VEKTORANALYSKursvecka 6

övningar

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

22

0 20

( ) 14

in cV RR

20

0

( ) ln2

out cV RR

2

02out cr

RE

0

( )2

in crE

R

r

R

V0

9