Download Lecture 3. Electric Field Flux, Gauss` Law From the concept of

Lecture 3. Electric Field Flux, Gauss’ Law
From the concept of electric field flux –
to the calculation of electric fields of complex charge
distributions.
1
Iclicker Question
Charged particles are fixed on grids having the same spacing. Each charge
has the same magnitude Q with signs given in the figure. Rank the
magnitude of the electric field (from strong to weak) at the location marked
with an “x”.
1
2
3
4
A. 1 > 2 > 3 > 4
B. 2 > 3 > 1 = 4
C. 1 = 4 > 3 > 2
D. 3 > 2 > 1 > 4
2
Iclicker Question
Charged particles are fixed on grids having the same spacing. Each charge
has the same magnitude Q with signs given in the figure. Rank the
magnitude of the electric field (from strong to weak) at the location marked
with an “x”.
1
2
3
4
A. 1 > 2 > 3 > 4
B. 2 > 3 > 1 = 4
C. 1 = 4 > 3 > 2
D. 3 > 2 > 1 > 4
3
Iclicker Question
A negative point charge -Q is released from rest in an electric
field. At this moment, the acceleration of the point charge
A. is in the direction of the electric field at the position
of the point charge.
B. is directly opposite the direction of the electric field
at the position of the point charge.
C. is perpendicular to the direction of the electric field
at the position of the point charge.
-Q
D. is zero.
E. not enough information given to decide
𝐸
4
Iclicker Question
A negative point charge -Q is released from rest in an electric
field. At this moment, the acceleration of the point charge
A. is in the direction of the electric field at the position
of the point charge.
B. is directly opposite the direction of the electric field
at the position of the point charge.
C. is perpendicular to the direction of the electric field
at the position of the point charge.
-Q
D. is zero.
E. not enough information given to decide
𝐸
5
Iclicker Question
A positive point charge +Q is released from rest in an electric
field. At any later time, the velocity of the point charge
A. is in the direction of the electric field at the position
of the point charge.
B. is directly opposite the direction of the electric field
at the position of the point charge.
C. is perpendicular to the direction of the electric field
at the position of the point charge.
𝑣
D. is zero.
E. not enough information given to decide
+Q
𝐸
6
Iclicker Question
A positive point charge +Q is released from rest in an electric
field. At any later time, the velocity of the point charge
A. is in the direction of the electric field at the position
of the point charge.
B. is directly opposite the direction of the electric field
at the position of the point charge.
C. is perpendicular to the direction of the electric field
at the position of the point charge.
𝑣
D. is zero.
E. not enough information given to decide
+Q
𝐸
7
Flow of Fluid
Flow of fluid: the speed of water, v, and the pipe cross
section, A, are known. Calculate how much water you
added to the bucket in 1 sec.
∆𝑉 = 𝐴 ∙ 𝑣 ∙ 1𝑠
length (water displacement in 1s)
∆𝑉 = 𝐴 ∙ 𝑣 ∙ 1𝑠
𝜙
scalar product, 𝐴 ∙ 𝑣 ∙ cos 𝜙
∆𝑉/∆𝑉𝑚𝑎𝑥
𝜙
8
Concept of Flux
small element
of surface
- applies to any vector field
𝑎 𝑟 (the field of velocities
in a fluid flow, the E field,
etc.).
vector field
The flux of the field 𝑎 𝑟
through a small element of
surface 𝑑𝐴 :
𝐸
Φ𝑎 ≡ 𝑎 𝑟 ∙ 𝑑𝐴 = 𝑎 𝑟 𝐴 ∙ 𝑐𝑜𝑠𝜙
scalar !
𝐸
𝐴
𝐴
(a)
(b)
(c)
9
Units of Flux
The flux of
electrostatic
field:
Φ𝐸
𝑬 𝒓
The flux of
gravitational
field:
𝑁 2
→ 𝑚
𝐶
(𝐹 = 𝑞𝐸)
𝑁 2 𝑚3
Φ𝑔 →
𝑚 = 2
𝑘𝑔
𝑠
𝒈 𝒓
(𝐹 = 𝑚𝑔)
10
Iclicker Question
Calculate the flux of uniform electric field through the surface shown in the Figure.
