PDF Radio Technology Laboratory 1 - Seoul National University

(1)

Ex 4 6) the potential distribution enclosed by the electrodes Ex 4-6) the potential distribution enclosed by the electrodes

c Find B.C. first and apply to D.E.

2

0, ( , , ) ( , )

( )

V z V x y z V x y

z d Z z

→ ∂ = =

a) independent of ∂ and

2 2

0 0 0

2

( ) _z ( ) 0, _z 0 ( )

d Z z

k Z z k Z z A z B B

dz x

+ = = = + =

−

and

b) direction

( , ) 0 x

x→ ∞ 일때V ∞ y = V

b) direction

: is decreasing

0

2 2 2 2 2 2 2

0 (0, )

0 0

x

x V y V

k k k k k k k k jk k

= =

∴ + = ⇒ = = ∴ = = ± >

일때

function in the direction

therefore

1 2 2

0

0 0

( ) ) (0, )

x y y x x x

kx kx kx

k k k k k k k k jk k

X x C e D e D e cf V y V

− −

∴ + = ⇒ = − = ∴ = − = ± >

= + =

=

therefore, ,

then

will be applied later

(2)

Ex 4 6) the potential distribution enclosed by the electrodes

) di ti

Ex 4-6) the potential distribution enclosed by the electrodes

1 1

( , 0) 0, ( , ) 0

( ) sin cos

y

Y x Y x b

Y y A ky B ky

−

= = ⇒

= +

c) direction

implies sine function,

from

(0) 0, ( ) 1sin 0

( ) i

Y Y b A kb k n

b Y A n

π π

∴ = = = ∴ =

th ( ) ₁sin n Y y ₌ A π then

( ), ( ), ( )

n

b y X x Y y Z z

π

combine d

1 0 2

( , ) ( ) ( ) ( ) sin , 1, 2, 3,...

) ( )

b x n

V x y X x Y y Z z A B D e y n b

n X x

π

= = − =

cf cannot be negative since should be decaying function

(3)

Ex 4 6) the potential distribution enclosed by the electrodes

( ) 0

V x y x =

e alone cannot satisfy the B C that at

Ex 4-6) the potential distribution enclosed by the electrodes

0

( , ) 0,

(0, )

( , ) ( , ), ( , )

n

V x y x

V y V y b

V x y V x y V x y

=

= < <

=

∑

e alone cannot satisfy the B.C that at

for 0

let then is a solution of Laplace equation

n

∞

f apply the last B.C

∞

1 1

(0, ) _n(0, ) _n sin

n n

V y V y C n y

b π

∞

= =

=

∑

=

V₀, 0 y b

∞ = < <

∑

^for

0 1

sin ,

i i i

n n

n

b b

C n y V C

b

n m m

C d V d

π

π π π

∞

=

∞

∑

=

∑∫ ∫

i.e) to find

0 0 0

1

sin sin sin

n n

C y ydy V ydy

b b b

=

∑∫

=

∫

(4)

Ex 4 6) the potential distribution enclosed by the electrodes

left hand side¾

Ex 4-6) the potential distribution enclosed by the electrodes

0 0

( ) ( )

sin sin [cos cos ]

2

b b

n n

n

C

n m n m n m

C y ydy y y dy

b b b b

C b m n

π π = − π − + π

⎧ =

⎪

∫ ∫

for ^b _nπ _mπ

∫

2 0

b m n

m n

⎪ =

⎨⎪ ≠

⎩

¾

, for =

, for right hand side

For , 0^bsinn sinm 0

n m y ydy

b b

π π

≠

∫

⋅ =

0 0

0^b sin m b [ cosm ]

V ydy V y

b m b

π π

= π −

∫

⁰

0

[1 cos ] 0

b V b

m m m even

π π

= −

⎧ =

⎪ , for

0

2

4

V b m odd

m

V n odd

π

⇒ ⎨⎪⎪⎩ =

⎧ =

⎪⎨

, for , for

0

( , ) sin 4 sin ,

n

n n

x x

b b

n odd

C n

n even

V

n n

V x y C e y e y

π π

π

π π

∞ − ∞ −

∴ = ⎨⎪

⎪ =

⎩

= =

∴

∑ ∑

, for

for x >0 and 0< <y b

( , )y

∑

y

∑

y, ^y

(5)

Ex 4 7) the potential distribution enclosed by conducting planes Ex 4-7) the potential distribution enclosed by conducting planes

