Download Lecture: Gauss Law

Chapter 22
Gauss’s Law
Lecture 4
Dr. Armen Kocharian
Field Due to a Plane of
Charge
„
„
E must be
perpendicular to the
plane and must have
the same magnitude at
all points equidistant
from the plane
Choose a small cylinder
whose axis is
perpendicular to the
plane for the gaussian
surface
qin σA
Φ E = 2EA =
=
εo
εo
σA
σ
2EA =
⇒E =
2εo
εo
Field Due to a Plane of
Charge, cont
„
„
E is parallel to the curved surface and
there is no contribution to the surface
area from this curved part of the
cylinder
The flux through each end of the
cylinder is EA and so the total flux is
2EA
Field Due to a Plane of
Charge, final
„
„
The total charge in the surface is σA
Applying Gauss’s law
σA
σ
Φ E = 2EA =
and E =
2εo
εo
„
„
Note, this does not depend on r
Therefore, the field is uniform
everywhere
Electrostatic Equilibrium
„
When there is no net motion of charge
within a conductor, the conductor is said
to be in electrostatic equilibrium
Properties of a Conductor in
Electrostatic Equilibrium
„
„
„
„
The electric field is zero everywhere inside
the conductor
If an isolated conductor carries a charge, the
charge resides on its surface
The electric field just outside a charged
conductor is perpendicular to the surface and
has a magnitude of σ/εo
On an irregularly shaped conductor, the
surface charge density is greatest at locations
where the radius of curvature is the smallest
Property 1: Einside = 0
„
„
„
„
„
Consider a conducting slab in
an external field E
If the field inside the conductor
were not zero, free electrons in
the conductor would
experience an electrical force
These electrons would
accelerate
These electrons would not be
in equilibrium
Therefore, there cannot be a
field inside the conductor
Property 1: Einside = 0, cont.
„
„
„
„
Before the external field is applied, free
electrons are distributed throughout the
conductor
When the external field is applied, the
electrons redistribute until the magnitude of
the internal field equals the magnitude of the
external field
There is a net field of zero inside the
conductor
This redistribution takes about 10-15s and can
be considered instantaneous
Property 2: Charge Resides
on the Surface
„
„
„
„
Choose a gaussian surface
inside but close to the actual
surface
The electric field inside is
zero (prop. 1)
There is no net flux through
the gaussian surface
Because the gaussian
surface can be as close to
the actual surface as
desired, there can be no
charge inside the surface
Property 2: Charge Resides
on the Surface, cont
„
„
Since no net charge can be inside the
surface, any net charge must reside on
the surface
Gauss’s law does not indicate the
distribution of these charges, only that it
must be on the surface of the conductor
Property 3: Field’s Magnitude
and Direction
„
„
Choose a cylinder as
the gaussian surface
The field must be
perpendicular to the
surface
„
If there were a parallel
component to E,
charges would
experience a force and
accelerate along the
surface and it would
not be in equilibrium
Property 3: Field’s Magnitude
and Direction, cont.