600
E= 2 N/C
A. 𝜋 0.01 𝑁
𝑚2
∙
𝐶
B. −𝜋 0.01 𝑁
C. 𝜋 0.02 𝑁
𝑚2
∙
𝐶
𝑚2
∙
𝐶
D. −𝜋 0.02 𝑁 ∙
𝑚2
𝐶
11
Iclicker Question
Calculate the flux of uniform electric field through the surface shown in the Figure.
600
E= 2 N/C
A. 𝜋 0.01 𝑁
𝑚2
∙
𝐶
B. −𝜋 0.01 𝑁
C. 𝜋 0.02 𝑁
𝑚2
∙
𝐶
𝑚2
∙
𝐶
D. −𝜋 0.02 𝑁 ∙
𝑚2
𝐶
Φ𝐸 = 𝐸 𝑟 ∙ 𝐴 ∙ 𝑐𝑜𝑠𝜙
𝜙 = 1200
cos 1200 = −0.5
12
Flux through Closed Surfaces
We integrate over
this “Gaussian”
surface
The net flux of the electric field
through the closed surface:
Φ𝐸 =
𝐸 𝑟 ∙ 𝑑𝐴
𝑑𝐴
the integral is taken over
the whole surface
𝐸
Convention:
the vector normal to the
closed surface points
outward.
𝑑𝐴
𝑑𝐴
𝑑𝐴
13
Example: Point Charge at the Center of a Spherical Surface
At each point of the surface 𝐸 ∥ 𝑑𝐴 ( cos𝜙=1 )
Φ𝐸 =
𝐸 𝑟 ∙ 𝑑𝐴 = 𝐸 𝑟 4𝜋𝑟 2
1 𝑞
𝑞
2
=
4𝜋𝑟 =
4𝜋𝜀0 𝑟 2
𝜀0
- doesn’t
depend
on r
!
- holds for any vector field whose strength  1/𝑟 2 .
(electrostatic, gravitational)
The flux of the gravitational field of Earth through Earth’s surface?
Φ𝑔 = − 𝐺
𝑀
2 = −4𝜋𝐺𝑀
4𝜋𝑅
𝑅2
14
Point Charge at the Center of a Spherical Surface (cont’d)
What happens if we place the charge outside?
Φ𝐸 = 0
!
consequence of 1/𝑟 2 field dependence = continuity of field lines
The flux magnitudes are the same for these two surfaces.
15
Iclicker Question
Two point charges, +q
(in red) and –q (in
blue), are arranged as
shown.
Through which closed
surface(s) is the net
electric flux equal to
zero?
A. surface A
B. surface B
C. surface C
D. surface D
E. both surface C and surface D
16
Iclicker Question
Two point charges, +q
(in red) and –q (in
blue), are arranged as
shown.
Through which closed
surface(s) is the net
electric flux equal to
zero?
A. surface A
B. surface B
C. surface C
D. surface D
E. both surface C and surface D
17
Gauss’ Law
The total flux of the electric field though
any closed surface is proportional to the
net charge inside the surface.
charges
the same
fluxes for
all three
surfaces
Φ𝐸 ≡
𝐸 𝑟 ∙ 𝑑𝐴 =
1
𝜀0
𝑞𝑖
𝑖
The net charge:
𝑄=
𝑞𝑖
𝑖
𝑄=
𝜌 𝑟 𝑑𝜏
𝑣𝑜𝑙𝑢𝑚𝑒
charge
density
element of
volume
18
Applications of Gauss Law
𝑸𝒏𝒆𝒕
𝑬 𝒓 ∙ 𝒅𝑨 =
𝜺𝟎
Gauss’ Law (GL) in its integral form (one scalar equation) is not sufficient for
finding three components of a vector field E.
However, GL is very useful whenever it is possible to reduce a 3D (vector)
problem to a 1D (scalar) problem. The key is the proper symmetry of a
problem.
«Useful» symmetries:
- spherical
- cylindrical + translational
- plane
An intelligent choice of a Gaussian
surface is crucial!
Example of an
«insufficient»
symmetry
(cylindrical but
without
translational):
an electric
dipole.
19
Spherical Symmetry
Example: electric field of a uniformly charged ball. Radius R, total charge Q.
charge
density
𝑄
𝜌=
4 3
3 𝜋𝑅
R
O
Gaussian surface: a sphere centered at () O.
