Gauss’s Law
• The Faraday Experiment
• Charge, Flux, and Flux Density
• Field Lines
• Why D and E vectors?
• Gauss’s Law
• Gauss’s Law Examples – Enclosed charge from field
• Gauss’s Law Examples – Field for point, line, sheet
charges
• Gauss’s Law for differential element
• Divergence Theorem
That Faraday Experiment
• Concentric Spheres
– Put +Q on inner sphere
– Pack dielectric around
inner sphere
– Assemble outer sphere
– Discharge outer sphere
– Disassemble and measure
induced charge on outer sphere (not sure how you do this?)
• Result
– Induced charge outer sphere equal to charge on inner sphere
– Regardless of dielectric (which modifies local electric field)
– Continuity of Displacement, or Displacement Flux, or Electric Flux
𝑄𝑖𝑛𝑛𝑒𝑟 = Ψ = 𝑄𝑜𝑢𝑡𝑒𝑟
Charge, Flux, and Flux Density
1.
Charge equals flux
𝑄𝑖𝑛𝑛𝑒𝑟 = Ψ = 𝑄𝑜𝑢𝑡𝑒𝑟
2.
Flux density equals flux/area
Ψ
Ψ
𝐷 = 𝐴𝑟𝑒𝑎 = 4𝜋𝑟 2
• Since 𝑄𝑖𝑛𝑛𝑒𝑟 = Ψ = 𝑄𝑜𝑢𝑡𝑒𝑟 = 𝑄
𝑫𝑟=𝑎 =
Ψ
4𝜋𝑎2
=
𝑄
𝒂
4𝜋𝑎2 𝒓
𝑫𝑟=𝑏 =
Ψ
4𝜋𝑏2
=
𝑄
𝒂
4𝜋𝑏2 𝒓
𝑄
𝑫 = 4𝜋𝑏2 𝒂𝒓
𝑎<𝑟<𝑏
• For point charge
𝑫=
𝑄
𝒂
4𝜋𝑏2 𝒓
(gauss)
𝑬=
𝑄
𝒂
4𝜋𝜺𝒐 𝑏2 𝒓
(coulomb)
𝒔𝒐 𝑫 = 𝜺𝒐 𝑬
Field Lines
• Field Lines
1.
Field lines begin on (+) charge and end on (–) charge.
2.
Number of Field lines proportional to Ψ and Q.
3.
Point in direction of D.
4.
D magnitude equal to (field lines)/(area).
• Gauss’s Law for sphere
𝑄 = Ψ = 𝑓𝑖𝑒𝑙𝑑 𝑙𝑖𝑛𝑒𝑠 =
𝑓𝑖𝑒𝑙𝑑 𝑙𝑖𝑛𝑒𝑠
𝑎𝑟𝑒𝑎
∙ 𝑎𝑟𝑒𝑎 = 𝐷 ∙ 4𝜋𝑟 2
• Similar to Coulomb’s Law
𝑫=
𝜌𝑣 𝑑𝑣
𝒂
4𝜋𝑅 2 𝒓
𝑬=
𝜌𝑣 𝑑𝑣
𝒂
4𝜋𝜀𝑜 𝑅 2 𝒓
𝑫 = 𝜺𝒐 𝑬
Why the D and E Vector?
Charged plates with dielectric
• Displacement (D).
– Only free charge density.
𝐷=
Ψ
𝐴
𝑄
=𝐴
– Independent of dielectric.
• Polarization (P)
–
–
–
–
Dipole moment of aligned charge.
𝐩 = 𝐐𝐝 𝑷 = 𝒑
Points (-) to (+) for aligned charge.
Same direction as D
𝜺𝒐 E
𝐏
𝐃
• Electric Field (E)
– Reduced by aligned charge
– Add E and P
𝑫 = 𝜀𝑜 𝑬 + 𝑷
𝑷 = 𝜒𝑜 𝜀𝑜 𝑬
(all units C/m2)
𝑫 = 𝜀𝑜 𝑬 + 𝜒𝑜 𝜀𝑜 𝑬 = 𝜀𝑟 𝜀𝑜 𝑬
𝜀𝑟 = 𝜒𝑜 + 1
Gauss’s Law
The electric flux passing through any closed surface equals the total charge
enclosed by that surface.
𝑄𝑒𝑛𝑐𝑙 =
𝑑Ψ =
𝑫 ∙ 𝒅𝑺
• Universal flux/charge “balance”
– (+) Net flux equals (+) charge enclosed.
– (-) Net flux equals (-) charge enclosed.
– (0) Net Flux equals (0) net charge enclosed.
• Define Gaussian Surface.
