2. Electric Flux – Case 1
Easiest case:
• The E-field is uniform
• The plane is perpendicular to
the field
r
ΦE ≡ E A
Electric Flux
3. Electric Flux – Case 1
Easiest case:
• The E-field is uniform
• The plane is perpendicular to
the field
r
ΦE ≡ E A
Electric Flux
Flux depends on how strong
the E-field is and how big the
area is.
4. Electric Flux – Case 2
r
E
most general case:
• The E-field is not uniform
r
E⊥
•
θ
∆A
A
The surface is curvy and is not
perpendicular to the field
5. Electric Flux – Case 2
r
E
Imagine the surface A is a mosaic of little
tiny surfaces ΔA.
r
E⊥
θ
Pretend that each little ΔA is so small that
it is essentially flat.
∆A
A
6. Electric Flux – Case 2
r
E
r
E⊥
θ
Then, the flux through each little
ΔA is just:
Φ E = E • ∆A
∆A
A
∆A is a special vector. It points
in the normal direction
and has a magnitude that
tells us the area of ΔA .
7. Electric Flux – Case 2
r
E
r
E⊥
θ
So… to get the flux through the
entire surface A, we just have
to add up the contributions
from each of the little ΔA’s
that compose A.
∆A
A
Φ E = ∑ E • ∆An
n
8. Electric Flux – Case 2
r
E
Φ E = ∑ E • ∆An
r
E⊥
θ
n
∆A
Φ E = ∫ E • dA
surface
A
Electric Flux through an
arbitrary surface caused by a
spatially varying E-field.
9. Electric Flux – Flux through a Closed Surface
•
The vectors dAi point in
different directions
– At each point, they are
perpendicular to the surface
– By convention, they point
outward
Af_2404.swf
11. Electric Flux: General Definition
Φ E = ∫ E • dA
surface
E-Flux through a surface depends on three things:
1. How strong the E-field is at each infinitesimal area.
2. How big the overall area A is after integration.
3. The orientation between the E-field and each
infinitesimal area.
12. Electric Flux: General Definition
Φ E = ∫ E • dA
surface
Flux can be negative, positive or zero!
-The sign of the flux depends on the convention you
assign. It’s up to you, but once you choose, stick
with it.
14. Electric Flux – Calculating E-Flux
• The surface integral means the integral must be evaluated
over the surface in question… more in a moment.
• The value of the flux will depend both on the field pattern
and on the surface
• The units of electric flux are N.m2/C
15. Electric Flux
• The net electric flux through a surface is directly
proportional to the number of electric field lines passing
through the surface.
16. Flux through a Cube
Assume a uniform E-field
pointing only in +x direction
Find the net electric flux through
the surface of a cube of edgelength l, as shown in the
diagram.
17. Gauss’s Law
Gauss’s Law is just a flux calculation
We’re going to build imaginary surfaces – called Gaussian
surfaces – and calculate the E-flux.
Gauss’s Law only applies to closed surfaces.
Gauss’s Law directly relates electric flux to the charge
distribution that creates it.
19. Gauss’s Law
Gauss’s Law
Φ NET
qenclosed
= ∫ E • dA =
ε0
surface
The net E-flux through a closed
surface
Charge inside the surface
20. Gauss’s Law
In other words…
1. Draw a closed surface around a some
charge.
2.
Set up Gauss’s Law for the surface
you’ve drawn.
3.
Use Gauss’s Law to find the E-field.
Φ NET
qenclosed
= ∫ E • dA =
ε0
surface
You get to choose the surface –
it’s a purely imaginary thing.
21. Which surface – S1, S2 or S3 –
experiences the most electric
flux?
22. Gauss’s Law – confirming Gauss’s Law
Let’s calculate the net flux through a Gaussian
surface.
Assume a single positive point charge of
magnitude q sits at the center of our imaginary
Gaussian surface, which we choose to be a
sphere of radius r.
Φ NET
= ∫ E • dA
surface
23. Gauss’s Law – confirming Gauss’s Law
At every point on the sphere’s surface, the
electric field from the charge points normal to
the sphere… why?
