2.2 Divergence and Curl of Electrostatic Fields
2.2.1 Field Lines, Flux, and Gauss’s Law
Tools for avoiding integral…
Field Lines:
Connect vectors to form field lines.
The magnitude of the field is indicated by the density of the field.
Field lines begin on positive charge and end on negative ones.
Field lines cannot simply terminate in midair.
Field lines can never cross.
Electric Flux:
It is a measure of the number of field lines passing through a surface.
r r
Φ E = ∫ E ⋅ da
S
The flux through any closed surface is a measure of the total charge.
Equipotential surface:
It is easy to understand the image method that introduced in Chapter 3 by
equipotential surface.
Gauss’s Law:
r
For a point charge q at the origin, the flux of E through a sphere of radius r is:
π 2π
r
1 q
1 q
q
2
ˆ
ˆ
ˆ
r
n
da
E
⋅
n
da
=
⋅
=
∫
∫ 4πε 0 r 2
∫0 ∫0 4πε 0 r 2 rˆ ⋅ rˆr sin θdφdθ = ε 0
The closed surface didn’t have to be a sphere.
The flux through any surface enclosing the charge is
q
ε0
.
According to the principle of superposition:
r r
r
r
∫ ∑ Ei ⋅ da = ∫ Etotal ⋅ da =
S
i
S
∑q
i
i
ε0
=
Qenc
ε0
By applying the vector calculus:
r
r r
r
Qenc 1
∇
⋅
E
=
∫ E ⋅ da = ∫ ∇ ⋅ Edτ = = ∫ ρdτ Æ
ε0
S
corollary?
ε0
ρ
(Do you believe this crude
ε
You might have to prove it in the next section.)
Gauss’s Law:
r r Qenc
E
∫ ⋅ da =
S
ε0
r ρ
Gauss’s law in differential form: ∇ ⋅ E =
ε0
Example: A charge q sits at the back corner of a cube.
r
the flux of E through the shaded area?
Guess it from the consideration of symmetry.
q 11
ΦE =
ε0 8 3
What is
Electric Field Due to an Electric Dipole
  r dr   r dr  
 r −  r +  
r r
2  
2  
q  

E (r ) =
−
r3 
r
4πε 0  r d 3
r d 
r+
 r −

2
2 

z
r
r
r
d +q
-q
r −3
r
−3
−3 / 2
r r
 3 r ⋅d 
 rˆ ⋅ d d 2 
r d
d2 r r
−3
−3
2

r−
~ r 1 +
= r +
− r ⋅ d = r 1 −
+ 2 
2 
2
4
r
r
r
4
2




r
r
r
r
 r d  3 rr ⋅ d   r d  3 rr ⋅ d 

 −  r + 1 −
 r − 1 +
2  2 r 2 
4πε 0 r 3 
2  2 r 2  
r r
q  r ⋅ d r r
=
r − d
3
4πε 0 r 3  r 2

r r
E (r ) ≅
q
2.2.2 The Divergence of E
To find out the Gauss’s law in differential form, we start from the electric field from a
volume charge.
 rr − rr ' 

