Poisson’s and Laplace’s Equations Poisson s and Laplace s Equations

(1)

Field and Wave Electromagnetic Chapter.4

Solution of electrostatic Problems

(2)

⎛

0

D

E

⎛ ∇ = ρ

⎜⎜ ∇× =

⎝

i : Two fundamental equations for electrostatic problem

E V

D ε E

⎛ = −∇

⎜⎜ =

⎝

Where, V is scalar electric potential V

( ε V ) ρ

∇ ⋅ ∇ = −

∴

D ε E

⎜ =

⎝

(3)

ε

Assuming constant for simple and homogeneous media

2

V

ρ :

∇ = −

ε

∇

g p g

Poisson's equation.

Laplacian operator

2

2 2 2

2

2 2 2

:

V V V

,

V x y z

∇

∂ ∂ ∂

∇ = + +

∂ ∂ ∂

Laplacian operator

for cartesian coordinate

2 2

1 1

( )

y

V V

r r r r r

∂ ∂ ∂

= +

∂ ∂ ∂

2

2 2

2

,

1 1 1

V z

V V V

φ

+ ∂

∂

∂ ∂ ∂ ∂ ∂

for cylindrical coordinate

2 2

2 2 2 2

1 1 1

= ( ) (sin ) ,

sin sin

V V V

R R R R R

θ

R

θ θ θ θ φ

ρ

∂ ∂ + ∂ ∂ + ∂

∂ ∂ ∂ ∂ ∂

for spherical coordinate

2

0 :

V

= −

ε

∇ =

note : No free charge

Laplace's equation

∇ V =

0 : Laplace s equation

(4)

ex) Spherical cloud problem

_ρ ₌ ₀

ρ = −ρ0

b

0

2 0

0 0

(a) 0 , =

_i

R b V

ρ ρ

ρ ρ ε ε

≤ ≤ −

∇ = − =

∵

ρ ρ0

2

2 2 0

2 2 2 2

0

1 1 1

= ( ) (sin )

sin sin

, 0

By symmetry

V V V

V R

R R R R R

θ ρ

θ θ θ θ φ ε

∂ ∂ ∂ ∂ ∂

∇ + + =

∂ ∂ ∂ ∂ ∂

∂ = ∂ =

2 0

2

, 0

1 ( )

By symmetry dVi

d R R dR dR

θ φ

ρ

∂ ∂

= ε0

2 0 2

0

0 1

2 1

( )

0 is singular point unless = 0

i

d dV

R R

dR dR

dV C

R R C

ρ ε

ρ

=

= + → =

∴ ₂ ₁

3 g p

dR ε R

(5)

Poisson’s and Laplace’s Equations

∂V

Poisson s and Laplace s Equations

1

0

( )

=0 , since can not be infinite at 0 0

i

i i

E V R V

R

C E R

ρ b

= −∇ = − ∂

∂

=

0 0 0 2

1 0

,0 3

6

i

E R R R b

V R C

ρ ε ρ

ε

∴ = − ≤ ≤

= +

0 2

0

2

6

( ) , =0 0

1 (

b R b V

R R R

ε

ρ ∴

≥ ∇ =

∂

2 2

) 0 2

o o

dV dV C

dR = ∴ dR = R R ∂R

2 2 0

0 0 3

2

_o _o ^o at , _i (homogeneous medium)

dR dR R

dV C

E V R R R b E E

dR R

C ρ b C ρ b

= −∇ = − = − = =

0 0 3

2 2 2

0 0

0 3

0 2

0

= =

3 3

= , 3

b C b

b

E R b R b

R

ρ ρ

ε ε

ρ ε

→

≥

∴ −

0

(6)

3

0 2

T

4 =

3 ^o 4

Q b E R Q

R ρ π

= − ∴ πε

otal charge in the cloud

0

3 3

' '

0 0

2 2

0 0

3 4

, As 0 =0

3 3

o

o o

R

dV b b

V C R V C

dR R R

πε

ρ ρ

ε ε

= ∴ = − + →∞ = ∴

B d diti t R b

Boundary condition

2 2 '

0 0

1

0 0

3 6

R b

b b C

ρ ρ

ε ε

=

− = +

at

' 0 2 0 2 0 2 0 2

1

0 0 0 0

2 2

0

1 1

( )

