Chapter 23 Guass’ Law
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林鼎荃 朱珈聖 蔡子翔
23-1 Electric Flux
(A) A problem
<sol 1> Integral (complex!)
𝑞 → 𝑑𝑞 → 𝑑𝐸 →
𝐸
+
<sol 2> Guass’ Law
Simple algebra
Try: Find the electric field
which the source establish.
The source is complex, don’t
have symmetry.
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23-1 Electric Flux
(B) Electric Flux
(a)Definition
∆𝜙𝐸 ≡ (𝐸 cos 𝜃)∆𝐴…… for a super small area
(b)Total Flux
∆𝜙𝐸 = 𝐸 ∆𝐴cos 𝜃……(length length cos, dot product)
∴ ∆𝜙𝐸 = 𝐸 ∙ ∆𝐴……dot product form
∴ 𝜙𝐸 =
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𝐸 ∙ 𝑑𝐴
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23-1 Electric Flux
NOTE : Integral dot product formula
(B) Electric Flux
(b)Total Flux
ϕE =
1) Transform to some x and some y
2) Use length length cos
E ∙ dA ……we use 2)
ϕE =
E cos(θ) dA = 𝐸 cos 𝜃
dA
(c) The area vector
Direction: Outside is positive
Value: area
(d) Closed surface integral
→
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23-1 Electric Flux
(C) EXAMPLE 01
𝜙 =
E ∙ dA = ( E ∙ dA)𝑎 +( E ∙ dA)𝑏 +( E ∙ dA)𝑐
For a:
( E ∙ dA)𝑎 =
𝐸 (cos 2𝜋)𝑑𝐴 = −𝐸
𝑑𝐴 = −𝐸𝐴
For c:
( E ∙ dA)𝑐 =
Try: Find the total flux of the cylinder.
𝐸 (cos 0)𝑑𝐴 = 𝐸
𝑑𝐴 = 𝐸𝐴
For b: Directly look
∵ 𝑑𝐴 ⊥ 𝐸
∴ 𝐸 ∙ dA = 0
The total flux:
𝜙 = −𝐸𝐴 + 0 + 𝐸𝐴 = 0
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23-1 Electric Flux
(C) EXAMPLE 02
Take right side as example
𝑑 𝐴 = 𝑑𝐴𝑖 ……vector is equal to value and direction
𝜙𝐸𝑅 =
Try: Find the flux of right, left and top.
The electric field is not uniform,
𝐸 = 3𝑥 𝑖 + 4𝑗
𝐸 ∙ 𝑑𝐴 =
=
(3𝑥 𝑖) ∙ 𝑑𝐴𝑖 +
=
(3𝑥 𝑖) ∙ 𝑑𝐴𝑖 +
∵ 𝑥 = 3 ∴ 𝜙𝐸𝑅 = 3𝑥
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3𝑥 𝑖 + 4𝑗 ∙ (dA𝑖)
(4𝑗) ∙ 𝑑𝐴𝑖
𝑑𝐴 = 9
4𝑗 ∙ 𝑑𝐴𝑖 = 3
𝑥𝑑𝐴
𝑑𝐴 = 9𝐴 = 36 𝑁 ∙ 𝑚2 /𝐶
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23-2 Guass’ Law
(A) Guass’ Law
• Idea: Guass’ Law relates the electric
field at point on a closed
Gaussian surface to the net
charge enclosed by that surface.
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• Definition: 𝜀0 𝜙𝐸 = 𝑞𝑒𝑛𝑐 𝑤ℎ𝑒𝑟𝑒 𝜀0 ≡
𝜇0 𝐶 2
𝜀0 : Vacuum permittivity
𝜇0 : vacuum magnetic permeability
• Transformation: 𝜀0 E ∙ dA = 𝑞𝑒𝑛𝑐
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23-2 Guass’ Law
< 𝑝. 𝑓. >
(B) Extra electric field
• Extra electric field (out of the
Gaussian surface) doesn’t make flux.