This helps make our calculation easy.
Φ NET
= ∫ E • dA
surface
24. Gauss’s Law – confirming Gauss’s Law
At every point on the sphere’s surface, the
electric field from the charge points normal to
the sphere… why?
This helps make our calculation easy.
Φ NET
= ∫ E • dA =
surface
∫ EdA cos θ
surface
=
∫ EdA
surface
25. Gauss’s Law – confirming Gauss’s Law
Now we have:
Φ NET =
∫ EdA
surface
But, because of our choice for the Gaussian
surface, symmetry works in our favor.
The electric field due to the point charge is
constant all over the sphere’s surface. So…
Φ NET = E
∫ dA
surface
26. Gauss’s Law – confirming Gauss’s Law
This, we can work with.
Φ NET = E
∫ dA
surface
We know how to find the magnitude of the
electric field at the sphere’s surface.
Just use Coulomb’s law to calculate the E-field
due to a point charge a distance r away from the
charge.
E po int ch arg e
ke q
= 2
r
27. Gauss’s Law – confirming Gauss’s Law
Thus:
Φ NET
ke q
= 2 ∫ dA
r surface
28. Gauss’s Law – confirming Gauss’s Law
Thus:
Φ NET
ke q
= 2 ∫ dA
r surface
And, this surface integral is easy.
dA = 4π ⋅ r 2
∫
sphere
29. Gauss’s Law – confirming Gauss’s Law
Therefore:
Φ NET
ke q
= 2 ⋅ (4π ⋅ r 2 )
r
But, we can rewrite Coulomb’s constant.
1
ke =
4πε 0
30. Gauss’s Law – confirming Gauss’s Law
Therefore:
Φ NET
ke q
= 2 ⋅ (4π ⋅ r 2 )
r
But, we can rewrite Coulomb’s constant.
1
ke =
4πε 0
Thus, we have confirmed Gauss’s law:
⇒ Φ NET
q
=
ε0
31. A few more questions
+Q +
–
– 3Q
• If the electric field is zero for all points on the surface, is
the electric flux through the surface zero?
• If the electric flux is zero for a closed surface, can there be
charges inside the surface?
• What is the flux through the surface shown? Why?
32. Flux due to a Point Charge
A spherical surface surrounds a point charge.
What happens to the total flux through the surface if:
(A)
(B)
(C)
(D)
the charge is tripled,
the radius of the sphere is doubled,
the surface is changed to a cube, and
the charge moves to another location inside the surface?
33. Applying Gauss’s Law
Gauss’s Law can be used to
(1) find the E-field at some position relative to a known charge
distribution, or
(2) to find the charge distribution caused by a known E-field.
In either case, you must choose a Gaussian surface to use.
34. Applying Gauss’s Law
Choose a surface such that…
1.
Symmetry helps: the E-field is constant over the surface (or
some part of the surface)
2.
The E-field is zero over the surface (or some portion of the
surface)
3.
The dot product reduces to EdA (the E-field and the dA vectors
are parallel)
4.
The dot product reduces to zero (the E-field and the dA vectors
are perpendicular)
35. Spherical Charge Distribution
An insulating solid sphere of radius a has a uniform volume charge density
ρ and carries total charge Q.
(A) Find the magnitude of the E-field at a point outside the sphere
(B) Find the magnitude of the E-field at a point inside the sphere
37. Spherical Charge Distribution
Find the E-field a distance r from a
line of positive charge of infinite
length and constant charge per
unit length λ.
39. Conductors in Electrostatic Equilibrium
• In an insulator, excess charge stays put.
• Conductors have free electrons and, correspondingly, have different
electrostatic characteristics.
• You will learn four critical characteristics of a conductor in
electrostatic equilibrium.
• Electrostatic Equilibrium – no net motion of charge.
40. Conductors in Electrostatic Equilibrium
• Most conductors, on their own, are in electrostatic equilibrium.
• That is, in a piece of metal sitting by itself, there is no “current.”
41. Conductors in Electrostatic Equilibrium
Four key characteristics
1.
