Example 1: THE ELECTRIC DIPOLE 1 The Electric Dipole: z

r+
+ d
θ
P

r

r−
_ Q
Q
Q ⎛ 1 1⎞
Φ=
−
=
− ⎟
⎜
4πε r+ 4πε r− 4πε ⎝ r+ r− ⎠
2 The Electric Dipole: z

r+
+ θ
d
P

r
Law of Cosines: B
A
α
C
A2 = B 2 + C 2 − 2 ABcos α

r−
_ ( 2)
r± = r + d
2
2
d
 2r cosθ
2
3 The Electric Dipole: z

r+
+ θ
d

r
P
Important pracBcal approximaBon: d << r

r−
_ ( 2 )  rd cosθ
r± = r 2 + d
2
4 The Electric Dipole: d << r
( 2 )  rd cosθ
r± = r + d
2
2
d2 d
= r 1+ 2  cosθ
r
4r
d
≈ r 1 cosθ
r
⎛
⎞
d
d
≈ r ⎜ 1 cosθ ⎟ = r  cosθ
2
⎝ 2r
⎠
x << 1
x
1± x ≈ 1±
2
5 d << r
The Electric Dipole: z
+ d
d
r − cosθ
2
r
θ
_ d
r + cosθ
2
6 The Electric Dipole: x << 1
Q ⎛ 1 1⎞
Φ=
− ⎟
⎜
4πε ⎝ r+ r− ⎠
1
≈ 1 x
1± x
⎛
⎞
d<<r
⎟
Q ⎜
1
1
=
−
⎜
⎟
d
d
4πε r
⎜ 1− cosθ 1+ cosθ ⎟
⎝
⎠
2r
2r
⎛
⎞⎞
Q ⎛
d
d
Qd
≈
1+ cosθ − ⎜ 1− cosθ ⎟ ⎟ =
cosθ
2
⎜
4πε r ⎝ 2r
⎝ 2r
⎠ ⎠ 4πε r
7 The Electric Dipole: 
Define p ≈ Qd ẑ
and note cos θ = ẑ i r̂
( )
Qd
Φ≈
cosθ
2
4πε r
ẑ
θ
r̂

p i r̂
Φ≈
2
4πε r
8 The Electric Dipole: 
p i r̂
Φ≈
4πε r 2

∂Φ ˆ 1 ∂Φ
E = −∇Φ = − r̂
−θ
∂r
r ∂θ
⎛ ∂
⎞
⎜⎝ ∂ϕ = 0⎟⎠
⎞ ˆ 1 ∂ ⎛ Qd
⎞
∂ ⎛ Qd
= − r̂ ⎜
cosθ ⎟ − θ
cosθ ⎟
2
2
⎜
∂r ⎝ 4πε r
r ∂θ ⎝ 4πε r
⎠
⎠
⎛
⎞ ˆ 1 ⎛ Qd
⎞
Qd
= − r̂ ⎜ −2
cosθ ⎟ − θ ⎜ −
sin θ ⎟
3
2
r ⎝ 4πε r
⎝ 4πε r
⎠
⎠
(
Qd
ˆ sin θ
=
2
r̂
cos
θ
+
θ
4πε r 3
)
9 The Electric Dipole: 
Qd
ˆ sin θ
E=
2
r̂
cos
θ
+
θ
3
4πε r

Qd
Qd
2
2
2
E =
4cos
θ
+
sin
θ
=
1+
3cos
θ
3
3
4πε r
4πε r
(
)
10 Example 2: FINITE LENGTH LINE OF CHARGE (again) Earlier we found the E-­‐field on the z-­‐axis. Doing anything else would have required difficult integraBons. Here is a case where it is easier to find the potenBal and then compute the electric field. 11 a
dQ = ρ dz ′

r ′ = z ′ẑ
0
ρ
−a
z
Note the φ-­‐independence (0,0, z′ )
  
2
2
R = r − r ′, R = r + ( z − z ′ )
r
z

r
r
dΦ =
P dE
r
dEz

dE P
ρ dz ′
 
4πε o r − r ′
12 ρ  dz ′
dΦ =
 
4πε o r − r ′
ρ dz ′
Φ= ∫
 
4
πε
r
− r′
−a
o
a
a
=
∫
−a
∫
ρ dz ′
4πε o r + ( z − z ′ )
2
dx
x +a
2
2
{
= ln x + x 2 + a 2
}
2
2
⎧
2
z
−
a
+
r
+
z
−
a
(
)
ρ
⎪
=−
ln ⎨
4πε o ⎪ z + a + r 2 + z − a 2
( )
⎩
⎫
⎪
⎬
⎪
⎭
13 2
⎧
2
z
−
a
+
r
+
z
−
a
(
)
ρ
⎪
Φ=−
ln ⎨
4πε o ⎪ z + a + r 2 + z − a 2
( )
⎩

