Download 9/6/16 1 Continuous Charge Distributions: Electric

9/6/16
Continuous Charge Distributions:
Electric Field and Electric Flux III
Lecture 5
Gauss’ Law: Cylindrical Symmetry
Two long, charged concentric cylinders have radii of r1=3.0 cm
and r2=6.0 cm. The charge per unit length is 5.0 x 10-6 C/m
on the inner cylinder and -7.0 x 10-6 C/m on the outer cylinder.
Find the electric field at (a) r = 4.0 cm and (b) r = 8.0 cm,
where r is the radial distance from the common axis.
r
(a)
r1
b
(b)
! !
ε o "∫ E • dA = qenc
!
ε oE "∫ dA = q
E=
ra
r2
(λ + λ )h (λ + λ )
q
= 1 2 = 1 2
2ε o π rh
2ε o π rh
2ε o π r
2π h
ε oE ∫
∫ r dθ dz = ε E2π rh = q
o
E=
0 0
E=
E=
q
λh
λ
=
=
2ε o π rh 2ε o π rh 2ε o π r
−2 x10−6
2(8.55x10−12 ) π (0.08)
= -4.654 x105 N / C
5 x10−6
= 2.327 x10 6 N / C
2(8.55x10−12 ) π (0.04)
Gauss’ Law: Spherical Symmetry
A spherically symmetric charge distribution has a charge
density given by ρ=a/r where a is a constant. Find the electric
field as a function of r.
! !
ε o "∫ E • dA = qenc
2π
π
ε oEr 2 ∫ d φ ∫ sinθ dθ = ε oE4π r 2 = q
0
0
2
r
ε oE4π r = q =
∫ ρ 4π r
0
2
r
dr = a4π
∫
0
r2
4π ar 2
dr =
r
2
a
E=
2ε o
Figure from: http://www.oceanopticsbook.info/view/light_and_radiometry/geometry
1
9/6/16
Thin Sheet of Any Charge Distribution
!" !" !"
ΔE = (E '+ E disk+ )
!" !"
− (E '+ E disk− )
!"
!"
= E disk+ − E disk−
⎛ σ ⎛ σ ⎞⎞
= ⎜⎜
− ⎜−
⎟⎟⎟ n#
⎝ 2ε 0 ⎝ 2ε 0 ⎠⎠
σ
= n#
ε0
tiny disk
!" !" !"
E L = E '+ E disk−
Just to left of disk
Discontinuity of magnitude:
!"
!" !"
E R = E '+ E disk+
ΔEn =
Just to right of disk
σ
ε0
From Lecture 1
Conductors vs. Insulators
•  Insulators: material in which electric charges are
“frozen” in place
•  Conductors: material in which electric charges can
move around “freely
Demo
Conductors
5A-12 Gauss' Law: Charge Within a Conductor
2
9/6/16
Charges on a Conductor
•  Why do charges always move to the surface of a
conductor?
–  Gauss’ Law
Charges on a Conductor
•  Why do charges always move to the surface of a
conductor?
–  Gauss’ Law
•  Charge redistributes inside conductor
•  Cancels external field inside
•  No field across surface => no flux through the surface
•  No net charge density inside conductor
–  Charge must reside on surface
Electric Field in a Conductor
Conducting cylinder
•  Electric field lines for an
oppositely charged metal
cylinder and metal plate.
•  Note that:
1.  Electric field lines are
perpendicular to the conductors
2.  No electric field lines inside the
cylinder
•  Interior shielded from the
outside
Conducting plate
3
9/6/16
Electric Field in a Conductor
•  Key points
–  An external electric field exerts a force on charge carriers which
causes them to move
–  They will continue to move until their electric field negates the
external field
–  The motion stops when the net electric field is zero
= Electrostatic equilibrium
•  Main point to remember
–  The electric field is always zero inside a conductor
•  IF it is in a state of electrostatic equilibrium
Gauss’ Law: Spherical Symmetry
Electric Field inside and outside a shell
of uniform charge distribution
Demo
Conductors: Shielding from an
Electrical Field
student
sensor
sparks
screened cage
Van de Graaff
generator
4
9/6/16
Demo
Charge Distribution on a Conductor
http://cnx.org/content/m42317/latest/Figure_19_07_07a.jpg
Demo
Charge Distribution on a Conductor
Infinite Sheet of Charge
Planar
E=
Qenc
Aε 0
A = 2π r 2
σ
E=
2ε 0
•  An infinite plane sheet of charge creates a
CONSTANT electric field
–  Note: 2x smaller than for conducting plates
5
9/6/16
Conductors: External Electric Field
•  General case
•  Choose a cylindrical Gaussian
surface with one end inside the
conductor
•  Ignore the curvature
–  Cross section of the cylinder is chosen
to be very small
•  The electric field must be
perpendicular to the surface
–  Otherwise, there would be perpetual
currents flowing along the surface
Conductors: External Electric Field
σ = surface charge density
•  Left end cap
!
∫ E dA = 0 because E = 0
n
in a conductor
•  Right end cap
!
E || n̂ ⇒ ∫ EndA = EA
•  Cylinder
∫ EndA = 0!
!
because E ⊥ n̂ and E ⋅ n̂ = 0
Conductors: External Electric Field
∴ !∫ En dA = 0 + EA + 0 = EA
S
qinside = σ A
Gauss' law:
⇒ EA =
!∫
S
En dA =
1
q
ε 0 inside
1
σA
ε0
E=
σ
ε0
6
9/6/16
Two Conducting Plates
•  The electric field near a
charged conducting plate is
given by
-
σ1 σ1 σ1
+
=
2ε 0 2ε 0 ε 0
E=
•  Similarly
E =−
σ1 σ1
σ
−
=− 1
2ε 0 2ε 0
ε0
•  What happens if the two plates
are pulled toward each other?
Two Conducting Plates
•  Excess charge on one plate
attracts the excess charge on
the other plate
–  Surface charge density
•  Vanishes on the outer faces
-
•  Doubled on the inner faces
•  Electric field magnitude
–  2x that near an isolated plate
between the plates
–  0 everywhere else
E=
2σ 1
ε0
•  (c) not a superposition of (a)
and (b)
–  Charges redistributed
Uniform Charge Density: Summary
Non-conductor
Cylindrical
symmetry
Planar
Spherical
symmetry
ρr
E=
2ε 0
E=
E =0
2
R ρ
2ε 0r
σ
2ε 0
1 Q
E=
r
4πε 0 R 3
E=
E=
Conductor
1 Q
4πε 0 r 2
E=
λ
2πε 0r
E=
σ
ε0
E =0
E=
1 Q
4πε 0 r 2
inside
outside
inside
outside
7
9/6/16
Summary of Lectures 3, 4 & 5
•  Gauss’ law relates net flux, F, of an electric field
through a closed surface to the net charge that is
enclosed by the surface
! !
ε oφ = ε o "∫ E • dA = qenc
•  Takes advantage of certain symmetries (spherical,
cylindrical, planar)
•  Gauss’ Law proves electric fields vanish in
conductors while extra charges reside on surface
8
9/6/16
9