The E-field is zero at all points inside a conductor, whether hollow
or solid.
2.
If an isolated conductor carries excess charge, the excess charge
resides on its surface.
3.
The E-field just outside a charged conductor is perpendicular to the
surface and has magnitude σ/ε0, where σ is the surface charge
density at that point.
4.
Surface charge density is biggest where the conductor is most
pointy.
42. Conductors (cont.) – Justifications
Einside = 0
•
•
•
•
•
Place 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.
43. Conductors (cont.) – Justifications
Einside = 0
•
•
•
•
Before the external field is applied,
free electrons are distributed evenly
throughout the conductor.
When the external field is applied,
charges 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.
Redistribution takes about 10-15s.
44. Conductors (cont.) – Justifications
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
45. Conductors (cont.) – Justifications
Charge Resides on the Surface
•
•
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
46. Conductors (cont.) – Justifications
E-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
47. E-Field’s Magnitude and Direction
Conductors (cont.) – Justifications
E-Field’s Magnitude and Direction
•
•
The net flux through the surface
is through only the flat face
outside the conductor
– The field here is
perpendicular to the surface
Applying Gauss’s law
σA
Φ E = EA =
ε0
σ
E=
ε0
48. E-Field’s Magnitude Justifications
Conductors (cont.) –and Direction
E-Field’s Magnitude and Direction
•
The field lines are
perpendicular to both
conductors
•
There are no field lines inside
the cylinder
49. Sphere inside a Spherical Shell
A solid insulating sphere of radius a
carries a uniformly distributed charge,
Q.
A conducting shell of inner radius b and
outer radius c is concentric and carries a
net charge of -2Q.
Find the E-field in regions 1-4 using
Gauss’s Law.
50. Example 1
Consider a thin spherical shell of radius
14.0 cm with a total charge of 32.0
μC distributed uniformly on its
surface. Find the electric field
(a) 10.0 cm and
(b) 20.0 cm from the center of the
charge distribution.
k Q ( 8.99 × 10 ) ( 32.0 × 10 )
E=
=
= 7.19 MN C
9
e
r2
( 0.200) 2
−6
51. Example 2
A uniformly charged, straight filament 7.00 m in
length has a total positive charge of 2.00 μC. An
uncharged cardboard cylinder 2.00 cm in length
and 10.0 cm in radius surrounds the filament at its
center, with the filament as the axis of the
cylinder. Using reasonable approximations, find
(a) the electric field at the surface of the cylinder
and
(b) the total electric flux through the cylinder.
(
)(
)
9
2
2
−6
2ke λ 2 8.99 × 10 N ×m C 2.00 × 10 C 7.00 m
E=
=
r
0.100 m
Φ E = EA cos θ = E ( 2π r l ) cos 0°
(
)
E = 51.4 kN C , radially outward
Φ E = 5.14 × 10 4 N C 2π ( 0.100 m ) ( 0.020 0 m ) ( 1.00 ) = 646 N ×m 2 C
52. Example 3
A square plate of copper with 50.0-cm sides
has no net charge and is placed in a region
of uniform electric field of 80.0 kN/C
directed perpendicularly to the plate. Find
(a) the charge density of each face of the plate
and
(b) the total charge on each face.
(
)(
)
σ = 8.00 × 10 4 8.85 × 10 −12 = 7.08 × 10 −7 C m 2
(
)
Q = σ A = 7.08 × 10 −7 ( 0.500)
2
C
Q = 1.77 × 10 −7 C = 177 nC
53. Example 4
A long, straight wire is surrounded by a hollow
metal cylinder whose axis coincides with that of
the wire. The wire has a charge per unit length of λ,
and the cylinder has a net charge per unit length of
2λ. From this information, use Gauss’s law to find
(a) the charge per unit length on the inner and outer
surfaces of the cylinder and (b) the electric field
outside the cylinder, a distance r from the axis.
0 = λ l + qin
qin
= −λ
l
E=
2 k e ( 3λ ) 6 k e λ
3λ
=
=
radially outward
r
r
2π ∈0 r
3λ