) V x y z( , , ) V x y( , ), 0, k_z2 0 z

= ∂ = =

¾ ∂

sol

0

2 2 2

(0, ) ( , ) 0, ( , 0) ( , ) 0 z

V y V V a y V x V x b

∂

= = = =

¾ ,

2 2 2

0

1

( ) ,

0, ( , ) 0 ( ) sin

y x

Z z B k k k

y V x y Y y A ky

= = − =

= = → ∴ =

c d

for 0, ( , ) 0 ( ) ₁sin

( , ) 0

y V x y Y y A ky

y b V x y

→ ∴

= = → ∴

for d

for , ( ) ₁sin n . ) n

Y y A y i e k

b b

π π

= =

(6)

Ex 4 7) the potential distribution enclosed by conducting planes Ex 4-7) the potential distribution enclosed by conducting planes

(0 ) 0 ( ) 0

V y =V V a y = k = jk ⇒

e (0, ) ₀ and ( , ) 0, not exponentially decay . )

x

kx kx

V y V V a y k jk

i e e e⁻

⇒

e and not exponentially decay

both and should exist

2 2

2 2 2

( ) sinh cosh : , ( ) 0

( ) sinh cosh

X x A kx B kx x a X x y

X a A ka B ka B

∴ = + = =

∴ = + ∴

for for all

₂ sinh

cosh A ka

= − ka

2

( ) [sinh sinh cosh ]

cosh

X x A kx ka kx

ka A

= −

2

3

[sinh cosh cosh sinh ] cosh

sinh ( ) sinh ( ) )

cosh

A kx ka kx ka

ka

A n

k x a A k x a cf k

ka b

π

= −

= − = − =

'

cosh

( , ) sinh ( )

n n

ka b

V x y C n x a

b

∴ = π −

sin n , 1, 2, 3,....

y n

b

π _for ₌

(7)

Ex 4 7) the potential distribution enclosed by conducting planes

(0 ) V y =V f

Ex 4-7) the potential distribution enclosed by conducting planes

0

' '

0

1 1 1

(0, )

( )

(0, ) sinh sin sinh sin

n n n

V y V

n a n n a n

V V y C y C y

b b b b

π π π π

∞ ∞ ∞

= = =

=

∑

=

∑

− = −

∑

f

' 1

sin , sinh

4

n n n

n

n n

C y C C a

b b

V

π π

∞

=

= = −

⎧ −

∑

where

⎧

0

'

4 4

0

n n

V V

odd n

C n C

even n

π

⎧⎪

= ⎨ ∴ =

⎪⎩

, for

0

sinh 0

odd n

n n a

b

even n

π π

⎧⎪⎪

⎨⎪

⎪⎩

, for

for

' 1

0

( , ) _n sinh ( ) sin

n

even n

n n

V x y C x a y

b b

π π

∞

=

⎪⎩

= −

∴

∑

, for

0 1,

4 sinh ( ) sin , 0 ,

sinh

n odd

V n n

a x y x a y b

n b b

n a

b

π π

∞

=

∑

− < < < <

for 0

(8)

Boundary value problem in cylindrical coordinate Boundary value problem in cylindrical coordinate

2 2

2

1 1

V V V 0

V ∂ ⎛ ∂ ⎞ ∂ ∂

∇ ⎜ ⎟

9

General solution has a form of Bessel function

2

2 2 2

0

V r

r r r r φ z

⎛ ⎞

∇ = ∂ ⎜ ⎝ ∂ ⎟ ⎠ + ∂ + ∂ = 9

c for simplicity,

2 2

2 2 2

1 1

0 V V 0 ( , ) ( ) ( )

r V r R r

z r r r r

φ φ

φ

∂ = ∴ ∂ ⎛⎜ ∂ ⎞⎟+ ∂ = = Φ

∂ ∂ ⎝ ∂ ⎠ ∂

assume the lengthwise dimension the radius dimension

then let

2 2

( ) 1 ( )

( ) ( ) 0

z r r r r

r d dR r d

R r dr r dr d

φ φ

∂ ∂ ⎝ ∂ ⎠ ∂

⎛ ⎞ + Φ =

⎜ ⎟ Φ

⎝ ⎠

d then

( ) 1 2 ( )

r d ⎛ dR r ⎞ d Φ

φ

function of onlyr function of onlyφ

let ( ) ² 1 ( )₂ ²

( ) ( )

r d dR r d

r k k

R r dr dr d

φ

φ φ

φ

⎛ ⎞ = Φ = −

⎜ ⎟ Φ

⎝ ⎠

∵

then ( periodic in for the most of case)

