VEKTORANALYS Kursvecka 6 övningar. PROBLEM 1 SOLUTION A dipole is formed by two point sources with...
-
Upload
christa-shewbridge -
Category
Documents
-
view
213 -
download
0
Transcript of VEKTORANALYS Kursvecka 6 övningar. PROBLEM 1 SOLUTION A dipole is formed by two point sources with...
PROBLEM 1
SOLUTION
A dipole is formed by two point sources with charge +c and -cCalculate the flux of the dipole field on a closed surface S that
(a)encloses both poles(b)encloses only the plus pole(c) does not enclose any pole
-
+
03
0
( )r r
A r cr r
Vector field from a point sourcelocated in r0 :
1 23 3
1 2
( )r r r r
A r c cr r r r
Vector field from two point sources
located in r1 and r2
0
4 c
If the origin is outside V
If the origin is inside VFlux from
a point source:
1 23 3
1 2S
r r r rc c dS
r r r r
1 23 3
1 2S s
r r r rc dS c dS
r r r r
(a) 4c -4c =0 S
dS (b) 4c 0 = 4c
S
dS
(c) 0 0 = 0 S
dS
(a)
(b)
(c)
1
SV
n̂
PROBLEM 2
SOLUTION
Use the Gauss theorem to calculate the flux of the vector field:
on the surface S:
2 1
ˆ, ,A z z e
22 2 2 4x y z
x
y
z
The field is singular at =0 (the z-axis)The Gauss theorem cannot be applied on S!!!
We divide V into 2 volumes:V=V0+V
Thin cylinder with radius along the z-axis
( )S
A r dS 1
1 2 2
( )S S S S
S S S
A r dS
The boundary surface to V0 is: 1 2S S S S
1 2 1 2
( ) ( )S S S S S S S
A r dS A r dS
Does not contain the z-axis here we can apply Gauss!
0 1 2
( )V S S S
divAdV A r dS
V
S
S
S
n̂
1̂n
2n̂
2
0 1 2
( )V S S S
divAdV A r dS
0 1 2
( ) ( ) ( )V S S S
divAdV A r dS A r dS A r dS
1
( )S
A r dS 1
2 1ˆ ˆ 0z
S
z e e dS
0ˆlim zdS e dS
=0
2
( )S
A r dS 2
2 1ˆ ˆ 0z
S
z e e dS
0ˆlim zdS e dS
=0
Integrals III and IV are zero:
I II III IV
Integrals II is:
( )S
A r dS
2 1
ˆ ˆS
z e e dS
ˆdS e dS
2 1
S
z dS
2 1
S
z d dz
On S,
dS d dz
2
2
2 2 42
0 2 4
1z d dz
2
2
2 42
2 4
2 1 zdz
2 28 1 4
3
0lim
But our cylinder is very “thin”, which means that we are interested in
2 2
0lim8 1 4 16
0V
divAdV
Integrals I is:
0
2 1ˆ
V
div z e dV
0
21 1
V
z dV
0
2V
zdV
The volume is a sphere with centre in the point (0,0,2)Therefore, to use spherical coord. we must do a coordinate transformation: z’=z-2
0 0
2 2( ' 2)V V
zdV z dV 0 0 0
324 2
4 2 ' 4 2 cos sin3V V V
dV z dV r r drd d
24 2
0 0
128 cos 1282 2
3 4 2 3
r
0128 80
16 0 03 3
I II III IV
4
LAPLACE OPERATOR IN CURVILINEAR COORDINATESLAPLACE OPERATOR IN CURVILINEAR COORDINATES2 0 The Laplace equation is
The The Laplace operator is the divergence of the gradientLaplace operator is the divergence of the gradient::2 2
CARTESIAN COORDINATESCARTESIAN COORDINATES, ,
x y z
yz zAA A
Ax y z
•GRADIENT
•DIVERGENCE
2 2 22
2 2 2x y z
CYLINDRICAL COORDINATESCYLINDRICAL COORDINATES
1, ,
z
1 1 zA A A
Az
•GRADIENT
•DIVERGENCE
2 22
2 2 2
1 1
z
SPHERICAL COORDINATESSPHERICAL COORDINATES
1 1, ,
sinr r r
22
1 1 1sin
sin sinrA r A A Ar r r r
•GRADIENT
•DIVERGENCE
22 2
2 2 2 2 2
1 1 1sin
sin sinr
r r r r r
CARTESIAN
CYLINDRICAL
SPHERICAL
5
PROBLEM 3
SOLUTION
A cylinder with radius R has a charge density c.Calculate the potential and the electric field:
(a)Inside the cylinder(b)Outside the cylinder
Assume that the cylinder is infinitely long and that the potential on the surface is V0.
Due to the symmetry of the problem, the solution will depend on the radius only:
2
0
c
Inside the cylinder
2 22
2 2 20
1 1 c
z
=0 =0 Because the solution depends only on
cylindrical coord.
0
1 c
The equation becomes:
6
=0 =0 Because the solution depends only on
2
0 0 0 0
1
2 2c c c c a
a
0
1, ,
2c
r
d d d d aE E
d d dz d
Divergent at =0NOT physical! a=0
2
0 0
( )2 4
c c b
2 0
Outside the cylinder
10
And the equation becomes:
0 ( ) lnc
c c d
r
d cE
d
Multiplying by Integrating in Dividing by
Integrating in
Multiplying by Integrating in Dividing by Integrating in
There is no charge
7
2
0 0
( ) ( )2 2
in out c cr r
cE R E R R c R
R
2
0
( )4
in c b
( ) lnout c d outr
cE
Now we must determine the three integration constants b, c and d.We have three conditions:(1)Continuity of the electric field at =R(2) The potential at =R(3) Continuity of the potential at =R
0
( )2
in crE
2 20 0 0
0 0
( ) ln ln2 2
out c cR V R R d V d V R R
2
0
( ) ln2
out c R d
2 2 2 20 0
0 0 0 0
( ) ln ln ln ln2 2 2 2
out c c c cR d R V R R V RR
(1)
(2)
(3) 2 2 20 0
0 0 0
( ) ( ) ln2 4 4
out in c c cRR R V R R b b V R
R
=0
8