter 23
7
Gauss’ Law
ÎGeneral
statement of Gauss’ law
qenc
Integration over closed surface
E
⋅
A
=
d
v∫ S
qenc is charge inside the surface
ε
0
ÎCan
Charges outside surface have no effect
be used to calculate E fields. But remember
‹ Outward
E field, flux > 0
‹ Inward E field, flux < 0
ÎConsequences
of Gauss’ law (as we shall see)
‹E
= 0 inside conductor
‹ E is always normal to surface on conductor
‹ Excess charge on conductor is always on surface
PHY2049: Chapter 23
Conductor
8
Reading Quiz
ÎWhat
is the electric flux through a sphere of radius R
surrounding a charge +Q at the center?
‹ 1)
‹ 2)
‹ 3)
‹ 4)
‹ 5)
0
+Q/ε0
−Q/ε0
+Q
None of these
PHY2049: Chapter 23
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Question
dS
dS
+Q
+Q
2
1
How does the flux ФE through the entire surface
change when the charge +Q is moved from position
1 to position 2?
a) ФE increases
ФE decreases
c) ФE doesn’t change
b)
PHY2049: Chapter 23
Just depends on charge
not position
10
Power of Gauss’ Law: Calculating E Fields
ÎValuable
for cases with high symmetry
‹ E = constant, ⊥ surface
v∫ E ⋅ dA = EA
S
‹E
|| surface
ÎSpherical
symmetry
v∫ S E ⋅ dA = 0
‹E
field vs r for point charge
‹ 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
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Example
Î4
Gaussian surfaces: 2 cubes and 2 spheres
ÎRank
magnitudes of E field on surfaces
ÎWhich
ÎWhat
ones have variable E fields?
are the fluxes over each of the Gaussian surfaces
(a) E falls as radius increases
(b) E non-constant on cube (r changes)
(c) Fluxes are same, +Q/ε0
PHY2049: Chapter 23
12
Derive Coulomb’s Law From Gauss’ Law
ÎCharge
‹ By
ÎDraw
+Q at a point
symmetry, E must be radially symmetric
Gaussian’ surface around point
r
‹ Sphere
of radius r
‹ E field has constant mag., ⊥ to Gaussian surface
v∫ S E ⋅ dA = E ( 4π r
E=
Q
4πε 0 r
2
=
kQ
r
2
2
)=ε
Q
Gaussian surface
(sphere)
Gauss’ Law
0
Solve for E
PHY2049: Chapter 23
13
Example
ÎCharges
on shells are
‹ +Q
(ball at center)
‹ +3Q (middle shell)
‹ +5Q (outside shell)
ÎFind
fluxes on the three Gaussian surfaces
(a) Inner +Q/ε0
(b) Middle +4Q/ε0
(c) Outer +9Q/ε0
PHY2049: Chapter 23
14