(9)

Boundary value problem in cylindrical coordinate Boundary value problem in cylindrical coordinate

2 ( )

d Φ φ ₂

0 0 1 2

2

( ) ( ) 0, ^jk ^jk

d k A B C e D e

d

φ φ

φ φ φ

φ

Φ + Φ = ( Φ = + or Φ = + − )

but for circular cylindrical configurations, potential functions and

h f d d

e

( )φ φ k n

therefore Φ are periodic in and is an integer ∴Φ( )φ = A_φ sinnφ + B_φ cosnφ

2

2 2

2

( ) ( )

( ) 0 ( ) _r ⁿ _r ⁿ

d R r dR r

r r n R r R r A r B r

d d

+ − = → = + −

for radial function,

f

2

2 '' '

0

dr dr

x y +axy +by = cf) Cauchy equation(Euler equation) :

2 (

y xm

m a

=

⇒ +

then form

−1)m b+ =0 ∴m = m₁ or then m₂ y = C x₁ ^m¹ +C x₂ ^m²

(10)

Boundary value problem in cylindrical coordinate Boundary value problem in cylindrical coordinate

' '

( , ) ⁿ( sin cos ) ⁿ( sin cos ), 0

V r φ = r A nφ +B nφ +r⁻ A nφ + B nφ n ≠

g ( , ) ( sin cos ) ( sin cos ), 0

0,

n n n n n

n

V r r A n B n r A n B n n

r r

φ φ φ φ φ

−

+ + + ≠

=

g

cf) if region of interest includes the cylindrical axis where the terms containing the factor cannot exist.

,

( , ) ( , )

n

n n

r r

V r φ V r φ

= ∞

∑

if region include the terms cannot exist

depending on the boundary condition, may be is

2

0 0

2

0

0, ( ) 0 ( )

0

k d A B

d

B k

φ φ φ

φ

= Φ = → Φ = +

Φ =

h for

if no variation along , = g φ ₀ for ( .i e A ₀₀ = 0)

0 0

1 2

( )

[ ( )] 0, ( ) ln 0

( ) ( ) ( ) ( ) ln ,

d dR r

r R r C r D k

dr dr

V r R r φ V r C r C z φ

= = + =

= Φ ∴ = + →

for

independent of

1 2

( ) ( ) ( ) ( ) ln ,

V r R r Φ φ ∴V r C r +C → z φ

independent of

(11)

Ex 4 8) a very long coaxial cable Ex 4-8) a very long coaxial cable

z z

sol) c very long in : no variation

0

0 0

( ) 0

( ) ln

( ) 0

B k

R R r C r D

V b

φ → Φ φ = =

→ = +

⎧ d

e

) y g

by symmetry, no variation and potential is only a function of

0

( ) 0 ( )

( )

V b V a V

V φ C

⎧ =

⎨ =

f ⎩

boundary condition,

1 2 0

( ) ln

l V a C a C V

C ⎧ = + =

very long coaxial cable's cross-section ⎨

< > ∴V r( , )φ =C₁ ₂ ¹ ² ⁰

1 2

ln , ( )

( ) ln 0

r C

V b C b C

+ ⎨⎩ = + =

0 0

1lna 0 1 V V

C V C

a b

b = ∴ = = −

⎛ ⎞ ⎛ ⎞

¾

0

2 1

ln ln

a b

b

b a

V b

C C b

b

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= − =

¾ ⎛ ⎞

0 0 0

ln ( ln

, ) ( ) ln ln

l l l

b a

V V b V b

V r V r r

b b b r

φ

⎛ ⎞⎜ ⎟

⎝ ⎠

= = − ⎛ ⎞ + ⎛ ⎞ = ⎛ ⎞ ⎝ ⎠⎛ ⎞⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

∴

ln⎛ ⎞b ln⎛ ⎞b ln⎛ ⎞ ⎝ ⎠b r

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

(12)

Ex 4 9) Infinitely long thin conducting circular tube Ex 4-9) Infinitely long, thin, conducting circular tube.

d 0 d = sol)

( )

0 0

( , ) dz

V b V

V

φ φ π

π φ π

= ≤ ≤

= − ≤ ≤ 2

)

for 0

(

for

)