E = −∇Φ
⎫
⎪
⎬
⎪
⎭
⎛
⎞
ρ
∂Φ
1
1
⎟
Ez = −
=−  ⎜
−
2 ⎟
∂z
4πε o ⎜ r 2 + z − a 2
2
( ) r + ( z + a) ⎠
⎝
⎡
−r
⎢
2
2
2
2
ρ ⎢ r + ( z − a ) + ( z − a ) r + ( z − a )
∂Φ
Er = −
=−
⎢
∂r
4πε o ⎢
r
+
⎢
2
2
2
2
r + ( z + a) + ( z + a) r + ( z + a)
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
14 for z = 0
⎛
⎞
ρ ⎜
1
1
⎟ =0
Ez = −
−
2 ⎟
4πε o ⎜ r 2 + 0 − a 2
2
r
+
0
+
a
( )
( )⎠
⎝
⎡
−r
⎢
2
2
2
2
r
+
z
−
a
+
z
−
a
r
+
z
−
a
⎢
( ) ( )
( )
ρ
Er = −
⎢
4πε o ⎢
r
+
⎢
2
2
2
2
r
+
z
+
a
+
z
+
a
r
+
z
+
a
( ) ( )
( )
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
z=0
⎤
ρ ⎡
−r
r
=−
+
⎢ 2
⎥
2
2
2
2
2
2
2
4πε o ⎣ r + a − a r + a
r +a +a r +a ⎦
ρ a
ρ
1
=
→
2πε o r r 2 + a 2 a → ∞ 2πε o r Agrees with our earlier results 15 Example 3: INFINITELY LONG LINE OF CHARGE via Gauss’s Law 16 Infinitely long line charge: Note that the fields MUST be independent of both z and φ
r
z

No contribuBon over end caps since Gaussian Surface  
∫ D i da =
S
∫
r̂ i ẑ = 0

E = Er r̂
ε o Er r̂ i r̂r dϕ dz +
Cylinder
∫
ε o Er r̂
i ẑ r dr dϕ
End
Caps
=0
2π
=  ∫ ε o Er r dϕ = 2π ε o Er r = Qenc = ρ 
0
ρ
ρ
⇒ Er =
=
2π ε o r 2πε o r
When the necessary symmetry exists, Gauss’s Law is generally MUCH simpler than Coulomb’s Law. 17 Example 4: A SPHERICAL CLOUD OF CHARGE 18 Spherical cloud of (uniform) charge: Since the charge density is uniform: QTotal = ρ 43 π r 3
Gaussian Surface 2 Gaussian Surface 1 ρv
a
Note that the fields must be independent of both θ and φ, thus 
E = Er r̂
19 Spherical cloud of (uniform) charge: Gaussian Surface 2 For Gaussian Surface 1:  
2
ε
E
i
d
a
=
ε
E
r̂
i
r̂4
π
r
o r
∫ o
S1
Gaussian Surface 1 ρv
= QEnclosed
a
⎛ r⎞
= QTotal ⎜ ⎟
⎝ a⎠
3
QTotal r 3
QTotal
⇒ Er =
=
r
2
3
3
4πε o r a
4πε o a
r<a
QTotal = ρ 43 π r 3

E = Er r̂
20 Spherical cloud of (uniform) charge: Gaussian Surface 2 For Gaussian Surface 2:  
2
ε
E
i
d
a
=
4
πε
E
r
o r
∫ o
Gaussian Surface 1 ρv
QTotal = ρ 43 π r 3
S2
= QEnclosed = QTotal
QTotal
⇒ Er =
,r >a
2
4πε o r
a

E = Er r̂
21 Spherical cloud of (uniform) charge: Gaussian Surface 2 Er
QTotal
2
4πε o a
Gaussian Surface 1 ρv
a
a
r
What is the potenBal? 22 Spherical cloud of (uniform) charge: r
Φ = − ∫ Er r̂ i r̂ dr
∞
r
⎧
QTotal
⎪
−∫
dr
r>a
2
⎪
∞ 4πε o r
=⎨ a
r
Q
QTotal r
⎪
Total
⎪ − ∫ 4πε r 2 dr − ∫ 4πε a 3 dr r < a
a
o
o
⎩ ∞
⎧
QTotal
⎪
r>a
4πε o r
⎪
=⎨
2
2
Q
Q
a
−
r
⎪ Total + Total
r<a
3
⎪ 4πε o a 4πε a
2
o
⎩
23 Example 5: AN INFINITE SHEET OF CHARGE 24 Infinite sheet of charge: a simple yet important result for the study of the parallel plate capacitor Note how the fields must be independent of x, y, and z Gaussian Surface ρs
r
z
y
x
 ⎧⎪ ẑEz
E=⎨
⎪⎩ − ẑEz
z>0
z<0
25 Infinite sheet of charge:  
ε o ∫ E i da = Q = ρs π r 2
( )
S
 