3 6 3 6 2

(3 )

C b b b b

b R

V

ρ ρ ρ ρ

ε ε ε ε

ρ

= − − = − − = −

= −

∴

∴ −

0

( )

3 2 2

Vi

= ε −

∴ −

(7)

Uniqueness of Electrostatic Solution Uniqueness of Electrostatic Solution

U i th

Uniqueness theorem

A solution of Poisson's equation that satisfies the given boundary conditions is a unique solution

proof of uniqueness theorem

charged conducting bodies with surface

S S₁ ₂ S

charged conducting bodies with surface , ,...

S S

, at specified potential

Sn

2 2

1 2

,

1 2

Assume two solutions and of Poisson's equation in

V V

τ

ρ ρ

ε ε

∇ = − ∇ = −

Assume that both and satisfy

V₁ V₂

ε ε

1 2 n

, ,...,

the same boundary condition on

S S S

(8)

Uniqueness of Electrostatic Solution

Define

2

1 2

V_d = −V V , , ∇ V_d =0

Define

then from

O d ti b d i th t ti l ifi d d 0

( )

Vd

f A f A A f

=

∇ ⋅ = ∇ ⋅ + ⋅∇

On conducting boundaries, the potentials are specified and

( ) 2 ( ) ( )

( ) | |

d d d d d d

d d d

V V V V V V

V V dv V

τ

→ ∇ ⋅ ∇ = ∇ + ∇ ⋅ ∇

∴ ∇ ⋅

∫

∇ = ∇ ² ⁽ ⁾ ^,

( )

d d

dv s V V nds

n S S S S

τ

= ∇ ⋅

∫ ∫

when denote the unit normal outward from ₀ ₁ ₂

. ( , , ,..., )

1 1 1

n

d

n S S S S

V

τ when denote the unit normal outward from over the conducting boundaries, = 0

2

1, 2 2 3 0

1 1 1

, , .

d d d

V V V V V S R

R R R

∝ ∇ ∝ → ∇ ∝ but surface area ∝

lim ( _d _d) 0 | _d |2 0

s

R V V nds V dv

τ

→∞ ∇ ⋅ = ∇

∴

∫

∴

∫

=

(9)

Uniqueness of Electrostatic Solution

|∇V | 0≥ for everywhere

2

| | 0

| | 0 only for 0

d

d d

d

V

V dv V

V

τ

∇ ≥

∴ ∇

∫

= ∇ =

for everywhere

in other words has the same value at all points in τ as well as on the conducting

0 .

Vd τ

→ =

surface throughout the volume

1 2

V =V

∴ ₁ ₂ and there is only one possible solution.

∴V V and there is only one possible solution.

(10)

Method of images Method of images

2 2 2

2

2 2 2 0

( , , )

V V V

V y

x y z

V x y z

∂ ∂ ∂

∇ = + + =

∂ ∂ ∂ for > 0 except at the point charge.

The solution should satisfy the following conditions

0

( , 0, ) 0

as 0 4

V x z

V Q R R Q

πε R

=

→ →

where is the distance to

At p ( , , ), l 0

( , , ) ( , , ), ( , , ) ( , , )

x y or z V

x z

V x y z V x y z V x y z V x y z

→ ±∞ → ±∞ → ±∞ →

= − = −

oint P the potentia

The potential function is even with respect to the and coordinates

→ very difficult to satisfy all these condition

(11)

Method of images

From figure (b)

Method of images

0

1 1

2 2 2 2 2 2

1 1

( , , ) ( )

4

V x y z Q y

R R

πε ₊ ₋

= −

From figure (b),

for >0

1 1

2 2 2 2 2 2 2 2

[ ( ) ] [ ( ) ]

R₊ = x + y−d + z R₋ = x + y+d + z

where, ,

satisfy all the conditions mentioned above.