. . i e

a) Inside tube

( ) ( )

( )

0 sin

sin

n r

n

r b R r A r r

A n

V r A r n

φ φ φ φ

φ φ

< → = =

Φ =

∴ =

∵

is included.

odd function in

1 2

( )

^, ^sin

n n

V r φ A r nφ

∴ =

In order to satisfy the pe3

( )

^φ ⁼

∑

^∞ ⁿ ^φ

riodic boundary condition,

( )

^φ ⁼

∑

^φ

,

(13)

Boundary Value Problems in Cylindrical Coordinates Boundary-Value Problems in Cylindrical Coordinates

r =b i.e. at

0

1 0

sin ,

, 2

n n n

A b n V

V φ φ π

π φ π

∞

=

⎧ <

= ⎨⎩− <

∑

rectangular periodic wave can be represented by Fourier Sine Series for 0<

for <

( ) ( )

1 0 ,

: (

n

f x f x L

= ⎩ φ

= − −

odd function and period Non-radia

0

( )

2 2

L sin

n n

A b f x n x dx

L L

⎛ π ⎞

∴ =

∫

⎜⎝ ⎟⎠

n)

( )

0

0 0 0

0

2

2 cos

sin

L L

V n

V n d

n

π φ φ φ π

π π

⎜ ⎟

⎝ ⎠

⎡ ⎤

= = ⎢⎣− ⎥⎦

∫

[ ]

0

0 0

2 1 cos

4 4

, ,

n

V n

n

V V

A b A

π π

= −

⎧⎪

⎨

for odd n ⎧ for odd n

⎪⎨

, ,

,

n n

A b ⁼ ^⎪_⎨ nπ ^∴ A ⁼ n bπ

∴ ⎪⎩

0 for even n ,

⎪⎨

⎪⎩ 0 for even n

(14)

Boundary Value Problems in Cylindrical Coordinates Boundary-Value Problems in Cylindrical Coordinates

r =b i.e. at

sin 0 n n

A b n V

V φ φ π

π φ π

∞ ⎧ <

= ⎨⎩

∑

rectangular periodic wave is represented by Fourier Sine Series for 0<

for < <2

1 0

0 0 0

1

sin sin sin

n n

V

A b n m d V n d

π π

π φ π

φ φ φ φ φ

=

∞

=

⎩−

∑

=

∫ ∫

for < <2

sin sin

n

A bn nφ m dφ φ

=

( )

( ) ( )

0

1 0 0

0

cos 1 2

0

n

V n

n

A b d V if dd if

π π

π

φ

∞

=

∞

= −

⎡ ⎤

∑∫

0 ^≠⁰

(

^{for n=m}

) (

⁼⁰

)

⁰

1

0

cos cos , , 0

2

2 1

2 2 2

n n n

n n

n

n n n

A b n m n m d if n odd if n even

n V

b b

A b A A

n

π π φ

π π

π

=

⎡ ⎤

= ⋅ − − + = =

∴

⎣ ⎦

= ⋅ = ⋅ =

∑∫

,

0

2 2 2

4 ,

n n

n A V for odd n

n bπ

∴

⎧ =

⎪⎨

⎪ =

(15)

Boundary Value Problems in Cylindrical Coordinates Boundary-Value Problems in Cylindrical Coordinates

( )

⁴^V⁰ ¹ ^r ⁿ ⁱ

V

( ) φ ∑

^∞ ^{⎛ ⎞}_{⎜ ⎟}

φ

f b

( )

, 0 sin ,

n odd

n

V r n for r b

n b r b

R B

φ φ

π

₌

= ⎛ ⎞⎜ ⎟⎝ ⎠ <

>

∴

∑

b) out-side the tube

( )

^'

1

, sin

n n

n n n

R r B r

V r

φ

B r n

φ

−

∞ −

=

∴ =

∑

( )

1

' 0

1 0

, , sin

n

n n n

at r b V b B b n V

V

φ φ φ π

π φ π

=

∞ −

=

= = = ⎧

⎩−

∑

⎨ for 0< <

for < <2

ll h h d f d ff f

Following

0 '

4 ⁿ

n

V b

B n

π

⎧⎪

= ⎨

the same method to find coefficients of Fourier series.

