Gauss’ law - plane symmetry

Gauss’ law - plane symmetry
Submitted by: I.D. 039023262
The problem:
Find the electric field for:
1. infinite uniformly charged plane (charge density σ)
2. infinite uniformly charged layer (charge density ρ and width h)
3. two adjacent infinite layers of width h charged ρ+ and ρ− uniformly
4. two infinite planes perpendicular to each other, both charged σ uniformly.
The solution:
H
R
~ · d~s = 4πk ρdv, and knowing that for plane symmetry E
~ = Ez ẑ
1. According to Gauss’ law E
we construct a Gauss shell enclosing a piece of the plane of area S:
I
Z
~
E · d~s = 4πk ρdv
(1)
2Ez · S = 4πkσS
(2)
Ez = 2πkσ · sign(z)ẑ
(3)
2. For infinite layer of the width h (0 < z < h), if z > h or z < 0 the solution is identical to the one
for the infinite plane. In the case 0 < z < h we construct a shell of area S symmetrical relatively
to the middle of the layer - its its bottom plate is at z, the top plate is at h − z and the height,
therefore, is h − 2z.
2E · S = 4πk(h − 2z)Sρ
(4)
E = 2πkρ(h − 2z)
(5)


2πkρh,
z>h



2πkρ(2z − h), h/2 < z < h
=

2πkρ(h − 2z), 0 < z < h/2



−2πkρh,
z<0
(6)
Then
~z
E
3. Two infinite layers of thickness h charged ρ+ and ρ− uniformly. We choose the ρ+ charged layer
to be at 0 < z < h and the ρ− charged layer to be at −h < z < 0. Then using the result of for the
one layer and the superposition principle we obtain:


2πk(ρ+ + ρ− )h,
z>h




2πkρ+ (2z − h) + 2πkρ− h,
h/2 < z < h

~
(7)
Ez =
2πkρ+ (h − 2z) + 2πkρ− h,
0 < z < h/2



−2πkρ+ h + 2πkρ− (h + 2z), −h < z < 0




−2πk(ρ+ + ρ− )h,
z < −h
4. Two infinite plains perpendicular to each other, both charged σ uniformly.
By using the principle of superposition, we simply calculate the field intensity of each plane, and
~ = 2πkσ(sign(z)ẑ + sign(y)ŷ) (assuming that the plains in
combine them. For tow plane we get E
question are the XY and XZ)
1
Gauss’s law - plane symmetry
Submitted by: I.D. 039023262
The problem:
Find the electric field along the z-axis of an infinite uniformely charged plane at the x − y plane
(charge density σ) with a hole at the origin of a radius r0 .
The solution:
Using the principle of superposition we shall calculate the electric field induced by the missing hole
as being of a disk charged with −σ, then combine it with an infinite plane:
!
z
Ezdisk = −2πkσ sign(z) − p
(1)
z 2 + r02
Ezplane = 2πkσ sign(z)
z
Ez = 2πkσ p
2
z + r02
(2)
(3)
1
Gauss’ law - cylindrical symmetry
Submitted by: I.D. 039023262
The problem:
Given an infinite cylinder charged ρ(~r) = ρ0
1. Find the electric field.
2. What should be the charge density ρ(~r) so that the electric filed inside the cylinder would be
zero?
The solution:
~ = Er r̂ . The volume element is dv = rdrdθdz
1. For cylindrical symmetry E
For r < R
Z
Z r
Er · 2πrh = 4πh
ρdv = 4πh
ρ2πr0 dr0 h
V
(1)
0
~ = 2πkρrr̂
E
(2)
For r > R
Z
Er · 2πrh = 4πh
Z
V
~ =
E
R
ρdv = 4πh
ρ2πr0 dr0 h
(3)
0
2πkρR2
r̂
r
(4)
2. We solve the problem in two ways - using integral and diferential versions of the Gauss’ law:
a. integral:
I
Z
~ = 4πk ρdv
~ · ds
E
(5)
Z r
E · 2πrh = 4πk
ρ(r0 ) · 2πr0 dr0 h
(6)
0
ρ(r) ∼
1
r
(7)
Checking:
Z r
ρ(r0 ) · 2πr0 dr0 h = 2πrh
(8)
0
~ = 4πkr̂ = const
E
(9)
b. differential:
~ = 4πkρ
div E
(10)
~ = const
~ =E
~ ⇒ div E
E
r
1
E
4πk
ρ =
⇒
=
r
r
r
~ = 4πkr̂ = const
E
(11)
(12)
(13)
1
Gauss’ law - cylindrical symmetry
Submitted by: I.D. 039023262
The problem:
In an infinite cylinder of a radius R charged uniformly with a charge density ρ there is a cylindrical
infinite hole of a radius r0 which center positioned at ~b. Find the electric field.
The solution:
We use the principle of superposition: the problem is equal to a full cylinder of a radius R charged
with ρ plus a full cylinder of a radius r0 charged with −ρ.
1. outside the cylinder
E =
2πkρR2
−2πkρr02
~
r
+
(~r − ~b)
|~r|2
|~r − ~b|3
(1)
2. inside the cylinder, outside the hole
E = 2πkρ~r +
−2πkρr02
(~r − ~b)
|~r − ~b|3
(2)
3. inside the hole
E = 2πkρ~r − 2πkρ(~r − ~b) = 2πkρ~b = const!
1
(3)
Gauss’ law - spherical symmetry
Submitted by: I.D. 039023262
The problem:
1. Find the electric field of a point charge
2. Find the electric field from a spherical shell of a radius R charged with a charge Q uniformly.
The solution:
For a point source the charge density is ρ(~r) = Qδ(~r). Then
I
Z
~ · d~s = 4πk
ρdv
E
s
V
Z
2
Er · 4πr = 4πk Qδ(~r)dV = 4πkQ
Er =
kQ
r2
(1)
(2)
(3)
~ where R
~ = Rr̂.
For a spherical shell the charge density is ρ(~r) = Qδ(~r − R)
Z
Z
ρdV =
V
(
0,
~
Qδ(~r − R)dV
=
1,
r<R
r>R
(4)
Then
~ =
E
(
0,
kQ
r̂,
r2
r<R
r>R
(5)
1