Download Gauss` law - spherical symmetry

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)
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Gauss’s law - plane symmetry
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
~ · d~s = 4πk ρdv
E
(1)
2Ez · S = 4πkσS
~ z = 2πkσ · sign(z)ẑ
E
(2)
(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)


z>h
2πkρh,
=
2πkρ(2z − h), 0 < z < h


−2πkρh,
z<0
(6)
Then
Ez
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,
0<z<h
+
−
Ez =
(7)

−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 two planes we get E
question are the XY and XZ).
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The electric field
Submitted by: I.D. 313947202
The problem:
An electrical infinite straight wire with charge density λ is at the center of an infinite hollow cylinder
with charge density σ and a radius a.
What is the electrical field everywhere?
The solution:
The solution is superposition of two fields: one from wire and another one from cylinder. Both
fields we will find with the Gauss’ law. Because of symmetry,the field are in the radial direction
only.
Field of the wire:
E · 2πrl = 4πkQ = 4πλl
2λk
E =
r
(1)
(2)
Field of the cylinder:
inside:
E · 2πrl = 4πkQ = 4π0
(3)
E = 0
(4)
outside:
E · 2πrl = 4πkQ = 4πk(2πal)σ
4πkaσ
E =
r
(5)
(6)
The field from the whole system:
(
2λk
r<R
r r̂,
~ =
E
4πaσ
2λk
r r̂ + r r̂, r > R
(7)
1
Electric field
Submitted by: I.D. 060645009
The problem:
• Calculate directly the electric field flux through an infinite plane a distance d from a point
charge.
• Explain why your result does not depend on d.
The solution:
Let the plane to be at the x − y plane, so that any point on it is ~r = (r cos(ϕ), r sin(ϕ), 0) and the
point charge is at ~r0 = (0, 0, d). By the Coloumb law
kdq(~r − ~r0 )
|~r − ~r0 |3
~ =
dE
(1)
In our case
~r − ~r0 = (r cos(ϕ), r sin(ϕ), −d)
0
2
(2)
2 1/2
|~r − ~r | = (r + d )
(3)
So that
~ =
E
(r2
kq
(r cos(ϕ), r sin(ϕ), −d)
+ d2 )3/2
The flus is
I
Z
~ =
~ · dA
E
Ez dA
Z 2π
Z
=
dϕ
0
Z
= 2πkqd
0
(4)
(5)
∞
0
∞
rdr
kqd
3
(r2 + d2 ) 2
rdr
3 = 2πkq
(r2 + d2 ) 2
(6)
(7)
The total flux from a point charge if 4πkq. Since the plain is infinite, half of the electric field lines
pass through this plane (not depending on d), that is half of the flux 2πkq.
1