– Dot product of D and dS.
– (dS outward normal).
– dot product 𝑫 ∙ 𝒅𝑺
• Outward (+)
• Inward (-)
• Glancing (0)
– Integrate over closed surface.
– Exploit Symmetry.
Gauss’s Law for charge distributions
• Charge enclosed can be
– Discrete charge
– Line charge
– Surface charge
– Volume charge
• For volume Gauss’s Law becomes
Example 1 – Enclosed charge from field
•
Find charge enclosed by field at r
Integrate over sphere,
Take divergence in spherical
coordinates!
𝑄
𝑫=
𝒂
4𝜋𝑟 2 𝒓
•
Make symmetry assumption (sphere)
•
Choose vector area element
𝑑𝑺 = 𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 𝒂𝒓
•
Form dot product integrand
𝑫 ∙ 𝑑𝑺 =
•
𝑄
𝑟 2 𝑠𝑖𝑛𝜃
4𝜋𝑟 2
𝑑𝜃 𝑑𝜑 𝒂𝒓 ∙ 𝒂𝒓
Integrate over closed surface to get flux Ψ = Q
2𝜋 𝜋
0
0
𝑄
𝑄
𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 =
4𝜋
2
𝜋
𝑠𝑖𝑛𝜃 𝑑𝜃 =
0
𝑄
−𝑐𝑜𝑠𝜃0 𝜋 = 𝑄
2
Example 2 – Enclosed charge from field
•
Find E at 𝑟 = 2, 𝜃 = 25, 𝜑 = 90
𝐷
0.3 ∙ 22 𝑒 −9
𝐸=
=
= 135.5 𝑉 𝑚
−12
−12
8.85𝑒
8.85𝑒
•
Total Charge within Sphere r = 3
2𝜋 𝜋
0.3 ∙ 32 𝑒 −9 32 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 𝒂𝒓 ∙ 𝒂𝒓 = 0.3 ∙ 81 ∙ 4𝜋 ∙ 1𝑒 −9 = 305 𝑛𝐶
𝑄=
0
•
0
Total Electric Flux within Sphere r = 4
2𝜋 𝜋
0.3 ∙ 42 𝑒 −9 42 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 𝒂𝒓 ∙ 𝒂𝒓 = 0.3 ∙ 256 ∙ 4𝜋 ∙ 1𝑒 −9 = 965 𝑛𝐶
𝑄=
0
0
Gauss’s Law – Field for point charge
• Must evaluate Symmetry (spherical)
– Choose coordinate system and surface such that
Dot product 𝑫 ∙ 𝒅𝑺 is either 1 or 0.
– Choose surface such that magnitude D is constant (L.H.S becomes trivial)
• Point charge
– Choose sphere and spherical coordinates
2𝜋 𝜋
2𝜋 𝜋
𝐷 𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 𝒂𝒓 ∙ 𝒂𝒓 = 𝐷
0
0
𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝜑 = 𝑄𝑒𝑛𝑐𝑙𝑜𝑠𝑒
0
𝑄
𝐷=
4𝜋𝑟 2
0
𝑄
𝐸=
4𝜋𝜀𝑜 𝑟 2
Integrate over sphere,
Take divergence in spherical
coordinates!
Gauss’s Law – Field for line of charge (wire)
• Must evaluate Symmetry - (cylindrical)
– Choose coordinate system and surface such that
Dot product 𝑫 ∙ 𝒅𝑺 is either 1 or 0. (sides + 2 ends)
– Choose surface such that magnitude D is constant (sides + 2 ends)
𝐿 2𝜋
2𝜋 𝑟
𝐷𝜌 𝑑𝜑 𝑑𝑧 𝒂𝜌 ∙ 𝒂𝜌 +
0 0
2𝜋 𝑟
𝐷𝜌 𝑑𝜌 𝑑𝜑 𝒂𝜌 ∙ 𝒂𝑧 +
0
0
𝐷𝜌 𝑑𝜌 𝑑𝜑 𝒂𝜌 ∙ 𝒂𝑧 = 𝜌𝐿 𝐿
0
0
– Ends go to zero since 𝒂𝜌 ∙ 𝒂𝑧 = 0
𝐿 2𝜋
𝐷
𝜌 𝑑𝜑 𝑑𝑧 𝒂𝜌 ∙ 𝒂𝜌 + 0 + 0 = 𝜌𝐿 𝐿
0 0
𝐷2𝜋𝜌𝐿 = 𝜌𝐿 𝐿
𝜌𝐿
𝐷=
2𝜋𝜌
𝜌𝐿
𝐸=
2𝜋𝜀𝑜 𝜌
Integrate over cylinder,
Take divergence in cylindrical
coordinates!
