Class 05: Outline
Hour 1:
Gauss’ Law
Hour 2:
Gauss’ Law
P05 - 1
Six PRS Questions On
Pace and Preparation
P05 - 2
Last Time:
Potential and E Field
P05 - 3
E Field and Potential: Creating
A point charge q creates a field and potential around it:
G
q
q
E = ke 2 rˆ ; V = ke
r
r
Use superposition for
systems of charges
They are related:
G
B G
G
E = −∇ V ; ∆ V ≡ VB − VA = − ∫ E ⋅ d s
A
P05 - 4
E Field and Potential: Effects
G
G
F = qE
If you put a charged particle, q, in a field:
To move a charged particle, q, in a field:
W = ∆U = q∆V
P05 - 5
Two PRS Questions:
Potential & E Field
P05 - 6
Gauss’s Law
The first Maxwell Equation
A very useful computational technique
This is important!
P05 - 7
Gauss’s Law – The Idea
The total “flux” of field lines penetrating any of
these surfaces is the same and depends only
on the amount of charge inside
P05 - 8
Gauss’s Law – The Equation
G G qin
Φ E = ∫∫ E ⋅ dA =
closed
surfaceS
ε0
Electric flux ΦE (the surface integral of E over
closed surface S) is proportional to charge
inside the volume enclosed by S
P05 - 9
Now the Details
P05 -10
Electric Flux ΦE
Case I: E is constant vector field
perpendicular to planar surface S of area A
G G
Φ E = ∫∫ E ⋅ dA
Φ E = + EA
Our Goal: Always reduce
problem to this
P05 -11
Electric Flux ΦE
Case II: E is constant vector field directed
at angle θ to planar surface S of area A
G G
Φ E = ∫∫ E ⋅ dA
Φ E = EA cos θ
P05 -12
PRS Question:
Flux Thru Sheet
P05 -13
Gauss’s Law
G G qin
Φ E = ∫∫ E ⋅ dA =
closed
surfaceS
ε0
Note: Integral must be over closed surface
P05 -14
Open and Closed Surfaces
A rectangle is an open surface — it does NOT contain a volume
A sphere is a closed surface — it DOES contain a volume
P05 -15
Area Element dA: Closed Surface
For closed surface, dA is normal to surface
and points outward
( from inside to outside)
ΦE > 0 if E points out
ΦE < 0 if E points in
P05 -16
Electric Flux ΦE
Case III: E not constant, surface curved
dA
S
E
G G
dΦE = E⋅ dA
Φ E = ∫∫ dΦ E
P05 -17
Example: Point Charge
Open Surface
P05 -18
Example: Point Charge
Closed Surface
P05 -19
PRS Question:
Flux Thru Sphere
P05 -20
Electric Flux: Sphere
Point charge Q at center of sphere, radius r
E field at surface:
G
E=
Q
4π ε 0 r
2
rˆ
Electric flux through sphere:
G G
ΦE = w
∫∫
∫∫ E ⋅ dA = w
S
S
=
Q
4πε 0 r
2
w
∫∫ dA
S
=
Q
4πε 0 r
Q
4πε 0 r
2
rˆ ⋅ dA rˆ
4π r =
2
2
Q
ε0
G
dA = dA rˆ
P05 -21
Arbitrary Gaussian Surfaces
G G Q
Φ E = ∫∫ E ⋅ dA =
closed
surface S
ε0
For all surfaces such as S1, S2 or S3
P05 -22
Applying Gauss’s Law
1. Identify regions in which to calculate E field.
2. Choose Gaussian surfaces S: Symmetry
G G
3. Calculate Φ E = ∫∫ E ⋅ dA
S
4. Calculate qin, charge enclosed by surface S
5. Apply Gauss’s Law to calculate E:
G G qin
Φ E = ∫∫ E ⋅ dA =
closed
surfaceS
ε0
P05 -23
Choosing Gaussian Surface
Choose surfaces where E is perpendicular & constant.
Then flux is EA or -EA.
Choose surfaces where E is parallel.
Then flux is zero
OR
Example: Uniform Field
E
EA
Flux is EA on top
Flux is –EA on bottom
Flux is zero on sides
− EA
P05 -24
Symmetry & Gaussian Surfaces
Use Gauss’s Law to calculate E field from
highly symmetric sources
Symmetry
Gaussian Surface
Spherical
Concentric Sphere
Cylindrical
Coaxial Cylinder
Planar
Gaussian “Pillbox”
P05 -25
PRS Question:
Should we use Gauss’ Law?
P05 -26
Gauss: Spherical Symmetry
+Q uniformly distributed throughout non-conducting
solid sphere of radius a. Find E everywhere
P05 -27
Gauss: Spherical Symmetry
Symmetry is Spherical
G
E = E rˆ
Use Gaussian Spheres
P05 -28
Gauss: Spherical Symmetry
Region 1: r > a
Draw Gaussian Sphere in Region 1 (r > a)
Note: r is arbitrary
but is the radius for
which you will
calculate the E field!
