Download Polarized sphere

Polarized sphere
The problem:
Two spheres with radii r0 are homogeneously charged with q and −q. The distance between their
centers is the vector d~ such that d r0
1. Find the electric field in the whole space.
2. Find the charge distribution on the surface of the sphere
~ 0 . Find the
A dielectric sphere (of permittivity ε) is placed into an external constant electric field E
electric field inside the sphere.
The solution:
1. The field outside the spheres is the same as one of two point charges.
p~ = q d~
p · ~r)~r − p~r2
~ = k 3(~
E
r5
(1)
(2)
~ = k q3 ~r. Let ~r be the
Inside the spheres we need to sum the two fields. The field of one sphere is E
r0
~
position of a point inside the spheres relatively to d/2, than ~r+ , ~r− are the position of this point
relatively to the spheres centers.
d~
2
d~
= ~r +
2
q
= k 3 ~r −
r0
~r+ = ~r −
~r−
E~+
E~−
q
= −k 3
r0
(3)
(4)
d~
2
!
d~
~r +
2
(5)
!
(6)
3~
~
~
~ = −k q d = −k p~ = −k 4/3πr0 P = − P
E
3ε0
r03
r03
r03
(7)
where P~ is the polarization vector.
2. The discontinuity of the normal components of the electric fields on the surface of the sphere is
E~r+ − E~r− = 4πkσ. Let us choose the dipole moment to be in the ẑ direction p~ = pẑ and the angle
θ to be the angle to the z-axis.
p · ~r)r2 − (~
p · ~r)r2
2kp cos θ
2k~
p · ~r
2kpr0 cos θ
~ · r̂ = E
~ · ~r = k 3(~
E~r+ = E
=
=
=
4
6
4
r
r
r
r0
r03
p~ ~r
kpr0 cos θ
kp
E~r− = −k 3 · = −
= − 3 cos θ
4
r0 r
r0
r0
3kp
∆E = E~r+ − E~r− = 3 cos θ = 4πkσ
r0
3p
σ =
cos θ = σ0 cos θ
4πr03
1
(8)
(9)
(10)
(11)
3. Given that the polarization is
~
P~ = ε0 χE
and the susceptibility χ =
(12)
ε
ε0
− 1 we write
~
~
~ =E
~0 + E
~b = E
~0 − P = E
~ 0 − ε0 χE
E
3ε0
3ε0
(13)
Then
~0
~ 1 + 1χ
= E
E
3
~ = 3ε0 E
~0
E
2ε0 + ε
(14)
(15)
2
Displecment Current
Submitted by: I.D. 043423755
The problem:
A long hollow cylinder is made out of a non-conducting material with a radius R, length l, and a
charge surface density σ.
An external torque rotates the cylinder around its axis with a steady angular velocity of ω(t) = αt
1. Calculate the magnetic field inside the cylinder
2. What is the electrical field on the inner surface of the cylinder?
3. Calculate the Poynting vector on the inner surface of the cylinder.
4. Calculate the energy flux which enters the inner volume of the cylinder and deduce the conclusion that arises from this.
The solution:
The Ampere’s law:
I
B · d~l = µ0 I
(1)
The current per a unit length is
Q/l
Qω
=
T
2πl
Q = σ(2πlR)
I =
(2)
(3)
So the magnetic field here is like that of a coil
~ = µ0 I0 nẑ = µ0 σRωẑ = µ0 σαRtẑ
B
(4)
2. Using the Faraday’s law
Φ = B · A = µ0 σαRt · πR2
(5)
~ is in the tangential direction
Field E
I
~ · d~l = − dΦ
E
dt
Φ
= µ0 σαR3 π
dt
2πRE θ̂ = −µ0 σαR3 π
1
Eθ = − µ0 σαR2
2
(6)
(7)
(8)
(9)
3. The Poynting vector is
1 ~
~
E×B
µ0
1 1
1
= −
µ0 σαR2 θ̂ × µ0 σRαtẑ = − µ0 σ 2 α2 R3 tr̂
µ0 2
2
~ =
S
1
(10)
(11)
The meaning of the minus sign is that the vector pointing into the cylinder from outside, i.e. the
energy flow is from outside.
4. The energy entering inside the cylinder is:
Z
~ · d~a
(−S)
φ =
(12)
s
=
=
=
1
1 d(t)2
µ0 σ 2 α2 R3 t · 2πRl = πµ0 (σα)2 R3 l
2 2 dt
1
d 1
d
2
2
2 4 2
(µ0 σRαt) · (πR l)
πµ0 (σα) R lt =
dt 2
dt 2µ0
1
d
BV = P
dt 2µ0
(13)
(14)
(15)
where P is the rate of the magnetic enrgy produced inside the cylinder. The energy density of the
electric field inside the cylinder is constant, so that it does not contribute to the power.
The power of the electric force opposing to the rotation is
Peff = F~ · ~v = −EQωR
1
=
µ0 σαR2 (σ2πlR) (αtR)
2
d
1
d
1
2
2
=
((µ0)σαRt) (πR l) =
BV
dt 2µ0
dt 2µ0
(16)
(17)
(18)
We obtained the the electric field is doing work for stopping the rotation. This work consumes all
the electromagnetic field power entering from outside. Actually, this work of the electric field causes
all the consumption of energy into the system.
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