• Another example: I put 2 candies into the box,
then I take 2 candies out from the box.
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23-2 Guass’ Law
(C) Guass’ Law and Coulomb’s Law
𝐸
𝜀0 E ∙ dA = 𝜀0 𝐸 (cos 0) dA = 𝜀0 𝐸 dA = 𝑞𝑒𝑛𝑐 = 𝑞
𝜀0 𝐸
+𝑞
𝑑𝐴 = 𝑞 = 𝜀0 𝐸(4𝜋𝑟 2 )
1 𝑞
𝑘𝑞
𝐸=
= 2
2
4𝜋𝜀0 𝑟
𝑟
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23-2 Guass’ Law
𝑄
(D) EXAMPLE 03
1° Build a Gaussian ball (the purple one)
𝑞𝑒𝑛𝑐 is positive, so 𝐸1 is outward.
𝑞
𝑃1
𝑃2
𝜀0 E ∙ dA = 𝜀0 𝐸 (cos 0) dA = 𝜀0 𝐸 dA = 𝑞𝑒𝑛𝑐 = 𝑞
𝑑𝐴 = 𝑞𝑒𝑛𝑐 = 𝜀0 𝐸(4𝜋𝑟 2 )
𝜀0 𝐸
𝑄 = −16𝑒, 𝑅 = 10𝑐𝑚, 𝑞 = +5𝑒
𝑟1 = 6𝑐𝑚, 𝑟2 = 12𝑐𝑚
Try: Find the 𝐸 of 𝑃1 and 𝑃2
𝐸=
𝑞𝑒𝑛𝑐
𝜀0 4𝜋𝑟 2
= 2 × 10−6 𝑁/𝐶……𝐸1
2° Find 𝐸2
The only difference: This time, 𝑞𝑒𝑛𝑐
Is negative, so 𝐸2 is inward.
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23-2 Guass’ Law
(E) EXAMPLE 04
1° We already know the flux of right, left, and top
(EXAMPLE 02)
2° Use the same trick of EXAMPLE 02, we can find
the flux of other area.
3° Plus all the six fluxes, we find the total flux
𝜙𝐸 = 24 𝑁 ∙ 𝑚2 /𝐶
Try: Find the qenc of the box.
E = 3xi + 4j
4° Guass’ Law
𝑞𝑒𝑛𝑐 = 𝜀0 𝜙𝐸 = 2.1 × 10−10 𝐶
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23-3 A Charged Isolated Conductor
(A) A Charged Isolated Conductor
(a)Law
If an excess charge is placed on an
isolated C, that amount of charge will
move entirely to the surface of the
conductor. Inside the C won’t exist charge.
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23-3 A Charged Isolated Conductor
(A) A Charged Isolated Conductor
(b)proof
𝑞
1° Inside the C, there are no 𝐸.
If it does, there will be a current
in the C. It us ridicules.
2° Inside the C, there are no 𝐸.
Thus, ∀ closed Gaussian surface, the flux is zero.
Thus, inside the C, 𝑞𝑒𝑛𝑐 = 0
Gaussian surface
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23-3 A Charged Isolated Conductor
(A) A Charged Isolated Conductor
(c)proof(has cavity)
𝑞
Guassian surface is nor actually a “surface.”
In this case, it is a hollow donut.
Gaussian surface
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23-3 A Charged Isolated Conductor
(B) External Electric Field
For a charge, the electric field is easy to find by
integration.
However, the electric field which established by an
irregular
shape field source is hard to find by integration.