P05 -29
Gauss: Spherical Symmetry
Region 1: r > a
Total charge enclosed qin = +Q
G G
ΦE = w
∫∫ dA = EA
∫∫ E ⋅ dA = E w
S
(
= E 4π r 2
)
Φ E = 4π r E =
2
E=
Q
4πε 0 r
2
S
qin
ε0
=
G
⇒E=
Q
ε0
Q
4πε 0 r
2
rˆ
P05 -30
Gauss: Spherical Symmetry
Region 2: r < a
Total charge enclosed:
⎛4 3⎞
3
⎜ πr ⎟
⎞
⎛
r
3
⎟Q = ⎜⎜ 3 ⎟⎟Q
qin = ⎜
a ⎠
⎜ 4 πa 3 ⎟
⎝
⎜
⎟
⎝3
⎠
OR qin = ρV
Gauss’s law:
3
⎛
qin
r ⎞Q
2
Φ E = E ( 4π r ) =
=⎜ 3⎟
ε0 ⎝ a ⎠ ε0
G
Q r
Q r
E=
⇒E=
rˆ
3
3
4πε 0 a
4πε 0 a
P05 -31
PRS Question:
Field Inside Spherical Shell
P05 -32
Gauss: Cylindrical Symmetry
Infinitely long rod with uniform charge density λ
Find E outside the rod.
P05 -33
Gauss: Cylindrical Symmetry
Symmetry is Cylindrical
G
E = E rˆ
Use Gaussian Cylinder
Note: r is arbitrary but is
the radius for which you
will calculate the E field!
A is arbitrary and should
divide out
P05 -34
Gauss: Cylindrical Symmetry
Total charge enclosed: qin = λA
G G
ΦE = w
∫∫ E ⋅ dA = E w
∫∫ dA = EA
S
S
λA
= E ( 2π r A ) =
=
ε0 ε0
qin
G
λ
λ
E=
⇒E=
rˆ
2πε 0 r
2πε 0 r
P05 -35
Gauss: Planar Symmetry
Infinite slab with uniform charge density σ
Find E outside the plane
P05 -36
Gauss: Planar Symmetry
Symmetry is Planar
G
E = ± E xˆ
Use Gaussian Pillbox
x̂
Note: A is arbitrary (its
size and shape) and
should divide out
Gaussian
Pillbox
P05 -37
Gauss: Planar Symmetry
Total charge enclosed: qin = σA
NOTE: No flux through side of cylinder, only endcaps
G G
ΦE = w
∫∫ E ⋅ dA = E w
∫∫ dA = EAEndcaps
S
S
σA
= E ( 2 A) =
=
ε0 ε0
qin
{
G σ xˆ to right
σ
E=
⇒E=
2ε 0
2ε 0 -xˆ to left
}
G
E
+
+
+
+
+
+
+
+
+
+
+
+
x
G A
E
σ
P05 -38
PRS Question:
Slab of Charge
P05 -39
Group Problem: Charge Slab
Infinite slab with uniform charge density ρ
Thickness is 2d (from x=-d to x=d).
Find E everywhere.
x̂
P05 -40
PRS Question:
Slab of Charge
P05 -41
Potential from E
P05 -42
Potential for Uniformly Charged
Non-Conducting Solid Sphere
From Gauss’s Law
⎧ Q ˆ
r, r > R
2
⎪
G ⎪ 4πε 0 r
E=⎨
⎪ Qr rˆ , r < R
⎪⎩ 4πε 0 R 3
G G
Use VB − VA = − ∫ E ⋅ d s
B
A
Region 1: r > a
Point Charge!
VB − V ( ∞ ) = − ∫
∞
=0
r
Q
1
Q
dr =
2
4πε 0 r
4πε 0 r
P05 -43
Potential for Uniformly Charged
Non-Conducting Solid Sphere
Region 2: r < a
R
r
VD − V ( ∞ ) = − ∫∞ drE ( r > R ) − ∫R drE ( r < R )
=0
R
Q
Qr
= − ∫ dr
− ∫ dr
2
R
∞
4πε 0 r
4πε 0 R 3
1
r
(
Q
1 Q 1 2
2
=
−
−
r
R
4πε 0 R 4πε 0 R 3 2
)
1 Q⎛
r2 ⎞
=
⎜ 3− 2 ⎟
R ⎠
8πε 0 R ⎝
P05 -44
Potential for Uniformly Charged
Non-Conducting Solid Sphere
P05 -45
Group Problem: Charge Slab
Infinite slab with uniform charge density ρ
Thickness is 2d (from x=-d to x=d).
If V=0 at x=0 (definition) then what is V(x) for x>0?
x̂
P05 -46
Group Problem: Spherical Shells
These two spherical
shells have equal
but opposite charge.
Find E everywhere
Find V everywhere
(assume V(∞) = 0)
P05 -47