„
The net flux through the gaussian
surface is through only the flat face
outside the conductor
„
„
The field here is perpendicular to the
surface
Applying Gauss’s law
σA
σ
Φ E = EA =
and E =
εo
εo
Conductors in Equilibrium,
example
„
„
The field lines are
perpendicular to
both conductors
There are no field
lines inside the
cylinder
Example
„
Find the field in regions
1, 2, 3 and 4
Φ E = EA
„
Find in region 1
(
E1 A = E1 4π r
qin = 0
E1 = 0 ( r ≤ a )
2
„
)= ε
qin
0
,
Field in region 2
(
E2 A = E2 4π r
2
)= ε
qin
0
qin = 2Q
E2 =
2Q
4πε 0 r
2
=
2 ke Q
r
2
(a ≤ r ≤ b)
Example
„
Find the field in regions
1, 2, 3 and 4
Φ E = EA
„
Field in region 3
(
)
E3 A = E3 4π r 2 =
qin = 0
E3 = 0 ( b ≤ r ≤ c )
„
qin
ε0
,
Field in region 4
(
E4 A = E4 4π r
2
)= ε
qin
qin = 2Q + ( −Q ) = Q
E4 =
Q
4πε 0 r
2
=
ke Q
r
2
,
0
(r ≥ c)
Example
„
Electrical field is zero
inside conductors
E1 = 0 ( r ≤ a )
E3 = 0 ( b ≤ r ≤ c )
„
Field exist outside of
conductors
E2 =
E4 =
2 ke Q
2
r
Q
4πε 0 r
(a ≤ r ≤ b)
2
=
ke Q
r
2
(r ≥ c)
Derivation of Gauss’s Law
„
„
„
We will use a solid
angle, Ω
A spherical surface
of radius r contains
an area element ΔA
The solid angle
subtended at the
center of the sphere
ΔA
is defined to be Ω = 2
r
Some Notes About Solid
Angles
„
„
„
A= 4πr2 and r2 have the same units, so
Ω is a dimensionless ratio
We give the name steradian to this
dimensionless ratio
The total solid angle subtended by a
sphere is 4π steradians
Derivation of Gauss’s Law,
cont.
„
„
Consider a point
charge, q, surrounded
by a closed surface of
arbitrary shape
The total flux through
this surface can be
found by evaluating
E.ΔA for each small
area element and
summing over all the
elements
Derivation of Gauss’s Law,
final
„
„
The flux through each element is
ΔA cos θ
Φ E = E ⋅ ΔA = ( E cos θ ) ΔA = keq
r2
Relating to the solid angle
ΔA cos θ
ΔΩ =
r2
„ where this is the solid angle subtended by
ΔA
„
The total flux is
Φ E = keq v
∫
dA cos θ
q
=
k
q
d
Ω
=
e v
∫
r2
εo
Derivation of Gauss’s Law,
cont.
(a)
„
qin > 0
qin = +3Q − Q = +2Q
The charge distribution is
spherically symmetric and
directed radially outward.
.
E=
keqin
2
r
=
2keQ
2
r
r≥ c
Insulating sphere 3Q within
conducting shell -Q. Prob. 55
(d)
„
qin = 0
b < r< c
Since all points within this
region are located inside
conducting material, E=0.
.
Φ E = ∫ E ⋅ dA = 0 ⇒ qin =∈0 Φ E = 0
Insulating sphere 3Q within
conducting shell –Q, cont.
(f)
E=
„
qin = +3Q
keqin
2
r
=
3keQ
2.
a≤ r< b
r
Directed radially outward.
Insulating sphere 3Q within
conducting shell –Q, cont.
(h)
⎛ +3Q ⎞ ⎛ 4 3⎞
r3
qin = ρV = ⎜ 4 3 ⎟ ⎜ π r ⎟ = +3Q 3
a
⎝ 3π a ⎠ ⎝ 3 ⎠
.
E=
„
keqin
r2
ke ⎛
r3 ⎞
r
= 2 ⎜ +3Q 3 ⎟ = 3keQ 3
r ⎝
a ⎠
a
Directed radially outward.
Problem
(a)
The inner charge is
+q
on the inner surface of the conductor,
−q
where its surface density is:
„
Inner surface density
σa =
−q
4π a2
.
(b)
The outer surface carries charge
„
Q +q
Outer surface density
.
σb =
Q +q
4π b2
Derivation of Gauss’s Law –
Notes
„
„
„
This result matches the earlier result for
the equation for Gauss’s law
This derivation was independent of the
shape of the closed surface
This derivation was independent of the
position of the charge within the surface
Problem
Find the flux through
the circular cap
„
2π r = 2π R sin θ
Φ E = EA
ds = Rdθ
θ
θ
θ
A= ∫ 2π rds = ∫ ( 2π R sinθ ) Rdθ = 2π R ∫ sinθ dθ = 2π R ( − cosθ ) = 2π R2 (1− cosθ )
2
0
0
ΦE =
2
0
1 Q
Q
2
π
θ
⋅
2
R
1
−
cos
=
(
)
( 1− cosθ )
2
4π ∈0 R
2 ∈0
Electric Flux
„
„
„
Field lines
penetrating an area
A perpendicular to
the field
The product of EA is
the flux, Φ
In general:
„
ΦE = E A sin θ
Assessment Test
Chapter 24
A uniformly charged rod has a finite
length L. The rod is symmetric under
rotations about the axis and under
reflection in any plane containing the
axis. It is not symmetric under
translations or under reflections in a
plane perpendicular to the axis other
than the plane that bisects the rod.