Gauss’s Law – Field for line of coaxial wire
•
For 𝑏 > 𝜌 > 𝑎 Gauss’s Law is exactly the same
𝐿 2𝜋
2𝜋 𝑟
𝐷𝜌 𝑑𝜑 𝑑𝑧 𝒂𝜌 ∙ 𝒂𝜌 +
0 0
2𝜋 𝑟
𝐷𝜌 𝑑𝜌 𝑑𝜑 𝒂𝜌 ∙ 𝒂𝑧 +
0
0
𝐿 2𝜋
𝐷
𝜌 𝑑𝜑 𝑑𝑧 𝒂𝜌 ∙ 𝒂𝜌 + 0 + 0 = 𝜌𝐿 𝐿
0 0
𝐷=
•
𝜌𝐿
2𝜋𝑟
𝐸=
𝜌𝐿
𝑓𝑜𝑟 𝑏 > 𝜌 > 𝑎
2𝜋𝜀𝑜 𝑟
For 𝜌 < 𝑎 charge enclosed is zero
𝐷 = 0 𝐸 = 0 𝑓𝑜𝑟 𝜌 < 𝑎
•
For 𝜌 >b net charge enclosed is zero
𝐷 = 0 𝐸 = 0 𝑓𝑜𝑟 𝜌 >b
𝐷𝜌 𝑑𝜌 𝑑𝜑 𝒂𝜌 ∙ 𝒂𝑧 = 𝜌𝐿 𝐿
0
0
Example 3.2
•
Surface charge density on inner conductor
𝜌𝑆,𝑖𝑛𝑛𝑒𝑟
•
Since Qinner = flux = Qouter
𝜌𝑆,𝑖𝑛𝑛𝑒𝑟
•
𝑄𝑖𝑛𝑛𝑒𝑟 30 × 10−9
=
=
= 9.55𝜇𝐶/𝑚2
−3
2𝜋𝑎𝐿
2𝜋10 0.5
𝑄𝑜𝑢𝑡𝑒𝑟
30 × 10−9
=
=
= 9.55𝜇𝐶/𝑚2
−3
2𝜋𝑎𝐿
2𝜋 4 × 10 0.5
Since Qouter = Flux = Qinner
𝜌𝐿
2𝜋𝑎𝜌𝑆 𝑎𝜌𝑆 9.55𝑛𝐶/𝑚2
𝐷𝜌 =
=
=
=
2𝜋𝜌
2𝜋𝜌
𝜌
𝜌
•
Electric Field is
Integrate over cylinder,
Take divergence in cylindrical
coordinates!
𝐷𝜌
9.55 × 10−9 𝐶/𝑚2
1079
1079
𝐸=
=
=
𝑁
𝐶
=
𝑉 𝑚
𝜖𝑜 (8.85 × 10−12 𝐶 2 /𝑁−𝑚2 )𝜌
𝜌
𝜌
Gauss’s Law – Field between parallel conducting plates
• Must evaluate Symmetry - (Perpendicular to sheet)
– Choose coordinate system and surface such that
Dot product 𝑫 ∙ 𝒅𝑺 is either 1 or 0. (sides + 2 ends)
– Choose surface such that magnitude D is constant (sides + 2 ends)
𝐷𝑆 ∙ 𝑑𝑆 =
+
𝑙𝑒𝑓𝑡
+
𝑠𝑖𝑑𝑒
= 𝑄𝑒𝑛𝑐𝑙
𝑟𝑖𝑔ℎ𝑡
– Left (zero field) and side (perpendicular)
contributions goes to zero so
𝐷𝐴 = 𝑄𝑒𝑛𝑐𝑙 = 𝜎𝑠 𝐴
– Result
𝐷 = 𝜎𝑠
𝜎𝑠
𝐸=
𝜀𝑜
Integrate over box,
Take divergence in rectangular
coordinates!
Flux Within a Differential Volume Element I.
The value of D at cube center (point P) is expressed in rectangular coordinates as
𝑫𝒐 = 𝐷𝑥𝑜 𝒂𝒙 + 𝐷𝑦𝑜 𝒂𝒚 + 𝐷𝑧𝑜 𝒂𝒛
Flux leaving cube of lengths Δx, Δy, Δz is:
𝐷 ∙ 𝑑𝑆 =
+
𝑓𝑟𝑜𝑛𝑡
𝑆
+
𝑏𝑎𝑐𝑘
+
𝑙𝑒𝑓𝑡
+
𝑟𝑖𝑔ℎ𝑡
Front contribution is:
= 𝑫𝒇𝒓𝒐𝒏𝒕 ∙ ∆𝑺
𝑓𝑟𝑜𝑛𝑡
= 𝑫𝒇𝒓𝒐𝒏𝒕 ∙ ∆𝑦∆𝑧 𝒂𝒙 = 𝐷𝑥,𝑓𝑟𝑜𝑛𝑡 ∆𝑦∆𝑧
Writing as series expansion:
= 𝐷𝑥0 +
𝑓𝑟𝑜𝑛𝑡
∆𝑥 𝜕𝐷𝑥
∆𝑦∆𝑧
2 𝜕𝑥
+
𝑡𝑜𝑝
𝑏𝑜𝑡𝑡𝑜𝑚
Flux Within a Differential Volume Element II.
