Lecture 3. Electric Field Flux, Gauss’ Law
Attention: the list of unregistered
iclickers will be posted on our Web
page after this lecture.
From the concept of electric field flux –
to the calculation of electric fields of complex charge
distributions.
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Flow of Fluid
(Laminar) 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)
𝜙
∆𝑉 = 𝐴⃗ ∙ 𝑣⃗ ∙ 1𝑠
scalar product, 𝐴 ∙ 𝑣 ∙ cos 𝜙
∆𝑉/∆𝑉𝑚𝑚𝑚
𝜙
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Concept of Flux
small element
of surface
- applies to any vector field
𝑎⃗ 𝑟⃗ (the field of velocities
in a fluid flow, the E field,
etc.).
vector field
The flux of the field 𝑎⃗ 𝑟⃗
through a small element of
surface 𝑑𝐴⃗ :
𝐸
Φ𝑎 ≡ 𝑎⃗ 𝑟⃗ ∙ 𝑑𝐴⃗ = 𝑎 𝑟 𝐴 ∙ 𝑐𝑐𝑐𝜙
scalar !
𝐸
𝐴
𝐴
(a)
(b)
(c)
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Units of Flux
The flux of
electrostatic
field:
𝑬 𝒓
The flux of
gravitational
field:
𝒈 𝒓
Φ𝐸
𝑁 2
→ 𝑚
𝐶
(𝐹 = 𝑞𝑞)
𝑁 2 𝑚3
Φ𝑔 →
𝑚 = 2
𝑠
𝑘𝑘
(𝐹 = 𝑚𝑚)
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Flux through Closed Surfaces
We integrate over
this “Gaussian”
surface
Φ𝐸 = � 𝐸 𝑟⃗ ∙ 𝑑𝐴⃗
𝑑𝐴⃗
the integral is taken over
the whole surface
𝐸
Convention:
the vector normal to the
closed surface points
outward.
The net flux of the electric field
through the closed surface:
𝑑𝐴⃗
𝑑𝐴⃗
𝑑𝐴⃗
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Example: Point Charge at the Center of a Spherical Surface
At each point of the surface 𝐸 ∥ 𝑑𝐴⃗ ( cos𝜙=1 )
Φ𝐸 = � 𝐸 𝑟⃗ ∙ 𝑑𝐴⃗ = 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 ∙
4𝜋𝑅2
𝑀
= − 𝐺 2 4𝜋𝑅2 = −4𝜋𝐺𝐺
𝑅
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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.
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Example: gravitational field flux
Gaussian surface
mass 𝑚
Φ𝑔 = −4𝜋𝐺𝐺
Does the result of flux calculation depends
on
1.
the shape of the G. surface?
2.
The position of the mass within the
surface?
3.
Presence of the Earth nearby?
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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 𝑟⃗ ∙ 𝑑𝐴⃗ =
1
� 𝑞𝑖
𝜀0
𝑖
The net charge:
𝑄 = � 𝑞𝑖
𝑖
𝑄=
� 𝜌 𝑟 𝑑𝜏
𝑣𝑜𝑜𝑜𝑜𝑜
charge
density
element of
volume
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Gauss’ Law (cont’d)
 Electrostatics: Gauss’ Law ≡ Coulomb’s Law.
 Electrodynamics:
Gauss’ Law 
Coulomb’s Law
(the total flux doesn’t depend on the position of charges, so
they can move inside the closed surface!)
Gauss’ Law doesn’t imply action-at-a-distance, and we don’t
need to modify it in electrodynamics. In fact, this is
the first Maxwell’s equation.
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Applications of Gauss Law
𝑸𝒏𝒏𝒏
� 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.
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Spherical Symmetry
Example: electric field of a uniformly charged ball. Radius R, total charge Q.
charge
density
𝑄
𝜌=
4 3
𝜋𝑅
3
R
O
r
Φ𝐸
𝑅
𝐸 𝑟 = 𝐸 𝑟 𝑟̂
𝑟
𝐸 𝑟 =
Gaussian surface: a sphere centered at (⋅) O.
Its radius r can be smaller or larger than R.
enclosed
charge
𝑞 𝑟𝑟 ≤ 𝑟
2
𝐸 𝑟 ∙ 4𝜋𝑟 =
=
𝜀0
flux
𝑄𝑄
4𝜋𝜋0 𝑅3
𝑄
4𝜋𝜋0 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.
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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�
𝜎𝜎
𝐸 𝑥 ∙ 2𝐴 =
𝜀0
flux
σ - the surface
charge density
[C/m2]
𝐸 𝑥
𝜎
2𝜀0
𝜎
−
2𝜀0
𝑥
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Superposition
+
+
+
-
+
+
+
-
+
+
+
+
+
+
+
+
+
-
𝐸 𝑥
𝐸 𝑥
𝑬𝟏
𝑬𝒏𝒏𝒏 = 𝑬𝟏 + 𝑬𝟐
𝑬𝟐
𝑥
𝑬𝟏
𝑬𝒏𝒏𝒏 = 𝑬𝟏 + 𝑬𝟐
𝑬𝟐
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𝑥
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.
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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
𝜌𝜌
𝐸 𝑟 ∙ 2𝜋𝜋𝜋 =
𝜀0
ρ - the linear
density of charges
(the unit length
charge)
flux
𝑟≥𝑅
𝜌
𝐸 𝑟 =
2𝜋𝜋0 𝑟
𝐸 𝑟 = 𝐸 𝑟 𝑟̂
𝐸 𝑟
∝ 1/r
𝑅
𝑟
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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
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