ε o ∫ E i da = ε o
S
εo
∫
 
E i da + ε o
∫
Top
Surface
ẑEz i ẑ da + ε o
Top
Surface
+ε o
∫
Bottom
Surface
 
E i da + ε o
∫ ( − ẑE ) i ( − ẑda )
∫
 
E i da
Cylindrical
Side
z
Bottom
Surface
∫ ( ± ẑE ) i r̂a dϕ dz
z
Cylindrical
Side
da = rdrdϕ = π r 2
ρs
⇒ 2ε o Ezπ r = π r ρs ⇒ Ez =
2ε o
2
2
26 Infinite sheet of charge: ⎧
ρs
⎪ ẑ
z>0
 ⎪ 2ε o
E=⎨
ρs
⎪
− ẑ
z<0
⎪
2ε o
⎩
Since the sheet extends of infinity we would expect trouble finding the potenBal: ρs
Φ = − ∫ ẑEz i ẑ dz = − ∫
dz = ∞
2ε o
∞
∞
z
z
27 Infinite sheet of charge: However, Φ ab = Φ ( b) − Φ ( a )
ρs
ρs
= − ∫ ẑEz i ẑ dz = − ∫
dz = −
2ε o
2ε o
a
a
b
b
b
a
ρs
ρs
=−
b − a) =
a − b)
(
(
2ε o
2ε o
28 Example 6: TWO COAXIAL SHELLS OF CHARGE 29 Two coaxial shells of charge: Once again, neglecBng end effects, z
b

E = Er r̂
a
h
Q
−Q
ρsa
Note :
− ρsb
Q
−Q
ρsa =
, ρsb =
2π ah
2π bh
2π ahρsa = −2π bhρsb
a
⇒ ρsb = − ρsa
b
30 Two coaxial shells of charge: z
b
Gaussian Surface 1 Once again, neglecBng end effects, 
E = Er r̂
a
Gaussian Surface 3 Gaussian Surface 2 The charge enclosed by surfaces one and three is zero, hence Er = 0 inside the inner cylinder and outside the outer cylinder. Also, the top and bofom surfaces do not contribute to the integral as usual, since r̂ i ( ± ẑ ) = 0
31 Two coaxial shells of charge: Q
∫S ( Er r̂ ) i ( r̂ r dϕ dz ) = ε o
2
Q
∫S Er r dϕ dz = ε o
2
Er ( 2π rh ) =
ρs ( 2π ah )
εo
ρs a
⇒ Er =
ε or
32 Two coaxial shells of charge: ⎧ 0
⎪
 ⎪ ρs a
E=⎨
⎪ ε or
⎪ 0
⎩
r<a
a<r<b
r>b
ρs a
Φ ( r ) = − ∫ ( Er r̂ ) i ( r̂ dr ) = − ∫
dr
εr
∞
b o
r
r
r
ρs a
ρs a
ρs a r
=−
ln r ) = −
ln r − ln b) = −
ln , a < r < b
(
(
b
εo
εo
εo b
⎧ ρa r
⎪ − s ln
a<r<b
Φ(r ) = ⎨
εo b
⎪
0
otherwise
⎩
33 Two coaxial shells of charge: Also, ρs a a
Φ ba = Φ ( a ) − Φ ( b) = −
ln , a < r < b
εo b
⎧
r
ln
⎪
⎪ Φ ba b
Φ(r ) = ⎨
a
ln
⎪
b
⎪
0
⎩
a<r<b
otherwise
34 Two coaxial shells of charge: Gauss’s Law was derived from: 
 
∇ i D = ρ ⇒ D i da = Qenclosed


S
pointwise




over a volume in space
At a point where there is no charge (i.e., inside the cylinder) the divergence should equal zero. Let’s verify this for this example: 
 ρs a
D = εoE =
, a<r<b
ε or
 1 ∂
⎛ ρs a ⎞
1
∂
∇i D =
rDr =
r
≡0
⎜
⎟
r ∂r
r ∂r ⎝ ε o r ⎠
As an exercise, verify that the divergence of the dipole field found earlier is also zero. ∫
( )
35 Two coaxial shells of charge: As an exercise, verify that the divergence of the dipole field found earlier is also zero. 
 Qd
ˆ sin θ
D = εE =
2
r̂
cos
θ
+
θ
4π r 3
i.e., show that (
)

∇i D = 0
36