Back contribution is:
= − 𝑫𝒃𝒂𝒄𝒌 ∙ ∆𝑺
𝒃𝒂𝒄𝒌
= −𝑫𝒃𝒂𝒄𝒌 ∙ ∆𝑦∆𝑧 𝒂𝒙 = −𝐷𝑥,𝑏𝑎𝑐𝑘 ∆𝑦∆𝑧
minus sign because Dx0 is inward flux through the back surface.
Writing as series expansion:
= − 𝐷𝑥0 −
𝑏𝑎𝑐𝑘
∆𝑥 𝜕𝐷𝑥
∆𝑦∆𝑧
2 𝜕𝑥
minus sign because extrapolating to back surface.
Combining front and back surfaces:
Flux Within a Differential Volume Element III.
Now, by a similar process, we find that:
and
All results are assembled to yield:
= Q
v
where Q is the charge enclosed within volume v
(by Gauss’ Law)
Example 1 - Charge in differential volume element from
divergence times volume
Charge in 10-9 volume is 2nC
Sum of discontinuity in D,
times volume, equals charge
enclosed
Example 2 - Charge from flux crossing area and
divergence times volume element
•
At surface z = 2 (surface density integrated over area)
3
𝐷𝑧 ∙ 𝑑𝑆 =
16𝑥 2 𝑦𝑧 3 𝒂𝒛 ∙ 𝒂𝒛 = 128
𝑥 2 𝑦 = 128
𝑥 2 𝑦 𝑑𝑥𝑑𝑦 = 1365𝑝𝑐
1
•
0
Partial Derivatives
𝜕𝐷𝑥
= 8𝑦𝑧 4 = −648 𝑝𝐶/𝑚3
𝜕𝑥
•
2
𝜕𝐷𝑦
=0
𝜕𝑦
𝜕𝐷𝑧
= 48𝑥 2 𝑦𝑧 2 = −1728 𝑝𝐶/𝑚3
𝜕𝑧
Total 𝛻 ∙ 𝐷 volume density integrated over volume
𝜕𝐷𝑥 𝜕𝐷𝑦 𝜕𝐷𝑧
+
+
∆𝑉 = −648 𝑝𝐶/𝑚3 − 1728 𝑝𝐶/𝑚3 10−12 𝑚3 = −2.38 × 10−21 𝐶
𝜕𝑥
𝜕𝑦
𝜕𝑧
Divergence in 3 Coordinate Systems
Back fly of book, see appendix for derivations
Divergence in 3 coordinates examples
•
a) rectangular coordinates problem
𝑑𝑖𝑣 𝐷 = 2𝑦𝑧 − 2𝑥 = −6 − 4 = −10
•
b)
cylindrical coordinates problem
1 𝜕
1 𝜕
𝜕
2
2
2
𝑑𝑖𝑣 𝐷 =
𝜌 2𝜌𝑧 𝑠𝑖𝑛 𝜑 +
𝜌𝑧 𝑠𝑖𝑛2𝜑 + (2𝜌2 𝑧𝑠𝑖𝑛2 𝜑)
𝜌 𝜕𝜌
𝜌 𝜕𝜑
𝜕𝑧
= 4𝑧 2 𝑠𝑖𝑛2 𝜑 + 2𝑧 2 𝑐𝑜𝑠2𝜑 + 2𝜌2 𝑠𝑖𝑛2 𝜑 = 3.53 − 1.53 + 7.06 = 9.06
•
c)
spherical coordinates problem Do with class.
