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Gauss’s Law in Pictures
Branislav K. Nikolić
Department of Physics and Astronomy, University of Delaware, U.S.A.
PHYS 208 Honors: Fundamentals of Physics II
http://www.physics.udel.edu/~bnikolic/teaching/phys208/phys208.html
Symmetry Operations
A charge distribution is symmetric if there is a set of
geometrical transformation that do not cause any physical change.
PHYS 208 Honors: Gauss’s Law
Symmetry of Charge Distribution vs.
Symmetry of Electric Field
The symmetry of the electric field must match the symmetry of
the charge distribution
PHYS 208 Honors: Gauss’s Law
Symmetry of Charge Distribution vs.
Symmetry of Electric Field
The symmetry of the electric
field must match the
symmetry of the charge
distribution
PHYS 208 Honors: Gauss’s Law
Electric Field Pattern of Cylindrically
Symmetric Charge Distribution
PHYS 208 Honors: Gauss’s Law
Electric Field Pattern of Charge
Distributions with Basic Symmetries
PHYS 208 Honors: Gauss’s Law
Example: Problem 27.1
The electric field of a
cylindrically symmetric charge
distribution cannot have a
component parallel to the
cylinder axis. Also, the
electric field of a cylindrically
symmetric charge distribution
cannot have a component
tangent to the circular cross
section. The only shape for
the electric field that
matches the symmetry of the
charge distribution with
respect to (i) translation
parallel to the cylinder axis,
(ii) rotation by an angle
about the cylinder axis, and
(iii) reflections in any plane
containing or perpendicular
to the cylinder axis is the
one shown in the figure.
PHYS 208 Honors: Gauss’s Law
Example: Problem 27.2
PHYS 208 Honors: Gauss’s Law
Example: Problem 27.3
PHYS 208 Honors: Gauss’s Law
The Concept of Flux of Electric Field
PHYS 208 Honors: Gauss’s Law
Gaussian surface which does not match
symmetry of charge is not useful!
The electric field pattern through the surface is particularly simple if the
closed surface matches the symmetry of the charge distribution inside.
PHYS 208 Honors: Gauss’s Law
Calculus for the Flux of Vector Fields
Motivation for the definition: Flux of a fluid flow
Flux of uniform electric field passing through a surface:
PHYS 208 Honors: Gauss’s Law
Flux of a Nonuniform Electric Field
Φ e = ∑ δ Φ i = ∑ E ⋅ (δ A)i
N
N
i =1
i =1 N →∞ , δ A → dA
Φ e = ∫ E ⋅ dA
surface
Simplified situations where integration goes away:
PHYS 208 Honors: Gauss’s Law
Gauss’s Law For Point Charge
Φe = ∫ E ⋅ dA = EASphere = E 4π r 2
1
q
q
2
Φe =
4π r =
2
ε0
4πε 0 r
Electric flux
is
independent
of surface
shape:
PHYS 208 Honors: Gauss’s Law
dΩ
dΦe = EdAcosα =
=
1 q
q
cos
dA
α
=
dΩ
4πε0 r2
4πε0
Gauss’s Law For Charge Distribution =
First Maxwell Equation
q1 q2
Qenclosed
Φe = E
⋅
dA
=
+
+
…
=
∫
ε0 ε0
ε0
Unlike Coulomb’ law for static point charges, Gauss’s law is valid for
moving charges and fields that change with time.
PHYS 208 Honors: Gauss’s Law
Applications: Charged Sphere
1. Determine the symmetry of the electric field.
2. Select a Gauss surface (as imaginary surface in the space surrounding the
charge) to match the symmetry of the field.
3. If 1. and 2. are possible, flux integral should become algebraic product
electric field x Gauss surface area.
Q
Φe = ∫ E ⋅ dA = EAsphere = inside
Q
Φ e = ∫ E ⋅ dA = E outside Asphere =
ε0
ε0
E outside 4π r =
2
Q
ε0
⇒ E outside =
q
4πε 0 r
PHYS 208 Honors: Gauss’s Law
2
rˆ
4 r 3πρ 4 r 3π
Einside 4π r 2 = 3
= 3
ε0
ε0
Q
4 R3π
3
⇒ Einside =
Q
4πε 0 R
3
rrˆ
Applications: Charged Wire
Φe =
∫
E ⋅ d A = Φ to p +Φ b o tto m + Φ w a ll = 0 + 0 + E Acylin d er
E 2 π rL =
Q in
ε0
PHYS 208 Honors: Gauss’s Law
Q
=
λL
L = λ
⇒ E w ire =
ε0
2 πε 0 r
2 πε 0 r
Applications: Charged Plane
Φ
EA =
Q in
ε0
PHYS 208 Honors: Gauss’s Law
e
=
∫
E ⋅ d A = 2E A
ηA
η
=
⇒ E plane =
ε0
2ε 0
Conductors in Electrostatic Equilibrium
BASIC FACT: Electric field is zero at all points within the conductor – otherwise
charges will flow, thereby violating electrostatic equilibrium.
Φe =
∫
Q inside
E ⋅ dA =
⇒ Q inside ≡ 0
ε0
Any excess charge resides
entirely on the exterior surface.
PHYS 208 Honors: Gauss’s Law
Any component of the electric field
tangent to the conductor surface
would cause surface charges to move
thereby violating the assumuption
that all charges are at rest.
Gauss’s Law Determines Electric Field
Outside of Metal
ηA
η
Φe = Esurface A =
=
⇒ Esurface = nˆ
ε0 ε0
ε0
Qin
PHYS 208 Honors: Gauss’s Law
Electric Field of Conducting Shells
+
- - -
∫
+ +
closed path
PHYS 208 Honors: Gauss’s Law
qE ⋅ ds ≠ 0
Applications: Faraday’s Cage
PHYS 208 Honors: Gauss’s Law
Example: Electric Force Acting on
the Surface of Metallic Conductor
Esurface
Erest
Erest
Eδ A
Eδ A
E = 0 ⇒ Eδ A = Erest
δA
Eδ A
E =0
η
=
2ε 0
Esurface = Eδ A + Erest = 2Erest ⇒ Erest =
Fδ A = ηδ A
E surface
2
Esurface
2
Fδ A 1 ⇔ f =
= η E surface
δA 2
1
ε0
2
2
f = ε 0 E surface nˆ ⇒ Fconductor =
E surface dA
∫
2
2
NOTE: Regardless
of the sign of surface charge density η and corresponding direction of the
electric field Esurface
, the forceFsurface is always directed outside of the conductor tending to stretch it.
PHYS 208 Honors: Gauss’s Law