< 𝑠𝑜𝑙 > Guass’ Law
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23-3 A Charged Isolated Conductor
(B) External Electric Field
1° Consider a super small area and a cylinder
2° For the right area of the cylinder:
𝜙𝐸 = 𝐸𝐴……(EXAMPLE 01)
3° 𝜎 ≡ 𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎
Thus, 𝑞𝑒𝑛𝑐 = 𝜎𝐴
Thus, Gauss’ Law becomes: 𝜀0 𝐸𝐴 = 𝜎𝐴
𝜎
𝐸=
𝜀0
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23-3 A Charged Isolated Conductor
(C) EXAMPLE 05
Inside the Gaussian surface: +5𝜇𝐶……stay neutrality.
Inside the Gaussian surface= +5𝜇𝐶: outside= −5μ𝐶
Inside the shell is not distributed uniformly inside the
shell, but it does for those outside the shell.
Try: Find the charges on its surface.
Are they distributed uniformly?
What is the field pattern inside
and outside the shell?
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23-4 APPLYING GAUSS’ LAW: CYLINDRICAL
SYMMETRY
1. Consider an infinitely long cylindrical plastic rod
with a uniform charge density 𝜆.
2. Use cylindrical symmetry:
𝜙𝐸 = 𝐸𝐴 cos 𝜃 = 𝐸2𝜋𝑟ℎ
3.𝜆 ≡ 𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
Thus, 𝑞𝑒𝑛𝑐 = 𝜆ℎ, and use Gauss’ Law
𝜆
⇒𝐸=
𝜀0 2𝜋𝑟
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23-4 APPLYING GAUSS’ LAW: CYLINDRICAL
SYMMETRY
EXAMPLE 06
Q: the critical value 𝐸𝑐 = 2.4𝑀𝑁/𝐶. What value of Q would
have put the air along her body on the verge of breakdown?
0.1m
1∵ 𝑅 ≪ 𝐿, ∴ 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑎 𝑙𝑜𝑛𝑔 𝑙𝑖𝑛𝑒
2.Use 𝐸 = 𝜆𝜀0 2𝜋𝑟
1.8m
3.𝜆 = 𝑄/𝐿, 𝑟 = 𝑅, 𝐸 = 𝐸𝑐
⇒ 𝐸𝑐 =
𝑄
2𝜋𝜀0 𝑅𝐿
⇒ 𝑄 = 𝐸𝑐 2𝜋𝜀0 𝑅𝐿 = 2𝜋 8.85 × 10−12 𝐶 2 /𝑁𝑚2 0.1𝑚
× (1.8𝑚)(2.4 × 106 𝑁/𝐶)
= 2.402 × 10−5 𝐶 ≈ 24𝜇𝐶
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23-5 APPLYING GAUSS’ LAW: Planar Symmetry
(A) Nonconducting sheet
1. Consider a thin, infinite, nonconducting sheet
with a uniform charge density 𝜎.
2. Use planar symmetry and the flux only through
two end faces:
𝜀0
𝐸 ∙ 𝑑 𝐴 = 𝑞𝑛𝑒𝑐 ⇒ 𝜀0 (𝐸𝐴 + 𝐸𝐴) = 𝜎𝐴
3. The electric field of any point at a finite
distance from this sheet is
𝜎
⇒𝐸=
2𝜀0
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23-5 APPLYING GAUSS’ LAW: Planar Symmetry
(B) Two Conducting Plates
1. Consider two thin, infinite, conducting sheets with
a uniform charge density 𝜎1 in the two face of plates.
𝐸 = 𝜎1 /𝜀0
2. One of them is positive charge, another is
negative.
3. Close two sheets, the charge is attractive
together, the charge density become twice 𝜎1 .
𝜎 = 2𝜎1
𝜎
⇒𝐸=
𝜀0
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23-5 APPLYING GAUSS’ LAW: Planar Symmetry
(C) EXAPLE 07
Q: find the 𝐸𝐿 , 𝐸𝐵 , 𝑎𝑛𝑑 𝐸𝑅
1. Consider two thin, infinite, Nonconducting sheets
with a uniform charge density.