Simple Divergence Examples
• For point charge in spherical coordinates*
𝐷=
𝑄
4𝜋𝑟 2
𝑑𝑖𝑣 𝐷 =
1 𝜕
𝑟 2 𝜕𝑟
𝑟2
𝑄
4𝜋𝑟 2
= 0 (𝑒𝑥𝑐𝑒𝑝𝑡 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛)
• For line charge in cylindrical coordinates*
𝐷=
𝜌𝐿
2𝜋𝜌
𝑑𝑖𝑣 𝐷 =
*fields also match surface charge
• Do in class
1 𝜕
𝜌 𝜕𝜌
𝜌
𝑄
2𝜋𝜌
= 0 (𝑒𝑥𝑐𝑒𝑝𝑡 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛)
Differential Gauss’s Law
• For differential element
𝐷 ∙ 𝑑𝑆 =
𝜕𝐷𝑥 𝜕𝐷𝑦 𝜕𝐷𝑧
+
+
∆𝑣 = 𝑄𝑒𝑛𝑐𝑙
𝜕𝑥
𝜕𝑦
𝜕𝑧
• Dividing both sides by Δv in the limit Δv -> 0
lim
∆𝑣→0
𝐷 ∙ 𝑑𝑆
∆𝑣
𝜕𝐷𝑥 𝜕𝐷𝑦 𝜕𝐷𝑧
𝑄𝑒𝑛𝑐𝑙
=
+
+
= lim
= 𝜌𝑣
∆𝑣→0 ∆𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
• Defining the divergence and del operators
𝜕𝐷𝑥 𝜕𝐷𝑦 𝜕𝐷𝑧
𝑑𝑖𝑣 𝑫 =
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝜕
𝛁=
𝒂 +
𝒂 + 𝒂
𝜕𝑥 𝒙 𝜕𝑦 𝒚 𝜕𝑧 𝒛
• Gives Gauss’s Law in Differential form
𝑑𝑖𝑣 𝑫 = 𝜌𝑣
𝜵 ∙ 𝑫 = 𝜌𝑣
“Physical” Gauss’s Law”
•
In general for any vector field
𝐴 ∙ 𝑑𝑆 =
Total flux entering
or leaving
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
+
+
∆𝑣 = 𝑆𝑜𝑢𝑟𝑐𝑒𝑒𝑛𝑐𝑙
𝜕𝑥
𝜕𝑦
𝜕𝑧
Source or sink
•
How field is “diverging” (outflow/inflow imbalance)
related to the local “sources” or “sinks”
•
Diverging – sum of gradients in each direction
(weighted by 1/ρ and 1/r2 in cylindrical and spherical)
•
Examples (flow related to sources and sinks)
– Water flow
– Current flow 𝛻 ∙ 𝐽 = −
– Heat flow
𝜕𝜌
𝜕𝑡
The Divergence Theorem
• Gauss’s Law in Integral form
𝑆
𝐷 ∙ 𝑑𝑆 = 𝑄𝑒𝑛𝑐𝑙
• Gauss’s Law in Differential form
𝜵 ∙ 𝑫 = 𝜌𝑣
• Combining
𝑆
𝐷 ∙ 𝑑𝑆 = 𝑄𝑒𝑛𝑐𝑙 =
𝑣𝑜𝑙
𝜌𝑣 𝑑𝑣 =
𝛻 ∙ 𝐷 𝑑𝑣
𝑣𝑜𝑙
• Divergence theorem for any vector field
𝐷 ∙ 𝑑𝑆 =
𝑆
𝛻 ∙ 𝐷 𝑑𝑣
𝑣𝑜𝑙
• Note: All “internal” fluxes cancel
Divergence theorem example
• Evaluate surface integral for 6 sides (top and bottom zero)
front
back
left
right
Outward normal points negative
– Back is zero for x=0
– Left and right cancel since 𝑫 ∙ 𝒂𝒚 = 𝐷𝑦 = 𝑥 2
Divergence theorem example (cont)
• Front is
(surface density integrated over area)
<< Left side Divergence Theorem
• Divergence is 𝛻 ∙ 𝐷 volume density integrated over volume
<< Right side Divergence Theorem
Quiz problem 2.26
•
•
A radially dependent surface charge is distributed on an infinite flat
sheet in the x-y plane and is characterized in cylindrical coordinates
by surface charge density 𝜌𝑠 = 𝜌𝑜 𝜌, where ρ is a constant.
Determine the electric field strength everywhere on z axis.
Setting up Coulomb’s Law in z direction:
2𝜋 ∞
𝐸=
0 0
𝜌𝑠 𝑧
4𝜋𝜀𝑜
𝜌𝑠
𝐸=
2𝜀𝑜
−𝜌𝑠
2
𝜌2
∞
0
+
3 𝒂𝒛
2
2
𝑧
𝑧
𝜌2
+
3 𝒂𝒛
2
2
𝑧
∞
1
𝜌2
+
1
𝑧2 2
𝜌𝑠
𝒂
2𝑧 𝒛
Doing the integral on my phone – www.mathstudio.net
𝒂𝒛
0