∴This is a solution of

E = −∇V y

this problems by uniquiness theorem.

for > 0

(12)

Method of images Method of images

1 2 3

2

1 2

0 2

4 (2 )

F F F F

F y Q πε d

= + +

= −

2

2 2

0 1

2

4 (2 )

( 2 2 )

F x Q

d

F Q d d

= − πε

3 3 1 2

2 2 2

0 1 2

2

1 2

( 2 2 )

4 [(2 ) (2 ) ]

1 1

F Q x d y d

d d

F Q x y

= πε +

+

⎧ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥

= − + −

∴

⎪ ⎫⎪

⎨ 3 2 3 2 ⎬

2 2 2 2 2 2

0 1 2

1 2 1 2

16 ( ) ( )

F x y

d d

d d d d

πε ^⎢_⎢_⎣ ₊ ^⎥_⎥_⎦⁺ ^⎢_⎢_⎣ ₊ ^⎥_⎥_⎦

∴

⎨ ⎬

⎪ ⎪

⎩ ⎭

Note : Explain whether the force is in the direction to the conductor or not?

(13)

Line Charge and Parallel Conducting Cylinder Line Charge and Parallel Conducting Cylinder

The image must be a parallel line charge inside the cylinder in order to

k th li d i l f t r = a i t ti l f

make the cylindrical surface at an equipotential surface.

Because of symmetry w.r.t. the line OP, the image line charge must lie somewhere along OP.

,

i i

P d

ρ d

→

g

At a point with a distance from the axis.

Two unknowns;

(14)

Line Charge and Parallel Conducting Cylinder

ρ = −ρ ⇒

Assume intelligent guess

Line Charge and Parallel Conducting Cylinder

i l

ρ = ρ ⇒

→

Assume intelligent guess.

proceed to check if this fails to satisfy the B.C. or not.

If this satisfy all B.C.'s, then it is the only solution by the uniquiness theo em

theorem.

Electric fiel _l

l

r

E r

ρ

= ρ

d intensity and potential at a distance from a line charge

0 0

0

0 0

2

1 ln

2 2

r r

l l

r r r

E r

r

V E dr dr r

r r

πε

ρ ρ

πε πε

= −

∫

= −

∫

=

note) reference point for zero potential can not be , since potential is finite.r₀ ∞

(15)

Line Charge and Parallel Conducting Cylinder

The potential at a point on or outside the cylindrical surface:

0 0

ln ln ln

2 2 2

l i

l l l i

M

r r r

V r r r

ρ ρ

ρ ρ ρ

πε πε πε

= − =

The potential at a point on or outside the cylindrical surface:

superposition of contributions by and , at a point

0 0 0

2πε r 2πε r_i 2πε r

where reference point for zero potential is at the same distance from and ρ_l ρ_i equipotential surface are specified by

ri

r =

constant

if an equipotential surface is to coinside with the cylindrical surface (OM= )a

(16)

Line Charge and Parallel Conducting Cylinder

i i

d a ∠ r

cf) If and are constant and MOP common then =constant

i i

a d r

r d a

a d d

∠

Δ Δ

= = ∵

i i

cf) If and are constant and MOP common, then =constant.

Since OMP and OMP is similar,

= constant ( _i fixed)

i

a d d r a d

r r

= = ∵

= constant ( , , fixed)

can be constant over the c _i a

= →d

ylindrical surface then, P Inverse point

r ri

cf) As the point M moves along the cylindrical surface, and changes but their ratio is constant

cf) By symmetry, parallel cylindrical surface sur

l a di P

ρ

rounding the original line charge with radius and its axis at a distance to the right of is also

equipotential surface.

(17)

Ex 4 4) Cross section of two wire transmission line Ex 4-4) Cross section of two-wire transmission line

Determine the capacitance per unit length between two long parallel, a

D

p p g g p ,

circular conducting wires of radius . the axes of the wire are seperated by a distance .

(18)

Ex 4 4) Cross section of two wire transmission line

→

Conducting surface equipotential surfaces

Ex 4-4) Cross section of two-wire transmission line

, ( 2 ) .

l l D di d di

ρ ρ

→

∴

+ − − = −

Conducting surface equipotential surfaces

equipotential surfaces can be generated by two parallel line charge seperated by a distance

Potential on 1

₁

0

2 ln

l a

V d

ρ

= −

πε

positive quantity.

0

2

l

ln

d

V a

πε

=

ρ Potential on 2

₂

negative quantity. because (a<d)

0

2

0

2

ln ) .