ˆ
r
Π
 = −∇'⋅

(
)
∇
⋅
'
'
ρ
r
d
τ
since
 rr − rr ' 3 

4πε 0 ∫ Π 2



r
r r
r r
r
1
ρ (r )
∇ ⋅ E (r ) =
4πδ 3 (r − r ')ρ (r ')dτ ' =
∫
4πε 0
ε0
r r
E (r ) =
1
r r
r
r − r ' 
3 r
r r 3  = 4πδ (r − r ')
r − r' 
Derive the integral form by using vector analysis:
r
r
r r Q
ρ (r )
∇
⋅
E
d
=
τ
∫
∫ dτ → ∫ E ⋅ da = enc
V
V
ε0
ε0
S
2.2.3 Applications of Gauss’s Law
All is to find the space symmetry.
1. Spherical symmetry. Make your Gaussian surface a concentric sphere.
2. Cylindrical symmetry. A coaxial cylinder.
3. Plane symmetry.
Example: Find the field of a solid sphere of radius R and of total charge q with
uniform volume charge ρ inside.
4π 3
Rρ
r r
q
=
r > R : 4πr 2 E = 3
Æ E (r ) =
ε0
ε0
4π 3
4π 3
r ρ
r
r < R : 4πr 2 E = 3
= 3
ε0
ε0
q
4π 3
R
3
1
q
rˆ
4πε 0 r 2
r r
Æ E (r ) =
1
qr
rˆ
4πε 0 R 3
Example: A long cylinder carries a charge density that is proportional to the distance
from the axis: ρ = kr , for some constant k. Find the electric field.
r < R : 2πrLE =
1
ε0
r 2π
L ∫ ∫ (kr ')r ' dφdr ' =
0 0
1
ε0
L 2πk
r
r3
kr 2
Æ E (r ,φ , z ) =
rˆ
3ε 0
3
Example: An infinite plane carries a uniform surface charge σ .
r
Aσ
σ
2 AE =
nˆ
Æ E ( x, y , z ) =
2ε 0
ε0
Example: Two infinite planes carries equal but opposite charge
densities ± σ .
Apply superposition rule:
r
σ
E = − xˆ
ε0
Example: Two spheres…
Apply superposition rule and vector addition.
-
+
2.2.4 The Curl of E
For a point charge q placing at the origin, the electric field is:
r r
1 q
E (r ) =
rˆ
4πε 0 r 2
The curl calculation by integration:
r
r
r
r r r
ˆrdθ + φˆr sin θdφ , E (rr ) ⋅ dl =
ˆ
(
)
E
r
⋅
d
l
,
d
l
=
r
dr
+
θ
∫
rb
ra
1
q
dr
4πε 0 r 2

q
q 
1  1  1 
∫ra 4πε 0 r 2 dr = − 4πε 0r  = 4πε 0  − rb −  − ra  


ra
rb
rb
1
r r
E
∇
×
⋅ da =
∫
surface
r r rb r r ra r r
∫ E ⋅ dl = ∫ E ⋅ dl + ∫ E ⋅ dl =
bounding _ loop
ra
rb
1 1 1
1 1 1
 −  +
 − =0
4πε 0  ra rb  4πε 0  rb ra 
r r
∇ r × E (r ) = 0
r
r r
r
r
r
r
By superposition rule: Etotal = E1 + E2 + E3 + ... , ∇ × Etotal = ∇ × E1 + ∇ × E2 + ... = 0
The curl calculation by differentiation:
rˆ
r
∂
1
∇× E = 2
r sin θ ∂r
Er
θˆ
∂
∂θ
rEθ
φˆ
∂
1
= 2
∂φ
r sin θ
r sin θEφ
rˆ
∂
∂r
1
4πε 0 r 2
θˆ
∂
∂θ
0
φˆ
∂
=0
∂φ
0
General form:
r r
E (r ) =
1
r
ˆ
ρ (r ')Π
4πε 0 V∫ Π 2
(rr − rr ')
dτ ' , r r 3 = ? rˆ + ?θˆ + ?φˆ , It’s difficult to calculate and we
r − r'
r
need to simplify the problem. Every time we have a source point at r ' , we can
r
r
reorient the coordinates to relocate the r ' at the z axis. Let r '= r ' zˆ to do simple
calculation.
(rr − rr ') = (rr − rr ') ⋅ rˆ rˆ + (rr − rr ') ⋅ θˆ θˆ + (rr − rr ') ⋅ φˆ φˆ
r r3
r − r'
r r3
r − r'
r r3
r − r'
r r3
r − r'
Remember that rˆ = sin θ cos φxˆ + sin θ sin φyˆ + cosθzˆ ,
θˆ = cosθ cos φxˆ + cosθ sin φyˆ − sin θzˆ , φˆ = − sin φxˆ + cos φyˆ
(rr − rr ') =
r r3
r − r'
(r
r − r ' cosθ
2
+ r ' −2rr ' cosθ
2
rˆ +
(r
r ' sin θ
2
+ r ' −2rr ' cosθ
2
rˆ
∂
∂r
r − r ' cosθ
r r
(
1
r − r ')
∇× r r 3 = 2
r sin θ
r − r'
(r
)
3/ 2
2
+ r '2 −2rr ' cosθ )
Exercise: 11, 12, 16, 18
3/ 2
r
(r
)
3/ 2
θˆ + 0φˆ
rθˆ
∂
∂θ
r ' sin θ
2
r sin θφˆ
∂
=0
∂φ
+ r '2 −2rr ' cosθ )
3/ 2
0