Its radius r can be smaller or larger than R.
enclosed
charge
𝑞 𝑟′ ≤ 𝑟
2
𝐸 𝑟 ∙ 4𝜋𝑟 =
=
𝜀0
r
Φ𝐸
𝑅
𝑟
flux
𝑄𝑟
4𝜋𝜀0 𝑅 3
𝐸 𝑟 =𝐸 𝑟 𝑟
𝐸 𝑟 =
𝑄
4𝜋𝜀0 𝑟 2
4
𝜌 3 𝜋𝑟 3
𝜀0
𝑄
𝜀0
𝑄 𝑟3
=
𝜀0 𝑅3
𝑟<𝑅
𝑟≥𝑅
𝑟<𝑅
𝑟≥𝑅
The field outside the sphere is the same as that
for a point charge Q at the sphere’s center.
20
Plane Symmetry
Example: the field of uniformly charged infinite plane.
Gaussian surface: a rectangular box (or
cylinder)centered at the plane.
enclosed
charge
𝜎
𝐸 𝑥 =
2𝜀0
𝐸 𝑥 = ±𝐸 𝑥 𝑥
𝜎𝐴
𝐸 𝑥 ∙ 2𝐴 =
𝜀0
flux
𝐸 𝑥
 - the surface
charge density
[C/m2]
𝜎
2𝜀0
𝜎
−
2𝜀0
𝑥
21
Superposition
+
+
+
-
+
+
+
-
+
+
+
+
+
+
+
+
+
-
𝐸 𝑥
𝐸 𝑥
𝑥
𝑥
𝑬𝟏
𝑬𝒏𝒆𝒕 = 𝑬𝟏 + 𝑬𝟐
𝑬𝟐
𝑬𝟏
𝑬𝟐
𝑬𝒏𝒆𝒕 = 𝑬𝟏 + 𝑬𝟐
22
Conclusion
Gauss’ Law: works in electrodynamics,
in electrostatics it is equivalent to
Coulomb’s Law.
Powerful tool for computing the electric fields if
a problem is essentially 1D due to symmetry.
Next time: Lecture 4. Applications of Gauss’ Law,
Conductors in Electrostatics. §§ 22.5
Read about Metals and Dielectrics.
23
Appendix I. Cylindrical Symmetry + Translational Symmetry
along the axis of symmetry
Example: the field of uniformly
charged infinitely long wire.
Gaussian surface: a cylinder of length L
centered at the wire.
enclosed
charge
𝐸 𝑟 ∙ 2𝜋𝑟𝑙 =
 - the linear
density of charges
(the unit length
charge)
𝜌𝑙
𝜀0
flux
𝑟≥𝑅
𝜌
𝐸 𝑟 =
2𝜋𝜀0 𝑟
𝐸 𝑟 =𝐸 𝑟 𝑟
𝐸 𝑟
 1/r
𝑅
𝑟
24
Appendix II: Symmetry Facilitates Flux Calculations
Example 1: A charge q is at the center of a cube. What’s the flux of E through the
shaded side?
𝑞 1
Φ𝐸 = ×
𝜖0 6
Example 2: A charge q is at the back corner of a cube. What’s the flux of E through the
shaded side?
Φ𝐸 =
𝑞
1
×
𝜖0 24
25
Iclicker Question
Rank the fluxes Φ𝑖 through
the five surfaces (from most
negative to most positive):
A. Φ1 < Φ2 < Φ3 < Φ4 < Φ5
B. Φ1 < Φ4 < Φ2 < Φ3 < Φ5
C. Φ4 < Φ5 < Φ2 < Φ3 < Φ1
D. Φ2 < Φ4 < Φ1 < Φ3 < Φ5
E. Φ5 < Φ4 < Φ3 < Φ2 < Φ1
26
Iclicker Question
Rank the fluxes Φ𝑖 through
the five surfaces (from
small to large):
A. Φ1 < Φ2 < Φ3 < Φ4 < Φ5
B. Φ1 < Φ4 < Φ2 < Φ3 < Φ5
C. Φ4 < Φ5 < Φ2 < Φ3 < Φ1
D. Φ2 < Φ4 < Φ1 < Φ3 < Φ5
E. Φ5 < Φ4 < Φ3 < Φ2 < Φ1
27