2

l i i

d r r a

V const

r r d

πε ρ

=

πε

= =

g q y ( )

(cf ,

0

r0

if i s at the same distance from both and ρ

_l −

ρ

_l

(19)

Ex 4 4) Cross section of two wire transmission line

ln per unit length =

l d

V ρ Q ρ

∴

Ex 4-4) Cross section of two-wire transmission line

12

0

0 12

ln

( [ / ]

ln( )

per unit length =

per unit length)

l

V Q l

a

C Q F m

V d

ρ ρ

πε

=

= =

∴

12 2

ln( )

, 1

i

a d D d D a

= − = − d

2 2

1( 4 ) (

2 choose only +sign

d = D± D − a

0 0

2 1

, )

cosh ( / 2 )

l ( ) ( ) 1

D d a

C D D D a

πε πε

= = −

⎡ ⎤

∴

∵

2

2 1

cosh ( / 2 )

ln ( ) ( ) 1

2 2

ln 1 cosh

cf)

D a

D D

a a

x x ⁻ x

⎡ + − ⎤

⎢ ⎥

⎣ ⎦

⎡ + − ⎤ =

⎣ ⎦

(20)

Point Charge and Conducting Sphere Point Charge and Conducting Sphere

i

Q Q d

Considering symmetry, assume the image charge (negative point charge) inside the sphere and on the line joining O and .

distance from origin O_i

Qi ≠ −Q R = a

g

to make the spherical surface a z

i i

d Q

ero-potential surface.

both and : unknowns

(21)

Point Charge and Conducting Sphere

At an arbitrary point M on the equipotential surface

Point Charge and Conducting Sphere

0

1 ( ) 0

4

i M

i

Q Q

V ⁼

πε

r ⁺ r ⁼

At an arbitrary point M on the equipotential surface

2

i i , i

r Q Q a

const

r Q Q d

a a

Q Q d

∴ = − = ∴−

∴

=

, ,

i i

Q Q d

d d

E

= − =

∴

, a

V Q Q

− d and can be calculated from two point charge

(22)

Boundary value problem Boundary value problem

Develop a method for solving three-dimensional problems where the boundaries, over which the potential or its normal derivative is specified, coincide with the coordinate surfaces of an orthogonal, curvilinear systemg , y

The method of seperation of variables

Boundary value problems

Dirichlet problems : the value of potential is specified everywhere on the boundaries

N bl

Neumann problems :

the normal derivative of the potential is specified everywhere on the boundaries Mixed boundary-value problems :

the potential is specified over the remaining ones.

(23)

Cartesian coordinate Cartesian coordinate

Laplace's equation for scalar electric potential V in Cartesian coordinates

2 2 2

2 2 2 0,

V V V

x y z

∂ ∂ ∂

+ + =

∂ ∂ ∂

( , , ) ( ) ( ) ( ), ( ), ( ), ( )

, , ,

V x y z X x Y y Z z X x Y y Z z

x y z

=

Assume where are functions of only

and respec

2 2 2

( ) ( ) ( )

( ) ( )d X x ( ) ( )d Y y ( ) ( )d Z z 0

Y Z + X Z +X Y

tively. Then,

2 2 2

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) 0

( , , ),

1 ( ) 1 ( ) 1 ( )

0

Y y Z z X x Z z y X x Y y

dx dy dz

V x y z

d X x d Y y d Z z

+ + =

divided by

2 2 2

2 2

1 ( ) 1 ( ) 1 ( )

( ) ( ) ( ) 0

1 ( ) 1

( ) ^x ( )

d X x d Y y d Z z

X x dx Y y dy Z z dz

d X x d

X x dx k Y y

+ + =

= −

∴ ∵

( )₂ 1 ² ( )₂

( )

Y y d Z z

dy + Z z dz is independent of x

( ) ( )

X x dx Y y

2 2 2

( )

( ) ( ) ( )

( ) 0, ( ) 0, ( ) 0

x y z

dy Z z dz

d X x d Y y d Z z

k X x k Y y k Z z

dx + = dy + = dz + =

separation constant

(24)

Cartesian coordinate Cartesian coordinate

Reminds

''( )

_x2

( ) 0 X x + k X x = Reminds

Possible solution of

0

,

0

,

1

,

1

,

2

,

2

A B A B A can be determined by the given boundary B