Which field shape or shapes match the
symmetry of the rod?
A. a and d
B. c and e
C. b only
D. e only
E. none of the above
A uniformly charged rod has a finite
length L. The rod is symmetric under
rotations about the axis and under
reflection in any plane containing the
axis. It is not symmetric under
translations or under reflections in a
plane perpendicular to the axis other
than the plane that bisects the rod.
Which field shape or shapes match the
symmetry of the rod?
A. a and d
B. c and e
C. b only
D. e only
E. none of the above
This box contains
A. a net positive charge.
B. no net charge.
C. a net negative charge.
D. a positive charge.
E. a negative charge.
This box contains
A. a net positive charge.
B. no net charge.
C. a net negative charge.
D. a positive charge.
E. a negative charge.
The total electric flux through this box is
A. 0 Nm2/C.
B. 1 Nm2/C.
C. 2 Nm2/C.
D. 4 Nm2/C.
E. 6 Nm2/C.
The total electric flux through this box is
A. 0 Nm2/C.
B. 1 Nm2/C.
C. 2 Nm2/C.
D. 4 Nm2/C.
E. 6 Nm2/C.
These are two-dimensional cross sections through threedimensional closed spheres and a cube. Rank order, from
largest to smallest, the electric fluxes a to e through
surfaces a to e.
A.
B.
C.
D.
E.
Φa > Φc > Φb > Φd > Φe
Φb = Φe > Φa = Φc = Φd
Φe > Φd > Φb > Φc > Φa
Φb > Φa > Φc > Φe > Φd
Φd = Φe > Φc > Φa = Φb
These are two-dimensional cross sections through threedimensional closed spheres and a cube. Rank order, from
largest to smallest, the electric fluxes a to e through
surfaces a to e.
Which Gaussian surface would allow you to use Gauss’s
law to determine the electric field outside a uniformly
charged cube?
A. A sphere whose center coincides with the center
of the charged cube.
B. A cube whose center coincides with the center of
the charged cube and which has parallel faces.
C. Either A or B.
D. Neither A nor B.
Which Gaussian surface would allow you to use Gauss’s
law to determine the electric field outside a uniformly
charged cube?
A. A sphere whose center coincides with the center
of the charged cube.
B. A cube whose center coincides with the center of
the charged cube and which has parallel faces.
C. Either A or B.
D. Neither A nor B.
Chapter 24
Reading Quiz
The amount of electric field passing through a
surface is called
A. Electric flux.
B. Gauss’s Law.
C. Electricity.
D. Charge surface density.
E. None of the above.
The amount of electric field passing through a
surface is called
A. Electric flux.
B. Gauss’s Law.
C. Electricity.
D. Charge surface density.
E. None of the above.
Gauss’s law is useful for calculating electric fields that are
A. due to point charges.
B. uniform.
C. symmetric.
D. due to continuous charges.
Gauss’s law is useful for calculating electric fields that are
A. due to point charges.
B. uniform.
C. symmetric.
D. due to continuous charges.
Gauss’s law applies to
A. lines.
B. flat surfaces.
C. spheres only.
D. closed surfaces.
Gauss’s law applies to
A. lines.
B. flat surfaces.
C. spheres only.
D. closed surfaces.
The electric field inside a conductor in
electrostatic equilibrium is
A. uniform.
B. zero.
C. radial.
D. symmetric.
The electric field inside a conductor in
electrostatic equilibrium is
A. uniform.
B. zero.
C. radial.
D. symmetric.