, for odd n

4 0 1

( , ) sin ,

n

V b

V r n r b

π

φ

^∞

φ

⎨⎪⎩

= ⎛ ⎞⎜ ⎟ >

∴ ⎠

∑

⎝

0 , for even n

( , ) sin ,

V r

φ

n

φ

r b

π

^{⎜ ⎟}_⎠ ^>

∴

∑

⎝

(16)

Boundary Value Problem in spherical Coordinates Boundary-Value Problem in spherical Coordinates

( )

²

2 1 2 1 1

i 0

V V V

θ φ ^∂ ^⎛ ^∂ ^⎞ ^∂ ^⎛ θ ^∂ ^⎞ ^∂

9 ²

(

^{, ,}

)

¹₂ ² ₂ ¹ ^sin ₂ ¹ ₂ ₂ ⁰

sin sin

0

V V V

V r r

r r r r r

θ φ θ

θ θ θ θ φ

φ φ

∂ ⎛ ∂ ⎞ ∂ ⎛ ∂ ⎞ ∂

∇ = ∂ ⎜⎝ ∂ ⎟⎠+ ∂ ⎜⎝ ∂ ⎟⎠+ ∂ =

∂ = Assuming symmetry in , then ∂

9

2 2

1 1

sin 0

sin

V V

V r

r r r r

φ

θ θ θ θ

∂

∂ ⎛ ∂ ⎞ ∂ ⎛ ∂ ⎞

∇ = ∂ ⎜⎝ ∂ ⎟⎠+ ∂ ⎜ ∂ ⎟ =

∴ ⎝ ⎠

( )

^,

( ) ( )

V r θ = R r Θ θ let

¾

( ) ( )

1 2 1

sin 0

( ) i

d dR d d

R d r d d d

θ θ

θ θ θ θ

⎡ Θ ⎤

⎡ ⎤ + ⎢ ⎥ =

⎢ ⎥ Θ

⎣ ⎦ ⎣ ⎦

then,

( )

² ²

( ) sin

1 1

:

R r dr dr d d

r dR k R r dr dr

θ θ θ ^⎢ θ ^⎥

⎢ ⎥ Θ

⎣ ⎦ ⎣ ⎦

⎡ ⎤ =

⎢ ⎥

⎣ ⎦

let separation constant

( )

2

2 2

2 2 0

d R dR

r r k R

dr dr

→ + − = ⇒ Cauchy's or

Euler's equation

(17)

Boundary Value Problem in spherical Coordinates Boundary-Value Problem in spherical Coordinates

2

2 2

sin ( ) ( )

0 ( 1)

d d

d d k

R ^α k k

θ θ

θ θ θ θ

⎡ Θ ⎤ = −

⎢ ⎥

Θ ⎣ ⎦

then 1

sin

cf) assume let

( )

2 2

, 0, ( 1),

1

R r k k n n

n or n

α α α

α

= + − = = +

= − +

cf) assume let

then

( )

⁽ ¹⁾

( )

n n

R r

R r = A r + B r^{− +}

General solution of will be as follows

where n

(

⁺¹

)

^k² ^d ^{0 1 2} positive integer

where

( )

²

2

2 2

1 0, 1, 2,

0 ( ) ⁿ ⁿ

n k and n

d R dR

r r k R R r A r B r⁻

+ = = ⋅⋅⋅

+ − = = + →

positive integer

Cauchy's or cf) r ₂ r k R 0, then R r( ) A r_n B r_n

dr + dr = = + →

cf) then,

Euler's equation

(18)

Boundary Value Problem in spherical Coordinates Boundary-Value Problem in spherical Coordinates

( )

n P cosθ

(

¹⁻^x²

)

^y^{'' 2}⁻ ^xy^'⁺^{n n}

(

⁺¹

)

^y ⁼ ⁰

cf) Legendre's D.E.

( )

n cos

0 1

1 cos

Pn θ

θ

( ) ( )

sin ( ) 1 sin 0

d d

d d n n

θ θ θ θ

θ θ

⎡ Θ ⎤

∴ ⎢⎣ ⎥⎦+ + Θ =

→

Turns out to be a Legendre's equation form

( )

2

3

2 1 3cos 1

2

3 1 5 cos 3cos

2

θ

θ θ

−

cos d

x d

θ θ

→

= =

Turns out to be a Legendre s equation form c.f.) let , then

( )

sin d

θ

dw

−

( )

2

( )

( ) ( )

cos

cos .