𝐸 = 𝜎/2𝜀0
2. 𝜎+ = 6.8𝜇𝐶/𝑚2 , 𝜎− = 4.3𝜇𝐶/𝑚2
𝜎+
6.8 × 10−6 𝐶/𝑚2
5 𝑁/𝐶
𝐸+ =
=
=
3.84
×
10
2𝜀0 2(8.85 × 10−12 𝐶 2 /𝑁 ∙ 𝑚2 )
𝜎−
4.8 × 10−6 𝐶/𝑚2
5 𝑁/𝐶
𝐸− =
=
=
2.43
×
10
2𝜀0 2(8.85 × 10−12 𝐶 2 /𝑁 ∙ 𝑚2 )
𝐸𝐿 = −𝐸𝑅 = 𝐸+ − 𝐸− = 1.4 × 105 𝑁/𝐶
𝐸𝐵 = 𝐸+ + 𝐸− = 6.3 × 105 𝑁/𝐶
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23-6 Spherical symmetry
In a charged spherical shell of total charge q and radius R, we can draw two Gaussian surfaces, one is inside
the spherical shell (𝑆1 ), and the other is inside the spherical shell (𝑆2 ).
Shell Theorem
1.
For 𝑆1 (𝑟1 < 𝑅), we can know that didn’t have any electric charge in the 𝑆1 𝐺𝑎𝑢𝑠𝑠𝑖𝑜𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒, so we
could know that there isn’t any electric field inside the spherical shell. Therefore, the spherical shell
won’t exert any force on any electric charge inside the shell.
𝜀0 𝐸 ∙ 𝑑𝐴 = 𝑞𝑒𝑛𝑐 = 0, 𝐸 = 0
𝑆1
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𝑆2
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23-6 Spherical symmetry
Shell Theorem
2. For 𝑆2 (𝑟2 > 𝑅), the Gaussian surface wrap around all charged spherical shell which means the
electric field which is created by a total electric charge on the spherical shell is the same as one set up
by a particle with charge q at the center of the shell of charge. Therefore, Coulomb’s law allows use in
this situation.
1 𝑞
𝐸=
4𝜋𝜀0 𝑟 2
𝑆1
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𝑆2
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23-6 Spherically symmetric charge distribution
There is a lot of fixed electric charge distributed symmetrically in space, and the total charge is q. R is
the radius of the imaginary circle outside these dots, and r is the radius of the Gaussian surface.
In picture 1 (𝑟 > 𝑅), according to the Gaussian law, the electric field of the Gaussian surface is
1 𝑞
𝐸=
4𝜋𝜀0 𝑟 2
picture 1
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23-6 Spherically symmetric charge distribution
Because the distribution of the charge is symmetrical, we can try to see it consists
with lots of different spherical shells, to use the Shell theorem, the volume charge
density 𝝆 should have been constant for each shell but need not be the same
from shell to shell.
In picture 2 (𝑟 < 𝑅), the 𝐸 created by the charge inside the shell is 𝐸 =
1 𝑞′
4𝜋𝜀0 𝑟 2
charge inside
the shell q’
picture 2
If the charge enclosed within radius R is uniform, just like in picture 3, then q enclosed
within radius r is proportional to q:
charge enclosed by
sphere of radius r
volume enclosed by =
sphere of radius r
𝑓𝑢𝑙𝑙 𝑐ℎ𝑎𝑟𝑔𝑒
𝑓𝑢𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑞′
4
𝜋𝑟 3
3
=4
𝑞
𝜋𝑅 3
3
→
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𝑞′
=
𝑟3
𝑞 3
𝑅
𝐸=
𝑞
4𝜋𝜀0 𝑅 3
𝑟
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picture 3
Summary
• Guass’ Law : To simplify the process of finding electric field.
• Problem solving strategy: Find flux first. Try to replace the dot formula
in order to simplify the integration.
• Static balance: Inside the conductor, there are no charge exist.
• Apply: Find the electric field from cylinder, flat and ball.
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