n n

P

n P Legendre Polynomials

θ

θ θ

Θ = →

Solution of Legendre's equation : Legendre's functions When is integer

( )

^, ⁿ ⁽ⁿ ¹⁾

(

^cos

)

n n n n

V r

θ

^⎡⎣A r + B r^{− +} ^⎤P

θ

∴ = ⎦

(19)

Ex 4 10) Conducting sphere in a uniform electric field Ex 4-10) Conducting sphere in a uniform electric field

( ) ( )

( )

0 ˆ 0. , , ?

b

E = zE V r

θ

E r

θ An uncharged conducting sphere of radius is placed

in an initially uniform electric field Find and

l) ( )

1

S f f th h i i t i d i t ti l

sol) a Surface of the sphere is maintained equipotential

1

E

⇒

i.e. Separation of charges and redistribution take place.

inside the sphere is zero

2 E

inside the sphere is zero.

Outside the sphere ;

the field line on the suface is normal t

2

3

o the surface

φ

⇒

The field intensity at points for away from the sphere No change.

Potential is independent of

4

5

φ

Potential is independent of

5

(20)

Boundary Value Problem in spherical Coordinates Boundary-Value Problem in spherical Coordinates

( )

^,

V r θ r ≥b

( )

for

( )

,

( , ) 0 ( 0

cos

V b V

V r E z E r r b

θ

θ θ

= =

B.C.

i.e. Assuming (may not be 0) in the equipotential plane)

( )

0 0 for

0

, cos

V r E z E r r b

E

θ = − = − θ

for

( is not distributed at points for awa

( )

^, n ⁿ n ⁽ⁿ ¹⁾ n

(

^cos

)

V r θ =

∑

^∞ ^⎡⎣A r + B r ^{− +} ^⎤⎦P θ

y from the sphere.)

6

( ) ( )

( )

⁰ 0

1 0

, cos

, , cos

n n n

n

V

r b V r E r

A E A

θ θ

=

⎡ ⎤

⎣ ⎦

= −

∴ = −

∑

and all other 's are zero.

( ) ( ) ( )

( )

( 1)

0 1

0

1 2 ( 1)

, cos _n ⁿ _n cos

n

V r E rP B r P

B B E B P

θ θ θ

θ

∞ − +

=

+

∴ = − +

∑

(

^θ

)

∑

∞

( )

1 2 ( 1)

0 1 0 cos _n ⁿ _n c

B r⁻ B r⁻ E r θ B r^{− +} P

= + − +

( )

2

0

os

0.

n

Q B

θ

πε

=

⇒ ∴ =

∑

form of Charged sphere form. But we have uncharged sphere πε

(21)

Boundary Value Problem in spherical Coordinates Boundary-Value Problem in spherical Coordinates

( )

, B2¹ 0 cos _n ⁽ⁿ ¹⁾ _n

(

cos

)

,

( )

, 0

V r E r B r P r b V b

θ =^⎛⎜ r − ^⎞⎟ θ + ^∞ ^{− +} θ = θ =

⎝ ⎠

∑

and at

( )

2 1 1

2 0

2 3

0 cos cos

n n

n

r

B E b B b P

b θ θ

=

∞ − +

=

⎝ ⎠

⎛ ⎞

∴ =⎜⎝ − ⎟⎠ +

∑

( )

3

1 0

0, 2

n

B E b

B n

b

∴ =

= ≥

⎡ ⎛ ⎞3⎤

⎢ ⎥

( )

, 0 1 b V r θ = −E

∴ −

( )

cos , ˆ ˆ

r r b

r

V V

E V r

θ θθ

⎛ ⎞ ≥

⎢ ⎜ ⎟ ⎥

⎢ ⎝ ⎠ ⎥

⎣ ⎦

∂ ∂

= −∇ = − −

∂ ∂

b

( )

3

0 1 2 cos ,

r

r r

V b

E E r b

r r

θ

∂ ∂

⎡ ⎤

∂ ⎛ ⎞

= − ∂ = ⎢⎢⎣ + ⎜ ⎟⎝ ⎠ ⎥⎥⎦ ≥

∴

3

0 1 sin ,

V b

E E

r r

θ θ

∴ = − ∂∂θ = − ⎡⎢⎢⎣ −⎛ ⎞⎜ ⎟⎝ ⎠ ⎤⎥⎥⎦

( )

³

r b

E E

θ θ

≥

( )

^E ³ ^E ^cos

ρ θ =ε = ε θ